isidore.co Infinitesimal... · Translator’s Preface...

Dennis M. Cates

Cauchy’s Calcul InfinitésimalAn Annotated English Translation

Cauchy’s Calcul Infinitésimal

An Annotated English Translation

Dennis M. Cates

Cauchy’s Calcul InfinitésimalAn Annotated English Translation

123

Dennis M. CatesSun City, AZ, USA

ISBN 978-3-030-11035-2 ISBN 978-3-030-11036-9 (eBook)https://doi.org/10.1007/978-3-030-11036-9

Library of Congress Control Number: 2018966111

Mathematics Subject Classification (2010): 01A05, 01A55, 01A75, 00B50, 26-03, 28-03, 30-03, 32-03,34-03, 40-03, 41-03, 45-03, 49-03, 97I10

© Springer Nature Switzerland AG 2019This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or partof the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmissionor information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilarmethodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a specific statement, that such names are exempt fromthe relevant protective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material contained herein orfor any errors or omissions that may have been made. The publisher remains neutral with regard tojurisdictional claims in published maps and institutional affiliations.

This Springer imprint is published by the registered company Springer Nature Switzerland AGThe registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

https://doi.org/10.1007/978-3-030-11036-9

For my wonderful wife, Dawne.

Translator’s Preface

Augustin-Louis Cauchy (1789–1857) is responsible for a great many publications, butamong his most important works is his textbook of 1823, R�esum�e des leçons donnéesa l’École royale polytechnique sur le calcul infinitésimal (Summary of Lec-tures given at the Royal Polytechnic School on the Infinitesimal Calculus). He wrotethe text, usually shortened to Calcul infinitésimal, while teaching at the �Ecolepolytechnique in Paris, one of the most prestigious institutes of higher education in allof Europe,1 to support the analysis course he taught at the school. Cauchy’s Calculinfinitésimal, translated here in its entirety, is his first full-length book devoted tocalculus. In a collection of lecture notes designed for his students at the École Poly-technique, Cauchy presents his differential and integral calculus in a highly efficientand well-organized manner. In this text, he builds a complete and solid foundationfor calculus from which the current subject is based.

The Calcul infinitésimal text is composed of forty relatively brief lectures whichbegin with his historic definition of the limit, then quickly progress in rigorous andmethodical fashion through all of the important topics and theorems necessary toconstruct what is essentially the modern version of this subject. His forty lecturespresent material in an order nearly identical to today’s calculus textbooks, begin-ning with differential calculus, stepping through integral calculus, and ending witha study of infinite series and Taylor’s Theorem. Calcul infinitésimal is importantbecause it presents a cohesive, consistent, and comprehensive package for the sys-tematic use and rigorous supporting theory of the calculus—a package based f irmlyand squarely upon reason. Cauchy methodically steps through result after result,introducing relevant precise definitions as needed. Basic theorems are rigorouslyproven one after the other, while more advanced theorems are derived in turn—al-ways basing the argument upon earlier results. His definitions for most all of theimportant concepts in the subject are the same modern definitions we use today,with only relatively minor modifications in some cases which occur later in the 19thand early 20th centuries.

1The École Polytechnique is still in operation today as a member of France’s elite Grande écoles.It was established in 1794 by the French Revolutionary government to train scientists and engineersand is still regarded as one of the most elite and selective institutions in Europe.

vii

Many of Cauchy’s ideas presented in Calcul infinitésimal were not new and hadbeen suggested in an ad hoc way by many mathematicians over the years long beforeCauchy was even born. The basic concept of the limit had been discussed amongleading mathematicians as far back as the 17th century, but without success. Jean-Baptiste le Rond d’Alembert (1717–1783) had pushed for the idea to be integrated intoa theory of the calculus as recently as the late 18th century, but once again his idea didnot catch on. As has occurred so many times through the history of science andmathematics, the early 19th century needed just the right brilliant individual to coa-lesce all the ideas then floating in the air. An individual who also possessed the abilityto combine these ideas with the strength of mathematics and proper reason to form aconsistent and rigorous theory. It was Augustin-Louis Cauchy who would be thisperson. Cauchy was the individual who had the insight and creativity to f inally bringin the precision of formulation of the limit concept with the correct mathematicalmuscle allowing him to place the calculus on a formal footing. It would be Cauchywho would manage to form a universal rigorous foundation for the calculus. Indeed,Cauchy’s work set a new standard of mathematical rigor in Europe for the remainderof the 19th century and would finally, after nearly 150 years of attempts, place thecalculus on firmly defendable ground.

The setting in which Cauchy found himself in this world is one of the most in-teresting in modern history. He was born at the dawn of the French Revolutioninto a family tightly associated with the monarchy of France. Although his familylived through a great deal of f inancial and political turmoil during his childhood, hisfather’s position in the government eventually allowed the young Cauchy access tohighly influential people. Two of these individuals were the prominent and importantFrench statesmen and scientists, Pierre-Simon de Laplace (1749–1827) and Joseph-Louis Lagrange (1736–1813). Both became family friends and recognized Cauchy’smathematical talent early. They helped place him in the best schools in Paris, ulti-mately to include the École Polytechnique, where Cauchy would then have André-Marie Ampère (1775–1836) as a tutor for his analysis and other courses. Knowingtoday the importance of the mathematical and scientific contributions from eventhese three mentors of Cauchy, the life into which this young man was launched issimply an incredible one.

A Few Notes Regarding Cauchy’s Original Works:Although real, single-variable functions are the primary focus of Cauchy’s Cal-

cul infinitésimal lectures, he spends time exploring how to extend his developmentto multiple variable and complex functions. Among many key results in the text,Cauchy will lay out all the common tools needed to utilize the calculus. Employinghis precise limit definition, Cauchy will define both the derivative and the def initeintegral as limits themselves—f irmly setting them both on a solid and rigorousfoundation. His definition of the definite integral will ultimately serve as a modelfor Bernhard Riemann’s (1826–1866) more modern version decades later. Alongwith this def inition, Cauchy will demonstrate the existence of the definite integralindependent of the derivative, a pioneering step forward in the f ield. Within Calculinf initésimal, Cauchy will provide rigorous, limit-based proofs for the General Mean

viii Translator’s Preface

Value Theorems of the Derivative and the Definite Integral, as well as the Funda-mental Theorem of Calculus and many others. Convergence properties of infiniteseries are covered in depth near the end of the text, including standard convergencetests and theorems. It is here where Cauchy will again seriously break from many ofhis contemporaries and require the convergence of an infinite series to be demon-strated before any further claims regarding its properties can be made. Cauchy willprove Taylor’s Theorem and show the Lagrangean attempt to rigorize the calculus isinadequate.2

As stated earlier, Cauchy published an enormous amount of mathematics, secondonly to Leonhard Euler (1707–1783). However, of main interest to us here are twoof the first full-length books Cauchy wrote in the early 1820s. The first of these twois, of course, our Calcul infinitésimal, published in 1823. The second is a bookpublished two years earlier, Cours d’analyse de l’École royale polytechnique (Anal-ysis Course at the Royal Polytechnic School). Within this earlier text, Cauchy laysout much of the analysis fundamentals required to develop his calculus fully, in-cluding his definitions of the limit and of continuity. But, the text stays away fromthe strictly calculus topics of the derivative and the integral. He initially plannedfor this earlier work to include two volumes, the first volume of which was pub-lished in 1821 with the title Analyse algébrique (Algebraic Analysis). For variousreasons, mostly related to disagreements with the Conseil d’instruction (Board ofInstruction)3 over the expected content of the analysis course to be taught at theÉcole Polytechnique, the second volume of this work was never produced. It is nowcustomary to refer to the published first volume simply as Cours d’analyse insteadof Analyse algébrique; but, the two titles are used interchangeably, even by Cauchyhimself. Due to the wealth of analysis results found in this text, Cauchy will refer-ence it often in Calcul infinitésimal and clearly expected his students to be familiarwith the text, or to refer to it heavily for themselves during the analysis course hewas teaching.

In the year 1882, well after Cauchy’s death, a monumental effort was begun tocompile all of Cauchy’s works into a single collection. The title of this compilationis Œuvres complètes d’Augustin Cauchy (Complete Works of Augustin Cauchy). Itconsists of 27 volumes, each published on different dates. Incredibly, this undertak-ing was not completed until the 1970s. Cauchy’s Cours d’analyse is located in Series

Translator’s Preface ix

2Many in the mathematics community of the early 19th century thought most any function could beexpanded into a Taylor series, and that the derivative of the function was simply the coefficientof the second (or linear) term of this expansion. Lagrange himself was a leading proponent of thistheory and taught this algebra-based theory of the calculus at the École Polytechnique during itsearly years. His important 1797 (and later editions) textbook, Théorie des fonctions analytiques(The Theory of Analytic Functions), was derived from these lectures. Indeed, Lagrange taught thesame analysis course that Cauchy would later inherit when young Augustin-Louis began his tenureat the school in 1815 at the age of 26.3This is the administrative body at the École Polytechnique overseeing the content and durationof each course in its curriculum. There was a near-constant struggle between Cauchy desiring arigorous class which was full of foundational analysis, and the Conseil d’Instruction who desired apractical course shorter in length and one with more applications. This struggle would cause aserious headache for Cauchy for much of his time at the École Polytechnique.

II, Tome III of this huge collection, published in 1897. His Calcul infinitésimal isfound in Series II, Tome IV, published in 1899.

A Few Technical Notes Regarding This Translation:This translation generally follows the overall page layout and formatting of the

1899 version, as the original 1823 version is quite cramped and unnecessarily dif-ficult to read, especially when it comes to Cauchy’s formulas and equations. Thecontent, emphasis, and special use of typeset of the two versions are nearly identical,but when any variations occur Cauchy’s original 1823 words and choice of text styleare almost always used here, the version under Cauchy’s control. The only rare ex-ceptions occur when the 1899 version more clearly presents Cauchy’s message andintent. Any footnotes included by Cauchy himself (of which there are only two) arefully replicated in this translation.

A short errata sheet was published with the original 1823 text. This sheet hasbeen included here; however, to improve the readability of the translation, each ofthe noted errors in the text has been corrected. Wherever one of these correctionsoccurs, a footnote has been included to identify the change. There are multiple addi-tional errors throughout the original 1823 text beyond those listed in its errata sheet.These, too, have been corrected and noted as they arise. The 1899 edition containsits share of duplicate and new errors of its own. To be complete in this translation,these have also been documented.

The 1899 reprint added several new footnotes throughout the text in places whereCauchy references a result from his 1821 Cours d’analyse. These new footnotesmark where in the Œuvres complètes d’Augustin Cauchy this reference canbe found. Cauchy makes a total of seventeen such direct references to his earlierwork throughout the Calcul infinitésimal text, twelve of which have added foot-notes in the 1899 edition. Since the original 1821 version of Cours d’analyse iswidely available,4 and once one understands where the 1897 reprint of this book islocated within the Œuvres complètes d’Augustin Cauchy, these new footnotes arenot usually very helpful. As a result, all have been excluded from this translation toreduce unnecessary clutter.

The editors of the Œuvres complètes d’Augustin Cauchy chose to append oneof Cauchy’s research papers, “On the Formulas of Taylor and Maclaurin,” to thevery end of the 1899 reprint of his Calcul infinitésimal. Although this new work isnot a part of Cauchy’s original Calcul infinitésimal, it too is translated and includedas Appendix D. Additional appendices to this book include translations of selectedexcerpts from Cauchy’s Cours d’analyse which hold particular interest.

In an effort to maintain the feel of an early 19th-century text, nearly all of theoriginal mathematical notation and symbolism have been retained in the translation,with only a small number of exceptions. Included in these exceptions is Cauchy’suse of a low-positioned dot to represent multiplication, as an example u: 1v ; or torepresent the differential of some functions, as an example d:sinx: Cauchy himselfis not consistent with the usage of this low-positioned dot in either situation, and its

x Translator’s Preface

4BnF Digital Library, Bibliotèque nationale de France—Gallica, www.bnf.fr.

http://www.bnf.fr

usage is abandoned in the 1899 reprint. There are a few other minor modifications,including an occasional use or removal of parentheses, but the number of changesfrom the original 1823 text is quite small.

Notes have been included throughout the translation to help the reader in lo-cations where Cauchy is particularly terse or unclear with his mathematics. As amodern reader progresses through these pages, one needs to keep in mind they area collection of his lecture notes, developed and published to benefit his students.It must be assumed there was plenty Cauchy would have been explaining in theclassroom as he went through this material in person from his blackboard that doesnot show up here in his text. These added footnotes help to fill that gap. Importantmathematical results Cauchy has achieved, or misconceptions he may have had atthe time are also noted to aid the reader get a sense of history in the making and afeel for how difficult this subject was to develop, even for one of the best mathe-maticians in the early 19th century. Occasional historical comments are included toremind the reader of the rich history surrounding the development of calculus andof the eventful time in Europe when Cauchy wrote the text in Paris.

Without losing Cauchy’s message or his characteristic style, punctuation hasbeen adjusted on occasion, but no attempt has been made to significantly mod-ify Cauchy’s early 19th-century wording. The primary goal of this translation hasbeen to present Cauchy’s original text in as true a form as possible, so the readerhas a sense of what it may have been like to actually sit in an analysis class ofMr. Cauchy’s in the 1820s.

A Few Final Thoughts:Cauchy’s lectures provide an excellent vehicle for the lifelong mathematics stu-

dent to gain an appreciation for the historical development of analysis and the calcu-lus. It may also serve as a particularly valuable supplement to a traditional calculusor real analysis text for those who desire a way to create more texture in a con-ventional calculus class by introducing original historical sources. As the modernreader will see, Cauchy leaves much for his students and future readers to discoverand demonstrate for themselves within the text.

It is hoped you will find the experience of full immersion into Cauchy’s worldfruitful and personally rewarding. Take your time and leisurely work throughCauchy’s lectures line by line to gain a complete appreciation for the developmentof his calculus. Witness firsthand how Cauchy struggles in spots and makes errorsin others. In this manner, you will experience the thrill and excitement one gains bywalking the same path as the master, Mr. Cauchy. Enjoy the journey.

—D. Cates, Ph.D.

Translator’s Preface xi

SUMMARY OF LECTURES

GIVEN

AT THE ROYAL POLYTECHNIC SCHOOL

ON

THE INFINITESIMAL CALCULUS,

By Mr. Augustin-Louis CAUCHY,

Engineer of Bridges and Roads,Professor of Analysis at the Royal Polytechnic School,

Member of the Academy of Sciences,Knight of the Legion of Honor.

FIRST VOLUME.

PARIS,THE ROYAL PRINTING OFFICE.

1823.

Original title page of 1823 text.

xiii

FOREWORD.

This work,5 undertaken on the request of the Board of Instruction of the Royal Poly-technic School,6 offers a summary of the lectures that I gave to this school on theinfinitesimal calculus. It will be composed of two volumes7 corresponding to thetwo years which form the duration of the course. I publish the first volume today di-vided into forty lectures, the first twenty of which comprise the differential calculus,and the last twenty a part of the integral calculus. The methods that I follow differin several respects from those which are found expressed in the works of similartype. My main goal has been to reconcile the rigor, which I have made a law in myAnalysis Course, with the simplicity which results from the direct consideration ofinfinitely small quantities. For this reason, I thought obliged to reject the expansionof functions by infinite series, whenever the series obtained are not convergent;8

and I saw myself forced to return the formula of Taylor to the integral calculus,and this formula can only be admitted as general so long as the series that it containsis found reduced to a finite number of terms and supplemented by a definite integral.I am aware that the illustrious author of the Analytical Mechanics has taken the for-mula in question9 for the basis of his theory of derived functions.10 But, despite all

5 This is the original AVERTISSEMENT included in the 1823 edition.6Known as the Conseil d’instruction.7Cauchy had been at constant odds with the Conseil d’Instruction over the content and durationof the analysis course for years. They much preferred a shorter course with more applications.In the summer of 1824, following the publication of Calcul infinitésimal (the first volume of thiswork), the printing of the lectures for the second volume was begun (the first thirteen lectures wereprinted), but then suddenly halted. The reason surrounding this interruption remains unclear, butlikely was due to the Conseil d’Instruction’s displeasure with the material Cauchy had thus fardeveloped. In any event, the printing was never resumed and the complete second volume wasnever published.8 Cauchy breaks from many mathematicians up to his time by requiring the convergence of aninfinite series be demonstrated before any claim of its properties could be made.9 The author Cauchy is referring to is one of his mentors, Joseph-Louis Lagrange (1736–1813).The formula is that of Taylor.10The phrase derived function had previously been used and established by Lagrange, but todayderivative function is more commonly employed.

xv

the respect that such a grand authority commands, the majority of mathematicians11

are now in accordance to recognize the uncertainty of the results which we can beled to by the employment of divergent series, and we add that in several cases, theTheorem of Taylor seems to provide the expansion of a function by convergentseries, even though the sum of the series differs fundamentally from the proposedfunction (see the end of the thirty-eighth lecture). Moreover, those who will read mywork, will be convinced, I hope, that the principles of the differential calculus andits most important applications, can be easily explained without the intervention ofseries.

In the integral calculus, it seemed necessary to demonstrate generally the exis-tence of integrals or primitive functions before making known their various proper-ties. To achieve this, it was first necessary to establish the notion of integrals takenbetween given limits, or definite integrals.12 These latter objects can sometimes beinfinite or indeterminate, it being essential to study in which cases they maintain aunique and finite value. The simplest means of resolving the question are the employ-ment of singular definite integrals which are the subject of the twenty-fifth lecture.Moreover, among the infinite number of values that we can attribute to an indeter-minate integral, there exists one which merits our particular attention and that wehave named principal value. The consideration of singular definite integrals andthose of the principal values of indeterminate integrals are very useful in the solu-tion of a large group of problems. We deduce a great number of general formulasthat work for the determination of def inite integrals and are similar to those that Igave in a report presented to the Institute in 1814. We will find in lectures thirty-fourand thirty-nine a formula of this type applied for the evaluation of several def initeintegrals, some of which were already known.

xvi FOREWORD.

11 Cauchy uses the term géomètres here, a term that would have referred to any mathematicianduring his time.12The integral was considered not nearly as important as the derivative by most mathematicians ofCauchy’s time, generally viewing it simply as the inverse of the derivative. Cauchy’s definiteintegral existence proof is important because it demonstrates the independence of the integral fromthe derivative.

ERRATA.

This is the original errata sheet included in the 1823 publication. All of these errors have beencorrected in this translation along with added footnotes wherever a change has been made. Thecorrections on pages 37 and 47 (these are page references to Cauchy’s original text) are due toslightly different characters or symbols being used. The variations between what was actually usedfor the printing and the corrected versions are slight—perhaps an odd typeset in use at the time.

PAGES. LINES. ERRORS. CORRECTIONS.

28. 27. LðxÞ lðxÞIbid. 28. Lðxþ hÞ � LðxÞ lðxþ hÞ � lðxÞ37. 28. L x

y

� �L x

y

� �

47. 23. , ,

Ibid. 29. d2x d2 x

56. 26. dnx u; vdnx u;

68. 2 and 12. Fðr; 0; 0; 0 : :Þ FðrÞIbid. Ibid. Fðr; s; 0; 0 : :Þ Fðr; sÞIbid. Ibid. Fðr; s; t; 0 : :Þ Fðr; s; tÞ

xvii

TABLE OF CONTENTS.

CONTAINED IN THIS VOLUME.

Translator’s Preface. vii

1823 Title Page. xiii

FOREWORD. xv

ERRATA. xvii

DIFFERENTIAL CALCULUS.

FIRST LECTURE. 3Of variables, their limits, and infinitely small quantities.

SECOND LECTURE. 7Of continuous and discontinuous functions. Geometric representation ofcontinuous functions.

THIRD LECTURE. 11Derivatives of functions of a single variable.

FOURTH LECTURE. 17Differentials of functions of a single variable.

FIFTH LECTURE. 21The differential of the sum of several functions is the sum of theirdifferentials. Consequences of this principle. Differentials of imaginaryfunctions.

xix

SIXTH LECTURE. 27Use of differentials and derived functions in the solution of severalproblems. Maxima and minima of functions of a single variable. Valuesof fractions which are presented under the form 0

0.

SEVENTH LECTURE. 31Values of some expressions which are presented under the indeterminateform 1

1 ;10; . . . A relationship which exists between the ratio offinite differences and the derived function.

EIGHTH LECTURE. 37Differentials of functions of several variables. Partial derivatives andpartial differentials.

NINTH LECTURE. 43Use of partial derivatives in the differentiation of composed functions.Differentials of implicit functions.

TENTH LECTURE. 49Theorem of homogeneous functions. Maxima and minima of functionsof several variables.

ELEVENTH LECTURE. 53Use of indeterminate factors in the study of maxima and minima.

TWELFTH LECTURE. 59Differentials and derivatives of various orders for functions of a singlevariable. Change of the independent variable.

THIRTEENTH LECTURE. 65Differentials of various orders for functions of several variables.

FOURTEENTH LECTURE. 71Methods that work to simplify the study of total differentials for functionsof several independent variables. Symbolic values of these differentials.

FIFTEENTH LECTURE. 77Relationships which exist between functions of a single variable and theirderivatives or differentials of various orders. Use of these differentials inthe study of maxima and minima.

xx TABLE OF CONTENTS.

SIXTEENTH LECTURE. 81Use of differentials of various orders in the study of maxima and minimaof functions of several variables.

SEVENTEENTH LECTURE. 85Conditions which must be fulfilled for a total differential to not changesign while we change the values attributed to the differentials of theindependent variables.

EIGHTEENTH LECTURE. 91Differentials of any function of several variables each of which is in itsturn a linear function of other supposed independent variables. Decom-position of entire functions into real factors of first or of second degree.

NINETEENTH LECTURE. 97Use of derivatives and differentials of various orders in the expansionof entire functions.

TWENTIETH LECTURE. 103Decomposition of rational fractions.

INTEGRAL CALCULUS.

TWENTY-FIRST LECTURE. 111Definite integrals.

TWENTY-SECOND LECTURE. 117Formulas for the determination of exact or approximate values ofdefinite integrals.

TWENTY-THIRD LECTURE. 123Decomposition of a definite integral into several others. Imaginarydefinite integrals. Geometric representation of real definite integrals.Decomposition of the function under the ʃ sign into two factors inwhich one always maintains the same sign.

TWENTY-FOURTH LECTURE. 129Of definite integrals whose values are infinite or indeterminate.Principal values of indeterminate integrals.

TABLE OF CONTENTS. xxi

xxii TABLE OF CONTENTS.

TWENTY-FIFTH LECTURE. 133Singular definite integrals.

TWENTY-SIXTH LECTURE. 137Indefinite integrals.

TWENTY-SEVENTH LECTURE. 143Various properties of indefinite integrals. Methods to determine thevalues of these same integrals.

TWENTY-EIGHTH LECTURE. 149On indefinite integrals which contain algebraic functions.

TWENTY-NINTH LECTURE. 155On the integration and the reduction of binomial differentials and ofany other differential formulas of the same type.

THIRTIETH LECTURE. 161On indefinite integrals which contain exponential, logarithmic, orcircular functions.

THIRTY-FIRST LECTURE. 167On the determination and the reduction of indefinite integrals in whichthe function under the ʃ sign is the product of two factors equal tocertain powers of sines and of cosines of the variable.

THIRTY-SECOND LECTURE. 173On the transition of indefinite integrals to definite integrals.

THIRTY-THIRD LECTURE. 179Differentiation and integration under the ʃ sign. Integration ofdifferential formulas which contain several independent variables.

THIRTY-FOURTH LECTURE. 185Comparison of two types of simple integrals which result in certaincases from a double integration.

THIRTY-FIFTH LECTURE. 191Differential of a definite integral with respect to a variable included inthe function under the ʃ sign and in the limits of integration. Integrals ofvarious orders for functions of a single variable.

THIRTY-SIXTH LECTURE. 197Transformation of any functions of x or of x + h into entire functions ofx or of h to which we append definite integrals. Equivalent expressionsto these same integrals.

THIRTY-SEVENTH LECTURE. 203Theorems of Taylor and of Maclaurin. Extension of these theorems tofunctions of several variables.

THIRTY-EIGHTH LECTURE. 209Rules on the convergence of series. Application of these rules to theseries of Maclaurin.

THIRTY-NINTH LECTURE. 215Imaginary exponentials and logarithms. Use of these exponentials andof these logarithms in the determination of definite or indefiniteintegrals.

FORTIETH LECTURE. 221Integration by series.

END OF THE TABLE.

ADDITION.General Mean Value Theorem. l’Hôpital’s Rule. Formula of Taylor.13 227

APPENDICES.

APPENDIX A: COURS D’ANALYSE – CHAPTER II, §III. 239Simple Convergence Theorems.

APPENDIX B: COURS D’ANALYSE – NOTE II. 245Average Theorems.

APPENDIX C: COURS D’ANALYSE – NOTE III. 249Intermediate Value Theorem.

TABLE OF CONTENTS. xxiii

13This addendum is not listed in the original 1823 TABLE OF CONTENTS but is included in thepublished text.

APPENDIX D: ON THE FORMULAS OF TAYLOR ANDMACLAURIN. 253Research paper addendum included in 1899 reprint.

APPENDIX E: PAGINATION OF THE 1823 & 1899 EDITIONS. 259Comparison of page numbers between the two versions.

REFERENCES 261

INDEX 263

xxiv TABLE OF CONTENTS.

Part IDIFFERENTIAL CALCULUS.

FIRST LECTURE.

OF VARIABLES, THEIR LIMITS,AND INFINITELY SMALL QUANTITIES.

We call a variable quantity that which we consider as having to successively receiveseveral different values, one from the other. On the contrary, we call a constantquantity any quantity that receives a fixed and determined value. When the valuessuccessively attributed to the same variable approach a fixed value indefinitely, insuch a manner as to end up differing from it by as little as we will wish, this finalvalue is called the limit of all the others.1 Thus, for example, the surface of a circleis the limit toward which the surfaces of regular polygons inscribed inside converge,while the number of their sides grows increasingly; and the radius vector, originatingfrom the center of a hyperbola to a point of the curve which moves further andfurther from the center, forms with the x-axis an angle which has for a limit, theangle formed by the asymptote with the same axis; …. We will indicate the limittoward which a given variable converges by the abbreviation lim placed in front ofthis variable.2

Often the limits toward which the variable expressions converge are presentedunder an indeterminant form, and nevertheless we can still determine, with the helpof particular methods, the actual values of these same limits. Thus, for example, thelimits which the two variable expressions3

sin α

α, (1 + α)

1α

1Incredibly, this is Cauchy’s historic definition of the limit. One of Cauchy’s most famous con-tributions, it looks very little like the modern definition. The words throw us off, but the mannerin which Cauchy ultimately uses this definition in several locations later in this text demonstrateshis definition is, in fact, the same one that will be more rigorously symbolized with δ’s and ε’s byKarl Weierstrass (1815–1897) later in the 19th century.2Others had used this notation for the limit concept before Cauchy. Simon Antoine Jean L’Huilier(1750–1840) is generally credited with the earliest use in the year 1786 in his winning entry to theRoyal French Academy of Sciences’ challenge of “A Clear and Precise Theory on the Nature ofInfinity,” a contest suggested and judged by Joseph-Louis Lagrange (1736–1813).3As Cauchy leads us through his solutions for the limits of these two examples, keep in mind theyare the first two challenging examples of a limit he chooses to provide his beginning infinitesimalcalculus students as their introduction into how to apply his limit concept.

© Springer Nature Switzerland AG 2019D. M. Cates, Cauchy’s Calcul Infinitésimal,https://doi.org/10.1007/978-3-030-11036-9_1

3

http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-030-11036-9_1&domain=pdf

https://doi.org/10.1007/978-3-030-11036-9_1

4 CALCUL INFINITÉSIMAL.

indefinitely approach, while α converges toward zero, are presented under the inde-terminant forms 0

0 , 1±∞; and yet, these two limits have fixed values that we are able

to calculate as follows.We obviously have, for very small numerical values of α,

sin α

sin α>

sin α

α>

sin α

tangα.

By consequence, the ratio sin αα

, always contained between the two quantities sin αsin α

=1, and sin α

tangα= cosα, the first of which serves to limit the second, will itself have

unity for a limit.4

Let us now look for the limit toward which the expression (1 + α)1α converges,

while α indefinitely approaches zero. If we first suppose the quantity α positiveand of the form 1

m , m designating a variable integer number5 and susceptible to anindefinite increment, we will have

(1 + α)1α =

(1 + 1

m

)m

= 1 + 1

1+ 1

1 · 2(1 − 1

m

)+ 1

1 · 2 · 3(1 − 1

m

) (1 − 2

m

)+ · · ·

+ 1

1 · 2 · 3 · · · m

(1 − 1

m

) (1 − 2

m

)· · ·

(1 − m − 1

m

).

Since, in the second member of this last formula,6 the terms which contain the quan-tity m are all positive, and grow in value and in number at the same time as this quan-tity, it is clear that the expression

(1 + 1

m

)mwill itself grow along with the integer

number m, while always remaining contained between the two sums

1 + 1

1= 2

and7

1 + 1

1+ 1

2+ 1

2 · 2 + 1

2 · 2 · 2 + · · · = 1 + 1 + 1 = 3;

4Cauchy argues in his 1821 Cours d’analyse Chapter II, §III, that α > sin α and tangα > α.

Presumably, his students would have been familiar with this earlier work, or perhaps Cauchy wouldhave quickly covered this in class. This may be one example explaining why Cauchy’s classestended to run overtime. He is making use of what we know today as the Sandwich (or Squeeze)Theorem, a result which likely seemed obvious to Cauchy and one not in need of a proof.5Generally, Cauchy uses the term number to denote a positive value and the term quantity todenote a signed number.6The “second member” of the formula is the right-hand side of the equation.7Cauchy is using an argument similar to that of Nicole Oresme (1325–1382), one of the first tosuggest the infinite series 1

2 + 12·2 + 1

2·2·2 + · · · = 1.

FIRST LECTURE. 5

therefore, it will indefinitely approach, for increasing values of m, a certain limitcontained between 2 and 3. This limit is a number which plays a great role in theinfinitesimal calculus and which we agree to designate by the letter e.8 If we takem = 10000, we will find for the approximate value of e, by making use of tables ofdecimal logarithms, (

10001

10000

)10000

= 2.7183.

This approximate value is exact to within a ten-thousandth,9 as we will see muchlater.

We now suppose that α, always positive, is no longer of the form 1m . We desig-

nate in this hypothesis by m and n = m + 1, the two integer numbers immediatelyless than and greater than 1

α, so that we have

1

α= m + μ = n − ν,

μ and ν being numbers contained between zero and unity. The expression (1+ α)1α

will obviously be contained between the following two

(1 + 1

m

) 1α

=[ (

1 + 1

m

)m ]1+ μ

m

,

(1 + 1

n

) 1α

=[(

1 + 1

n

)n ]1− νn

;

and since, for the infinitely decreasing values of α, or to what amounts to the samething, for the always increasing values of m and of n, the two quantities

(1 + 1

m

)m,(

1+ 1n

)nconverge, one and the other, toward the limit e, while 1+ μ

m , 1− νn indef-

initely approach the limit 1, it follows that each of the expressions

(1 + 1

m

) 1α

,

(1 + 1

n

) 1α

,

and as a result, the intermediate expression (1 + α)1α will also converge toward the

limit e.Finally, we suppose that α becomes a negative quantity. If we let in this hypoth-

esis

1 + α = 1

1 + β,

β will be a positive quantity which will itself converge toward zero, and we will find

(1 + α)1α = (1 + β)

1+β

β =[(1 + β)

1β

]1+β ;

8This is the first of multiple instances within Calcul infinitésimal in which Cauchy defines aquantity based on a limit.9The 1899 edition has the fraction 1

10000 .


then, by passing to the limits,10

lim (1 + α)1α = elim (1+β) = e.

When the successive numerical values of the same variable indefinitely decreasein such a manner as to drop below any given number, this variable becomes whatwe call an infinitesimal, or an infinitely small quantity.11 A variable of this kind haszero for a limit. Such is the variable α in the preceding calculations.

When the successive numerical values of the same variable grow increasinglyin such a manner as to rise above any given number, we say that this variable haspositive infinity for a limit, indicated by the sign∞, if the variable in question is pos-itive; and negative infinity, indicated by the notation −∞, if the variable in questionis negative. Such is the variable number m that we have employed above.

10Cauchy rarely includes in his notation the type of limit in question. As in this case, instead ofwriting

limα→0

(1 + α)1α and lim

β→0(1 + β),

as we would today, Cauchy uses the phrase, “passing to the limits.” Although Cauchy is not atall consistent as to how he deals with this notation throughout the remainder of these lectures, hisintent is generally clear. As will occur in many locations throughout the text, this second limitexample provided by Mr. Cauchy includes a wide bounty of beautiful mathematics.11This definition for an infinitesimal is also used earlier in Cauchy’s 1821 Cours d’analyse text;however, it was still relatively new at the time. An important point here, which was a break frommany of his contemporaries, is that Cauchy’s infinitesimal is not necessarily zero itself, only thatits limit is zero.

SECOND LECTURE.

OF CONTINUOUS AND DISCONTINUOUSFUNCTIONS. GEOMETRIC REPRESENTATIONOF CONTINUOUS FUNCTIONS.

When variable quantities are so related among themselves that, the value of one ofthem being given, we are able to deduce the values of all the others, we usuallyconsider these various quantities expressed by means of one among them, whichthen takes the name of the independent variable; and the other quantities, expressedby means of the independent variable, are what we call functions of this variable.1

When variable quantities are so related among themselves that, the values of afew being given, we are able to deduce those of all the others, we consider these vari-ous quantities expressed by means of several among them, which then take the nameof the independent variables; and the remaining quantities, expressed by means ofthe independent variables, are what we call functions of these same variables. Thevarious expressions that produce the algebra and the trigonometry, when they containvariables considered as independent, are such functions of these variables. Thus,for example,2

Lx, sin x, . . .

are functions of the variable x;x + y, x y, xyz, . . .

functions of the variables x and y, or x, y, and z, etc.When the functions of one or several variables are found, as in the previous ex-

amples, immediately expressed by means of these same variables, they are namedexplicit functions. But, when given only the relations between the functions and

1The concept of the function had been around for a long time. René Descartes (1596–1650) cer-tainly had the idea in mind as he wrote La géométrie in 1637, but the concept had been floating inthe air even before his time. However, in the middle of the 18th century, Leonhard Euler (1707–1783) made the function concept central to the calculus, shifting the focus of the discipline fromcurves to functions. This shift was instrumental in setting the stage for the rigorization of the cal-culus.2The notation Lx is the one Cauchy will use to denote our log x today. Additionally, lx is thenotation he will use to denote our ln x . Both will be used extensively later in the text. This notationhas been retained to maintain the feel of Cauchy’s original work.


7


https://doi.org/10.1007/978-3-030-11036-9_2


the variables, that is to say, the equations to which these quantities must satisfy,as long as these equations are not resolved algebraically, the functions, not beingimmediately expressed by means of the variables, are called implicit functions. Torender them explicit, it is sufficient to resolve, when these can, the equations whichdetermine them. For example, y being an implicit function of x determined by theequation

Ly = x,

if we call A the base of the system of logarithms that we consider, the same function,becoming explicit by the solution of the given equation, will be3

y = Ax .

When we want to represent an explicit function of a single variable x, or ofseveral variables x, y, z, . . . , without determining the nature of this function, weemploy one of the notations

f (x), F(x), ϕ(x), χ(x), ψ(x), ϖ(x), . . . ,

f (x, y, z, . . . ), F(x, y, z, . . . ), ϕ(x, y, z, . . . ), . . . .

Often, in the calculus, we make use of the character Δ to indicate the simultane-ous increments of two variables which depend one on the other. This granted, if thevariable y is expressed as a function of the variable x by the equation

y = f (x),(1)

Δy, or the increment of y corresponding to the increment Δx of the variable x, willbe determined by the formula

y + Δy = f (x + Δx).(2)

More generally, if we suppose

F(x, y) = 0,(3)

we will have

F(x + Δx, y + Δy) = 0.(4)

It is good to observe that, from combining equations (1) and (2), we infer

Δy = f (x + Δx) − f (x).(5)

Now let h and i be two distinct quantities, the first finite, the second infinitelysmall, and let α = i

h be the infinitely small ratio of these two quantities. If we

3Although Cauchy does not make this perfectly clear until later in this lecture, A is always apositive value. So, the function Ax is well defined.

SECOND LECTURE. 9

attribute toΔx the finite value h, the value ofΔy, given by equation (5), will becomewhat we call the finite difference of the function f (x), and will ordinarily be a finitequantity. If, on the contrary, we attribute to Δx an infinitely small value, if we let,for example,

Δx = i = αh,

the value of Δy, namely,

f (x + i) − f (x) or f (x + αh) − f (x),

will ordinarily be an infinitely small quantity. This is what we will easily verify withregard to the functions

Ax , sin x, cos x,

to which correspond the differences

Ax+i − Ax = (Ai − 1)Ax ,

sin (x + i) − sin x = 2 sini

2cos

(x + i

2

),

cos (x + i) − cos x = −2 sini

2sin

(x + i

2

),

which each contain a factor Ai − 1 or sin i2 that indefinitely converge with i toward

the limit zero.When, the function f (x) admitting a unique and finite value for all the values of

x contained between two given limits, the difference

f (x + i) − f (x)

is always, between these limits, an infinitely small quantity, we say that f (x) is acontinuous function of the variable x between the limits in question.4

We again say that the function f (x) is, in the vicinity of a particular value at-tributed to the variable x, a continuous function of this variable whenever it is con-tinuous between two limits, even very close together, which contain the value inquestion.

Finally, when a function ceases to be continuous in the vicinity of a particularvalue of the variable x, we say that it then becomes discontinuous, and there is for

4Cauchy’s critically important definition for a continuous function. His imprecise wording heredoes not clearly distinguish the difference between continuity at a point and uniform continuityover an interval. Debate over this issue goes on to the present day. Fortunately, the subtle conceptof uniform continuity does not create a serious issue in most of Cauchy’s work within his Calculinfinitésimal, as he does nearly all of his analysis in this text for well-behaved functions on closed,bounded intervals. It was shown much later in the 19th century that in this situation, a continuousfunction is also uniformly continuous – a result of the Heine–Borel Theorem, named after EduardHeine (1821–1881) and Émile Borel (1871–1956). However, Cauchy’s imprecision here will goon to haunt him in proofs of several theorems to follow.


this particular value a solution of continuity.5 Thus, for example, there is a solutionof continuity in the function 1

x , for x = 0; in the function tang x, for x = ± (2k+1)π2 ,

k being any integer number; etc.After these explanations, it will be easy to recognize between which limits a

given function of the variable x is continuous with respect to this variable. (See, formore ample developments, Chapter II of the 1st part of Analysis Course, publishedin 1821.)

Consider now that we construct the curve which has for an equation in rectan-gular coordinates, y = f (x). If the function f (x) is continuous between the limitsx = x0, x = X, to each abscissa x contained between these limits will correspond asingle ordinate; and moreover, as x comes to grow by an infinitely small quantityΔx, y will grow an infinitely small quantity Δy. As a result, two abscissas veryclose together, x, x +Δx, will correspond to two points very close, one to the other,since their distance

√Δx2 + Δy2 will itself be an infinitely small quantity. These

conditions can only be satisfied as long as the different points form a continuouscurve between the limits x = x0, x = X.

Examples. – Construct the curves represented by the equations

y = xm, y = 1

xm, y = Ax , y = Lx, y = sin x,

in which A denotes a positive constant and m an integer number.Determine the general forms of these same curves.

5Cauchy’s original text uses the phrase solution de continuité here, translated literally as solutionof continuity, which is an older term in use at his time meaning points where continuity dissolves,ceases to exist, or disappears. Today we would certainly call this a point of discontinuity. This olderphrase has been retained through this translation.

THIRD LECTURE.

DERIVATIVES OF FUNCTIONS OF A SINGLEVARIABLE.

When the function y = f (x) remains continuous between two given limits of thevariable x, and that we assign to this variable a value contained between the twolimits in question, an infinitely small increment attributed to the variable producesan infinitely small increment of the function itself. By consequence, if we then setΔx = i, the two terms of the ratio of differences

Δy

Δx= f (x + i) − f (x)

i(1)

will be infinitely small quantities. But, while these two terms indefinitely and simul-taneously will approach the limit of zero, the ratio itself may be able to convergetoward another limit, either positive or negative. This limit, when it exists, has a de-termined value for each particular value of x; but, it varies along with x . Thus, forexample, if we take f (x) = xm, m designating an integer number, the ratio betweenthe infinitely small differences will be

(x + i)m − xm

i= mxm−1 + m(m − 1)

1 · 2 xm−2i + · · · + im−1,

and it will have for a limit the quantity mxm−1, that is to say, a new function ofthe variable x . It will be the same in general, only the form of the new function,which will serve as the limit of the ratio f (x + i) − f (x)

i , will depend on the form ofthe proposed function y = f (x).1 To indicate this dependence, we give to the new

1Cauchy is quite aware of the difficulties involved in dealing with infinitesimal quantities. Fromthe beginnings of the calculus, Isaac Newton (1643–1727) and Gottfried Wilhelm Leibniz (1646–1716) developed similar approaches to finding the derivative but had trouble as they attemptedto explain the meaning of this ratio as i approached (or even became) zero. Cauchy avoids thisproblem by specifically not defining the derivative as the ratio of anything, as had been the basicapproach of Newton and Leibniz. Instead, Cauchy only considers the limit of the difference quo-tient as the two infinitesimals each approach zero. Notice also that he is very careful to only definethe derivative when this limit exists.


11


https://doi.org/10.1007/978-3-030-11036-9_3


function the name of derived function,2 and we represent it, with the help of anaccent mark, by the notation3

y′ or f ′(x).

In the study of derivatives of functions of a single variable x, it is useful to dis-tinguish the functions that we call simple and that we consider as resulting froma single operation performed on this variable, with the functions that we constructwith the help of several operations and that we call composed. The simple functionsthat produce the operations of the algebra and of the trigonometry (see the 1st partof Analysis Course, Chapter I) can be reduced to the following

a + x, a − x, ax,a

x, xa, Ax , Lx,

sin x, cos x, arcsin x, arccos x,

A designating a constant number, a = ±A a constant quantity,4 and the letter Lindicating a logarithm taken in the system whose base is A. If we take one of thesesimple functions for y, it will be easy in general to obtain the derived function y′.We will find, as an example, for y = a + x,

Δy

Δx= (a + x + i) − (a + x)

i= 1, y′ = 1;

for y = a − x,

Δy

Δx= (a − x − i) − (a − x)

i= −1, y′ = −1;

for y = ax,Δy

Δx= a(x + i) − ax

i= a, y′ = a;

for y = ax ,

Δy

Δx=

ax+i − a

x

i= − a

x(x + i), y′ = − a

x2;

for y = sin x,

Δy

Δx= sin

(12 i

)

12 i

cos(x + 1

2 i), y′ = cos x = sin

(x + π

2

);

2Cauchy’s original text uses the phrase fonction dérivée here, which has been translated literallyas derived function, a traditional phrase which Lagrange had used earlier to denote what today wewould call the derivative. Cauchy has defined his derived function to be the limit of the differencequotient, just as we do today.3Cauchy has also borrowed the prime derivative notation from his mentor, Lagrange.4Recall from the previous lecture Cauchy assumes A, a number, to be positive. Therefore, thequantity a can be positive or negative.

THIRD LECTURE. 13

for y = cos x,

Δy

Δx= − sin

(12 i

)

12 i

sin(x + 1

2 i), y′ = − sin x = cos

(x + π

2

).

Furthermore, by setting i = αx, Ai = 1 + β, and (1 + α)a = 1 + γ, we will find,for y = Lx,

Δy

Δx= L(x + i) − Lx

i= L(1 + α)

αx= L(1 + α)

1α

x, y′ = Le

x;

for y = Ax ,

Δy

Δx= Ax+i − Ax

i= Ai − 1

iAx = Ax

L(1 + β)1β

, y′ = Ax

Le;

for y = xa,

Δy

Δx= (x + i)a − xa

i= (1 + α)a − 1

αxa−1 = L(1 + α)

1α

L(1 + γ )1γ

axa−1, y′ = axa−1.

In these last formulas, the letter e represents the number 2.718 . . . which servesas the limit to the expression (1 + α)

1α. If we take this number for the base of a

system of logarithms, we will obtain the Napierian or hyperbolic logarithm, that wewill always indicate with the aid of the letter l.5 This granted, we will obviously have

le = 1, Le = Le

LA= le

lA= 1

lA;

and moreover, we will find, for y = lx,

y′ = 1

x;

for y = ex,

y′ = ex .

The various preceding formulas being established only for values of x which cor-respond to real values of y, we must suppose x positive in those of these formulaswhich relate to the functions Lx, lx, and also to the function xa when a denotes afraction of an even denominator or an irrational number.

Now, let z be a second function of x, related to the first, y = f (x), by the formula

z = F(y).(2)

5Our ln x, the natural logarithm.


z, or F[

f (x)],will be what we call a function of a function of the variable x; and, if

we designate by Δx,Δy,Δz the infinitely small and simultaneous increments of thethree variables x, y, z, we will find

Δz

Δx= F(y + Δy) − F(y)

Δx= F(y + Δy) − F(y)

Δy

Δy

Δx;


z′ = y′F ′(y) = f ′(x)F ′[ f (x)].(3)

For example, if we let z = ay, and y = lx, we will have

z′ = ay′ = a

x.

With the help of formula (3), we will easily determine the derivatives of the simplefunctions Ax , xa, arcsin x, arccos x by supposing those of the functions Lx, sin x,

cos x are known. We will find, in fact, for y = Ax ,

Ly = x, y′Le

y= 1, y′ = y

Le= Ax lA;

for y = xa,

ly = a lx, y′ 1y

= a

x, y′ = a

y

x= axa−1;

for y = arcsin x,

sin y = x, y′ cos y = 1, y′ = 1

cos y= 1√

1 − x2;

for y = arccos x,7

cos y = x, y′ × ( − sin y) = 1, y′ = −1

sin y= −1√

1 − x2.

Moreover, the derivatives of the composed functions

Ay, ey,1

y,

being respectively, by virtue of formula (3),

y′ Ay lA, y′ey, − y′

y2;

6Our modern-day Chain Rule. Cauchy is assuming a case where Δy cannot be zero.7Cauchy’s original 1823 text correctly has y′ = −1

sin y = −1√1−x2

, but the 1899 reprint does notinclude the second minus sign.

THIRD LECTURE. 15

the derivatives of the following

ABx, eex

, sec x = 1

cos x, cosec x = 1

sin x

will become

ABxBx lA lB, eex

ex ,sin x

cos2 x, − cos x

sin2 x.

In finishing, we will remark that the derivatives of composed functions are some-times determined as easily as those of simple functions. Thus, for example, we find,8

for y = tang x = sin xcos x ,

Δy

Δx= 1

i

[sin (x + i)

cos (x + i)− sin x

cos x

]= sin i

i cos x cos (x + i), y′ = 1

cos2 x;

for y = cot x = cos xsin x ,

Δy

Δx= 1

i

[cos (x + i)

sin (x + i)− cos x

sin x

]= − sin i

i sin x sin (x + i), y′ = − 1

sin2 x;

and we conclude, for y = arctang x,

tang y = x,y′

cos2 y= 1, y′ = cos2 y = 1

1 + x2;

for y = arccot x,

cot y = x,−y′

sin2 y= 1, y′ = − sin2 y = −1

1 + x2.

8Cauchy uses the notation tang x to symbolize the tangent function, tan x . In addition, he uses thenotation cosec x to denote the cosecant function. To maintain Cauchy’s style, these notations havebeen retained throughout this translation.

The idea of a limit-based theory for the calculus had been around since its infancy. Around theyear 1755, Jean le Rond d’Alembert (1717–1783) tried to revive the idea and suggested a newdevelopment of a theory based on limits. Although d’Alembert did not describe the limit con-cept with the algebraic precision that would follow by Cauchy, his idea was similar to Cauchy’s.D’Alembert’s idea once again did not catch on at the time, but other continental mathematicians,including Cauchy, would eventually take the reins of this development and provide the necessarymathematical muscle needed for a powerful rigorization of the calculus in the 19th century.

FOURTH LECTURE.

DIFFERENTIALS OF FUNCTIONS OF A SINGLEVARIABLE.

Let y = f (x) always be a function of the independent variable x, i an infinitely smallquantity, and h a finite quantity. If we set i = αh, α will also be an infinitely smallquantity, and we will have identically

f (x + i) − f (x)

i= f (x + αh) − f (x)

αh,

where we will deduce

f (x + αh) − f (x)

α= f (x + i) − f (x)

ih.(1)

The limit toward which the first member of equation (1) converges, while the vari-able α indefinitely approaches zero and the quantity h remaining constant, is whatwe call the differential of the function y = f (x). We indicate this differential by thecharacter d as follows

dy or d f (x).

It is easy to obtain its value when we know that of the derived function y′ or f ′(x).

In fact, by taking the limits of both members of equation (1), we will generally find

d f (x) = h f ′(x).(2)

In the particular case where f (x) = x, equation (2) is reduced to

dx = h.(3)

Thus, the differential of the independent variable x is nothing other than the finiteconstant h. This granted, equation (2) will become

d f (x) = f ′(x) dx,(4)

or to what amounts to the same thing,

dy = y′ dx .(5)


17


https://doi.org/10.1007/978-3-030-11036-9_4


It results from these last equations that the derivative, y′ = f ′(x), of any function,y = f (x), is precisely equal to dy

dx , that is to say, to the ratio between the differentialof the function and that of the variable, or if we want, to the coefficient by which itis necessary to multiply the second differential to obtain the first. It is for this reasonthat we sometimes give to the derived function the name of differential coefficient.

To differentiate a function is to find its differential. The operation by which wedifferentiate is called differentiation.

By virtue of formula (4), we will immediately obtain the differentials of func-tions from which we have calculated derivatives. If we first apply this formula tofunctions, we will find

d(a + x) = dx, d(a − x) = −dx, d(ax) = a dx,

da

x= −a

dx

x2, dxa = axa−1 dx,

d Ax = Ax lA dx, dex = ex dx,

dLx = Ledx

x, dlx = dx

x,

d sin x = cos x dx = sin(

x + π

2

)dx, d cos x = − sin x dx = cos

(x + π

2

)dx,

d arcsin x = dx√1 − x2

, d arccos x = − dx√1 − x2

.

We will also establish the equations

d tang x = dx

cos2 x, d cot x = − dx

sin2 x,

d arctang x = dx

1 + x2, d arccotx = − dx

1 + x2,

d sec x = sin x dx

cos2 x, d cosec x = −cos x dx

sin2 x.

These various equations, as well as those which we reached in the precedinglecture, must be considered up to the present as demonstrated only for values of xwhich correspond to real values of the functions for which we find the derivatives.

By consequence, among the simple functions, those for which the differentialscan be considered known for any real values of the variable x, are the following

a + x, a − x, ax,a

x, Ax , ex , sin x, cos x,

and the function xa when the numerical value of a is reduced to an integer number orto a fraction of odd denominator.1 But, we must suppose the variable x is included

1This requirement is necessary to ensure the function is well defined for negative values of x .Cauchy must also be assuming x �= 0 for the function a

x . He views this location as a solution ofcontinuity (a point of discontinuity).

FOURTH LECTURE. 19

between the two limits −1,+1 in the differentials found of the simple functionsarcsin x, arccos x, and between the limits 0,∞ in the differentials of the functionsLx, lx, and the same in that of the function xa whenever the numerical value of abecomes a fraction of even denominator or an irrational number.

It is still essential to observe that, in accordance with the conventions establishedin the 1st part of Analysis Course, we make use of one of the notations

arcsin x, arccos x, arctang x, arccot x, arcsec x, arccosec x

to represent, not just any of the arcs2 by which a certain trigonometric line3 is equalto x, but the one among them which has the smallest numerical value, or if these arcsare both equal and of opposite signs, that which has the smaller positive value; byconsequence, arcsin x, arctangx, arccotx, arccosecx are the arcs contained betweenthe limits −π

2 ,+π2 , and arccos x, arcsecx, the arcs contained between the limits 0

and π.

When we suppose y = f (x) andΔx = i = αh, equation (1), for which the secondmember has for a limit dy, can be presented under the form

Δy

α= dy + β,

β designating an infinitely small quantity and we infer

Δy = α(dy + β).(6)

Let z be a second function of the variable x . We will similarly have

Δz = α(dz + γ ),

γ designating again an infinitely small quantity. We will find, as a result,

Δz

Δy= dz + γ

dy + β;

then, by passing to the limits,

limΔz

Δy= dz

dy= z′ dx

y′ dx= z′

y′ .(7)

2When used in this manner, the term arc is an old trigonometric term referring to the signed lengthof the curve measured on a unit circle generated by a given angle at the center; therefore, at thesame time it is the signed value of the angle.3Cauchy uses the phrase ligne trigonométrique, translated here as trigonometric line, which isanother old trigonometric term used to denote the signed lengths of the right triangle line segmentswhich are represented by the various trigonometric functions. It is used to indicate circular functionvalues, today generally referred to as the values of the six trigonometric functions.


Thus, the ratio between the infinitely small differences of two functions of the vari-able x, has for a limit, the ratio of their differentials or of their derivatives.

We now suppose the functions y and z are related by the equation

z = F(y).(8)

We will deduceΔz

Δy= F(y + Δy) − F(y)

Δy;

then, by passing to the limits, and having regard to formula (7),

Δz

Δy= z′

y′ = F ′(y),

dz = F ′(y) dy, z′ = y′ F ′(y).(9)

The second of the equations in (9) coincides with equation (3) of the precedinglecture. Furthermore, if we write in the first equation, F(y) in place of z, we willobtain the following

d F(y) = F ′(y) dy,(10)

which is similar to the form of equation (4), and which serves to differentiate a func-tion of y, even when y is not the independent variable.

Examples. –

d(a + y) = dy, d(−y) = −dy, d(ay) = a dy, dey = ey dy,

dly = dy

y, dly2 = dy2

y2= 2 dy

y, d 1

2 ly2 = dy

y, . . . ,

d (axm) = a dxm = maxm−1 dx, d eex = eexdex = eex

ex dx,

dl(sin x) = d sin x

sin x= cos x dx

sin x= dx

tang x, dl(tang x) = dx

sin x cos x, . . . .

The first of these formulas proves that the addition of a constant to a function doesnot alter the differential, nor by consequence, the derivative.4

4One of our modern differentiation rules – the derivative of a constant is zero.

FIFTH LECTURE.

THE DIFFERENTIAL OF THE SUM OF SEVERALFUNCTIONS IS THE SUM OF THEIRDIFFERENTIALS. CONSEQUENCESOF THIS PRINCIPLE. DIFFERENTIALSOF IMAGINARY FUNCTIONS.

In previous lectures, we have shown how we form the derivatives and the differen-tials of functions of a single variable. We now add new developments to the studythat we have made to this subject.

Let x always be the independent variable, and Δx = αh = α dx an infinitelysmall increment attributed to this variable. If we denote by s, u, v, w, . . . severalfunctions of x, and by Δs,Δu, Δv, Δw, . . . the simultaneous increments that theyreceive while we allow x to grow by Δx, the differentials ds, du, dv, dw, . . . willbe, according to their own definitions, respectively equal to the limits of the ratios

Δs

α,

Δu

α,

Δv

α,

Δw

α, . . . .

This granted, first consider that the function s is the sum of all the others, so thatwe have

s = u + v + w + · · · .(1)

We will find successively,

Δs = Δu + Δv + Δw + · · · ,Δs

α= Δu

α+ Δv

α+ Δw

α+ · · · ;

then, by passing to the limits,

ds = du + dv + dw + · · · .(2)

When we divide the two members of this last equation by dx, it becomes

s ′ = u′ + v′ + w′ + · · · .(3)

From formula (2) or (3) compared to equation (1), it follows that the differentialor the derivative of the sum of several functions is the sum of their differentials or


21


https://doi.org/10.1007/978-3-030-11036-9_5


of their derivatives. From this principle, it follows, as we shall see, a number ofconsequences.

Firstly, if we denote by m an integer number, and by a, b, c, . . . , p, q, r constantquantities, we will find

⎧⎪⎨

⎪⎩

d(u + v) = du + dv,

d(u − v) = du − dv,

d(au + bv) = a du + b dv,

(4)

d(au + bv + cw + · · · ) = a du + b dv + c dw + · · · ,(5)

(6)

{d(axm + bxm−1 + cxm−2 + · · · + px2 + qx + r

)

= [maxm−1 + (m − 1)bxm−2 + (m − 2)cxm−3 + · · · + 2px + q

]dx .

The polynomial axm +bxm−1+cxm−2+· · ·+ px2+qx +r, in which all the terms areproportional to integer powers of the variable x, is what we call an entire functionof this variable.1 If we designate it by s, we will have by virtue of equation (6),

s ′ = maxm−1 + (m − 1)bxm−2 + (m − 2)cxm−3 + · · · + 2px + q.

So, to obtain the derivative of an entire function of x, it suffices to multiply each termby the exponent of the variable and to reduce each exponent by one unit.2 It is easyto see that this proposition remains valid in the case where the variable becomesimaginary.

Now, let

s = uvw · · · .(7)

Since we will have, by supposing the functions u, v, w, . . . all positive,

ls = lu + lv + lw + · · · ,(8)

and in all possible cases,3

s2 = u2v2w2 · · · ,

12 l

(s2

) = 12 l

(u2) + 1

2 l(v2

) + 12 l

(w2) + · · · ,(9)

1Cauchy uses this phrase throughout his text. His entire function is what we would call a polyno-mial with nonnegative integer powers. His usage is not to be confused with the modern definitiongenerally used in complex analysis.2Our common Power Rule of differentiation.3Meaning u, v, w, . . . are positive or negative.

FIFTH LECTURE. 23

the application of the principle stated in formula (8) or in formula (9) will producethe equation4

ds

s= du

u+ dv

v+ dw

w+ · · · ,(10)

from which we will deduce5

(11)

⎧⎪⎨

⎪⎩

d(uvw · · · ) = uvw · · ·(

du

u+ dv

v+ dw

w+ · · ·

)

= vw · · · du + uw · · · dv + uv · · · dw + · · · .

Examples. –

d(uv) = u dv + v du, d(uvw) = vw du + uw dv + uv dw,

d(x lx) = (1 + lx) dx, d(xae−x ) = xae−x(a

x− 1

), . . . .

Now, let

s = u

v.(12)

By differentiating ls or 12 l

(s2

), we will find

ds

s= du

u− dv

v, ds = u

v

(du

u− dv

v

)

,(13)

and as a result,6

d(u

v

)= v du − u dv

v2.(14)

We would arrive at the same result by observing that the differential of uv is equiva-

lent to

d

(

u1

v

)

= 1

vdu + u d

(1

v

)

= du

v− u dv

v2.

4Recall Cauchy has already shown in the previous lecture that d(lx) = dxx .

5Cauchy has just presented a clever derivation of what is essentially the multiple variable versionof our Product Rule for differentiation in differential form.6The differential form of our Quotient Rule for differentiation follows simply.


Examples. –

d tang x = dsin x

cos x= cos x d sin x − sin x d cos x

cos2 x= dx

cos2 x,

d cot x = − dx

sin2 x,

d(a

x

)= − a dx

x2, d

(eax

x

)

= eax

x

(

a − 1

x

)

dx,

d

(lxx

)

= 1 − lxx2

dx, d

(b

a + x

)

= −b dx

(a + x)2.

If the functions u, v reduce to entire functions, the ratio uv will become what we

call a rational fraction. We will easily determine its differential with the help offormulas (6) and (14).

After we have formed the differentials of the product uvw · · · , and of the quo-tient u

v , we will obtain without difficulty those of several other expressions such as

uv, u1v, uvw

, . . . . In fact, we will find, for s = uv,

ls = v lu,ds

s= v

du

u+ lu dv, ds = vuv−1 du + uvlu dv;

for s = u1v,

ls = 1

vlu,

ds

s= du

uv− lu

dv

v2, ds = u

1v −1 du

v− u

1v lu

dv

v2;

for s = uvw,

ls = vwlu, ds = uvwvw

(du

u+ w

vlu dv + lu lv dw

)

;

etc.

Examples. –

d(x x

) = x x (1 + lx) dx, d(

x1x

)= 1 − lx

x2x

1x dx, d

(x xx ) = · · · .

We will finish this lecture by investigating the differential of an imaginaryfunction. We name, in this way, any expression that can be reduced to the formu + v

√−1, u and v designating two real functions.7 This granted, if we call the

7Cauchy did not begin to use the symbol i to denote his√−1 until the 1850s. Although it had

been in print earlier, Carl Friedrich Gauss (1777–1855) is generally credited with the widespreaduse of the modern notation for i in use today.

FIFTH LECTURE. 25

limit of an imaginary variable expression that which becomes this expression whenwe replace the real part and the coefficient of

√−1 by their respective limits, andif, moreover, we extend to imaginary functions the definitions that we have givenfor the differences, the differentials, and the derivatives of real functions, we willrecognize that the equation

s = u + v√−1

leads to the following

Δs = Δu + Δv√−1,

Δs

Δx= Δu

Δx+ Δv

Δx

√−1,Δs

α= Δu

α+ Δv

α

√−1,

s ′ = u′ + v′√−1, ds = du + dv√−1.

We will have, by consequence,

d(

u + v√−1

)= du + dv

√−1.(15)

The form of this last equation is similar to those of the equations in (4).If we suppose, in particular,

s = cos x + √−1 sin x,

we will find

ds =[cos

(x + π

2

)+ √−1 sin

(x + π

2

)]dx = s

√−1 dx .

Add that the formulas in (4), (5), (6), (11), and (14) will remain the same, evenwhen the constants a, b, c, . . . , p, q, r or the functions u, v, w, . . . included inthese formulas will become imaginary.

SIXTH LECTURE.

USE OF DIFFERENTIALS AND DERIVEDFUNCTIONS IN THE SOLUTION OF SEVERALPROBLEMS. MAXIMA ANDMINIMA OFFUNCTIONS OF A SINGLE VARIABLE. VALUESOF FRACTIONS WHICH ARE PRESENTEDUNDER THE FORM 0

0.

After having learned to form the derivatives and differentials of functions of a singlevariable, we now indicate how we can make use of them for the solution of variousproblems.1

PROBLEM I.2 – The function y = f (x) being assumed continuous with respectto x in the vicinity of the particular value x = x0, we ask if, starting from this value,the function grows or diminishes while the variable itself is made to increase or todecrease.

Solution. – LetΔx, Δy be the infinitely small and simultaneous increments of thevariables x, y. The ratio Δy

Δx will have for a limit dydx = y′. We must conclude that,

for very small numerical values of Δx and for a particular value x0 of the variable x,the ratio Δy

Δx will be positive if the corresponding value of y′ is a positive and finitequantity, and negative if this value of y′ is a finite but negative quantity. In the firstcase, the infinitely small differences Δx, Δy being of the same sign, the function ywill grow or will diminish, starting from x = x0, at the same time as the variablex . In the second case, the infinitely small differences being of opposite signs, thefunction y will grow if the variable x diminishes and will decrease if the variableincreases.

These principles being accepted, consider that the function y = f (x) remainscontinuousbetween the twogiven limits3 x = x0, x = X. Ifwe cause thevariable x togrow by imperceptible steps from the first limit up to the second, the function y willgo on growing whenever its derivative, being finite, will have a positive value, anddecreasing whenever this same derivative will obtain a negative value. Therefore,the function y can only cease to grow and then to diminish, or to diminish and then

1Lecture Six is the only lecture within Cauchy’s Calcul infinitésimal textbook that is devoted toexploring applications of the calculus.2Problem I considers the classic application of the derivative in which the sign of the first derivativeindicates whether the function is increasing or decreasing.3Here, Cauchy is using the term “limits” to denote bounds.


27


https://doi.org/10.1007/978-3-030-11036-9_6


to grow, as long as the derivative y′ passes from positive to negative, or vice versa.It is essential to observe that, in this passage, the derived function will become nullif it does not cease to be continuous.

When a particular value of the function f (x) exceeds all of the neighboring val-ues, that is to say, all those that are obtained by varying x more or less a very smallamount, this particular value of the function is what we call a maximum.

When a particular value of the function f (x) is less than all the neighboringvalues, it takes the name of minimum.

This granted, it is clear that, if the two functions f (x), f ′(x) are continuous inthe neighborhood of a given value of the variable x, this value cannot produce amaximum nor a minimum of f (x), unless f ′(x) vanishes.4

PROBLEM II.5 – Find the maxima and minima of a function of the single vari-able x .

Solution. – Let f (x) be the proposed function. First, we will look for the valuesof x for which the function f (x) ceases to be continuous. At each of these values, ifany exist, will correspond a value of the function itself which will usually be eitheran infinite quantity, or a maximum, or a minimum.

In the second place, we will look for the roots of the equation

f ′(x) = 0,(1)

along with the values of x that render the function f ′(x) discontinuous, and amongwhich we should place at the forefront those that we deduce from the formula

f ′(x) = ±∞ or1

f ′(x)= 0.(2)

Let x = x0 be one of these roots or one of these values. The corresponding valueof f (x), namely f (x0), will be a maximum if, in the neighborhood of x = x0, thederived function f ′(x) is positive for x < x0 and negative for x > x0. On the otherhand, f (x0) will be a minimum if the derived function f ′(x) is negative for x < x0and positive for x > x0. Finally, if, in the neighborhood of x = x0, the derivedfunction f ′(x) being either constantly positive or negative, the quantity f (x0) willbe neither a maximum nor a minimum.

Examples. – The two functions x12 , 1

lx , which become discontinuous by passingfrom real to imaginary while the variable x diminishes by passing through zero,obtain for x = 0 a null value, which represents a minimum of the first function anda maximum of the second.

4Cauchy is setting the stage for his next problem by arguing extrema cannot occur unless thederived function is zero or it does not exist.5Problem II deals with using the derivative to locate extrema.

SIXTH LECTURE. 29

The two functions x2, x23 , whose derivatives pass from positive to negative by

reducing to zero or to infinity for a null value of x, have one and the other, zero fora minimum value. As for the two functions x3, x

13 , whose derivatives again become

null or infinite for x = 0, but remain positive for every other value of x, they admitneither a maximum nor a minimum.

The function x2 + px +q, whose derivative is 2x + p, obtains for x = − 12 p, the

minimum value q − 14 p

2, which is easily verified by putting the given function inthe form (x + 1

2 p)2 + q − 1

4 p2.

The function Ax

x , whose derivative is Ax

x

(1Le − 1

x

), obtains for x = Le, when A

exceeds unity, the minimum value eLe .

The function Lxx , whose derivative is 1

x2 (Le − Lx), obtains for x = e, the maxi-mum value Le

e .

The function xae−x , whose derivative is xae−x(ax − 1

), obtains for x = a, the

maximum value aae−a .

PROBLEM III.6 – Determine the slope of a curve at a given point.

Solution. – Consider the curve which has for an equation in rectangular co-ordinates, y = f (x). In this curve, the line segment from point (x, y)7 to point(x+Δx, y+Δy) forms, with the x-axis extending in the direction of positive x, twoangles, one acute and the other obtuse, the first of which measures the slope of theline segment with respect to the x-axis. If the second point is to get close to the firstby an infinitely small distance, the line segment will coincide noticeably with thetangent to the curve at the first point, and the slope of the line segment with respectto the x-axis will become the slope of the tangent, or what we call the slope of thecurve with respect to the same axis. This granted, as the slope of the line segmentwill have for the trigonometric tangent the numerical value of the ratio Δy

Δx , it is clearthat the slope of the curve will have for a trigonometric tangent the numerical valueof the limit toward which this ratio converges, that is to say, of the derived function,y′ = dy

dx .

If the value of y′ is null or infinite, the tangent to the curve will be parallel or per-pendicular to the x-axis. It is usually this that happens when the ordinate y becomesa maximum or a minimum.

6Problem III addresses the use of the derivative to find the slope of a line tangent to a curve at apoint. Cauchy has taken care to distance himself from anything geometric as well as possible thusfar in his text. This is the first time we see him directly connecting the derivative to this geometricinterpretation.7This is the placement of the first of the two footnotes Cauchy makes of his own in the original textof 1823. It reads, “We indicate here the points with the help of their coordinates included betweentwo parentheses, that we will always use hereafter. Often, we also indicate the curves or curvedsurfaces by their equations.”


Examples. –

y = x2, y = x3, y = xm, y = x23 ,

y = xa, y = Ax , y = sin x, . . . .

PROBLEM IV.8 –We ask the true value of a fraction whose two terms are func-tions of the variable x, in the case where we attribute to this variable a particularvalue, for which the fraction is presented under the indeterminate form 0

0 .

Solution. – Let s = zy be the proposed fraction, y and z denoting two functions of

the variable x, and suppose that the particular value x = x0 reduces this fraction tothe form 0

0 , that is to say, it makes y and z vanish. If we represent by Δx,Δy, Δz theinfinitely small and simultaneous increments of the three variables x, y, z, we willhave, for any value of x,

s = z

y= lim

z + Δz

y + Δy,

and for the particular value x = x0,

s = limΔz

Δy= dz

dy= z′

y′ .(3)

Thus, the value sought for the fraction s or zy will generally coincide with that of

the ratio dzdy or z′

y′ .9

Examples. – We will have, for x = 0,

sin x

x= cos x

1= 1,

l(1 + x)

x= 1

1 + x= 1;

for x = 1,lx

x − 1= 1

x= 1,

x − 1

xn − 1= 1

nxn−1= 1

n, . . . .

8Problem IV is the first of several times Cauchy deals with the important indeterminate form 00 .

9This result is essentially the weaker version of what today is known as l’Hôpital’s Rule. The ruleis named after Guillaume de l’Hôpital (1661–1704); however, the result is actually due to JohannBernoulli (1667–1748) from around the year 1694 while he was developing the content for a newtextbook. Interestingly, this textbook will eventually be published under l’Hôpital’s name due to amutually agreed arrangement the two had made allowing l’Hôpital to be credited for Bernoulli’swork in exchange for payment. The text, Analyse des infiniment petits pour l’intelligence des lignescourbes (Analysis of the Infinitely Small for the Understanding of Curved Lines), was publishedin 1696 and is recognized as the first major textbook on differential calculus. Cauchy will extendthe idea developed in this problem to a more general case at the very end of his Calcul infinitésimaltext.

SEVENTH LECTURE.

VALUES OF SOME EXPRESSIONS WHICH AREPRESENTED UNDER THE INDETERMINATEFORM ∞

∞,∞0, . . . . A RELATIONSHIP WHICHEXISTS BETWEEN THE RATIO OF FINITEDIFFERENCES AND THE DERIVED FUNCTION.

We have considered in the previous lecture the functions of the variable x, which,for a particular value of the variable, are presented under the indeterminant form00 . It often happens that this form is found replaced by one of the following∞∞ , ∞0, 0 × ∞, 00, . . . . Thus, when f (x) grows indefinitely with x, the particularvalues of the two functions

f (x)

x,

[f (x)

] 1x ,

for x = ∞, are presented under the indeterminant forms ∞∞ ,∞0. These same values

can, in general, be easily calculated with the help of two theorems that we haveestablished in Algebraic Analysis (Chapter II, pages 48 and 53).1 But, we will limitourselves here to show by some examples how we can resolve the questions of thiskind.

First, let the proposed function be Ax

x , A denoting a number greater than unity,

and let us consider that we are looking for the true value of this function for x = ∞.We will observe that, for values of x greater than 1

lA , the derived function alwaysbeing positive, the given function will always be increasing with x .2 Moreover, if werepresent by m an integer number capable of indefinite increment, the expression

Am

m= (1 + A − 1)m

m= 1

m+ (A− 1) + m − 1

2(A− 1)2 + (m − 1)(m − 2)

2 · 3 (A− 1)3 + · · ·

Acenterpiece of this lecture, and of the entire text, is Cauchy’s proof of theMeanValue Theoremfor Derivatives presented in this lecture. It is a fairly involved proof, illustrating the difficultyCauchy has at handling the complex concepts. His proof involves and illustrates his inequalitysqueezing technique resulting from his earlier “δ−ε like” definitions of the limit, continuity, andthe derivative. This lecture provides one of the clearest examples in Cauchy’s Calcul infinitésimaltext demonstrating how he translates his verbal definitions into these algebraic inequalities.

1A typographical error occurs in the 1899 text; it incorrectly references Chapter III. Because oftheir relevance to his Calcul infinitésimal text, translations of these two theorems from Cauchy’sCours d’analyse are presented in their entirety in Appendix A of this book.

2The derivative isAx

(x ln A−1

)

x2. As Cauchy notes, for values of x > 1

ln A the numerator of thederivative is positive. Hence, the entire expression is positive.


31


https://doi.org/10.1007/978-3-030-11036-9_7


will obviously have positive infinity for a limit. We will find, by consequence,

limAx

x= ∞.(1)

It follows from this last formula that the exponential Ax , when the number A exceedsunity, eventually grows much faster than the variable x .

In the second place, we look for the actual value of the function Lxx for x = ∞, the

base of the logarithm being a number A greater than unity. Since, by letting y = Lx,we will find

Lxx

= y

Ay,

and that the function yAy will have for a limit, 1

∞ = 0, we will deduce

limLxx

= 0.(2)

It follows that, in a system whose base is greater than unity, the logarithms of thenumbers grow much less rapidly than the numbers themselves.

We look again at the value of x1x for x = ∞. As we obviously have

x1x = A

Lxx ,

we will derive

lim x1x = A0 = 1.(3)

Whenwe replace, in formulas (2) and (3), x by 1x , we conclude that the functions

3

x Lx and xx converge respectively toward the limits 0 and 1, while we let x convergetoward the limit zero.

We now make known a relationship worthy of remark4 which exists between thederivative f ′(x) of any function f (x), and the ratio of the finite differences

f (x + h) − f (x)

h.

3The 1899 edition has a small typographical error here. It reads, x lx and xx .4This is where we find the second, and final, of Cauchy’s original footnotes from 1823. It reads,“We can consult on this subject a memoir by Mr. Ampère, inserted in the XIIIth notebook of Jour-nal of the Polytechnic School.” Cauchy is referring to his contemporary, André-Marie Ampère(1775–1836). Ampère was a close colleague during Cauchy’s tenure at the École Polytechnique.The two had met earlier in Cauchy’s career when Ampère helped Cauchy while a student at theÉcole Polytechnique. Ampère served as Cauchy’s analysis tutor during his early years but conti-nued to aid Cauchy throughout his career. Cauchy remained particularly grateful to Ampère duringhis entire adult life. As professors, the two of them would generally alternate their teaching of thefirst and second year curriculum (Analysis and Mechanics) every other year, each taking their ownset of students through both sequences. When Cauchy eventually leaves the school for political rea-sons in the year 1830, Ampère will carry on his analysis teachings. Ampère himself had a brilliantcareer and is best known for his pioneering work in the field of electromagnetism.

SEVENTH LECTURE. 33

If, in this ratio, we attribute to x a particular value x0, and if we make, in addition,x0 + h = X, it will take the form

f (X) − f (x0)

X − x0.

This granted, we will establish the following proposition without difficulty.5

THEOREM. – If, the function f (x) being continuous6 between the limits x =x0, x = X, we denote by A the smallest, and by B the largest values that the derivedfunction f ′(x) receives in this interval, the ratio of the finite differences,

f (X) − f (x0)

X − x0,(4)

will necessarily be contained between A and B.

Proof. – Denote by δ, ε two very small numbers, the first being selected so that,for the numerical values of i less than δ, and for any value of x contained betweenthe limits x0, X,7 the ratio

f (x + i) − f (x)

i

always remains greater than f ′(x) − ε, and less than f ′(x) + ε.8 If, between thelimits x0, X, we interpose n − 1 new values of the variable x, namely,

x1, x2, . . . , xn−1

in a manner to divide the difference X − x0 into elements

x1 − x0, x2 − x1, . . . , X − xn−1,

5Without a very definite indication, Cauchy begins his historic proof of the Mean Value Theoremfor Derivatives here. Cauchy’s proof is significantly different from the slick version of the theoremnormally found in one of today’s standard calculus textbooks due to Pierre Ossian Bonnet (1819–1892) from 1868. Cauchy’s version, although longer, is far more interesting. As a side note, Bonnetbecame a student at the École Polytechnique in 1838, a few years after Cauchy left the school.Bonnet would later take a teaching position at the École Polytechnique in 1844.6Cauchy requires f (x) to be continuous here and then immediately makes use of its derivative,implying the derivative must exist everywhere on the interval. To him, a continuous function isdifferentiable everywhere in some sense.7The reader may notice Cauchy assumes a uniformity condition here. He assumes that since f (x)is continuous, a single δ can be found regardless of which x is chosen. Today we draw a distinctionbetween a function which is continuous in a simple sense to one which is uniformly continuouson an interval. This is one example in the text illustrating why the debate over Cauchy’s earlywork continues. Does this wording suggest Cauchy had more of a uniform continuity in mind withhis initial definition for continuity back in Lecture Two, or does this show he lacked the clearunderstanding that this stronger form of continuity was even needed? We may never know his truethoughts on this important detail.8The 1899 text has a misprint. It reads, “…greater than f (x) − ε, and less than f ′(x) + ε.”


which, all being of the same sign and having numerical values less than δ, thefractions

f (x1) − f (x0)

x1 − x0,

f (x2) − f (x1)

x2 − x1, . . . ,

f (X) − f (xn−1)

X − xn−1(5)

are found contained, the first between the limits f ′(x0) − ε, f ′(x0) + ε, the secondbetween the limits f ′(x1) − ε, f ′(x1) + ε, . . . ,9 will all be greater than the quantityA − ε, and less than the quantity B + ε. Moreover, the fractions in (5) having de-nominators of the same sign, if we divide the sum of their numerators by the sum oftheir denominators, we will obtain an average fraction, that is to say, one containedbetween the smallest and the largest of those that we consider (see Algebraic Anal-ysis, Note II, Theorem XII).10 Expression (4), with which this average coincides,will therefore, itself be enclosed between the limits A − ε, B + ε; and, since thisconclusion remains valid however small the number ε, we can say that expression(4) will be contained between A and B.

Corollary. – If the derived function f ′(x)11 is itself continuous between the limitsx = x0, x = X, by passing from one limit to the other, this function will vary in amanner to always remain contained between the two values A and B, and to succes-sively take all the intermediate values.12 Therefore, any average quantity between Aand B will then be a value of f ′(x) corresponding to a value of x included betweenthe limits x0 and X = x0 + h, or to what amounts to the same thing, to a value of xof the form

x0 + θh = x0 + θ(X − x0),

θ denoting a number less than unity. By applying this remark to expression (4), wewill conclude that there exists between the limits 0 and 1,13 a value of θ that worksto satisfy the equation

f (X) − f (x0)

X − x0= f ′[x0 + θ(X − x0)

],

9Cauchy is once again assuming the same δ works for each individual x j to squeeze all of themwithin the same ε. This is only true if a uniformity condition holds.10A complete translation of this theorem, along with Cauchy’s brief definition for an average fromhis 1821 Cours d’analyse Note II, are presented in Appendix B of this book.11The 1899 text has f (x).12Cauchy is invoking the Intermediate Value Theorem here. This is why he needs f ′(x) to becontinuous. Cauchy proves the Intermediate Value Theorem in his Cours d’analyse Note III (atranslation of which is presented as Appendix C of this book). Cauchy’s frequent use of founda-tional results and repeated references to his 1821 Cours d’analyse text gives one a sense of whyhis lectures ran into overtime, and why Cauchy required more than the allotted number of meetingdays to complete his analysis course at the École Polytechnique—causing such student resentmenttoward their professor and dissatisfaction from the Conseil d’Instruction.13Cauchy continues to be vague when describing whether an interval is open or closed. Does hemean to include the endpoint values of 0 and 1? Cauchy’s wording here, and in many locationsthrough the text, is unclear in this regard.

SEVENTH LECTURE. 35

or to what amounts to the same thing, the following

f (x0 + h) − f (x0)

h= f ′(x0 + θh).(6)

This last formula above continues to be valid regardless of the value of x representedby x0,14 provided that the function f (x) and its derivative f ′(x) remain continuousbetween the extreme values x = x0, x = x0 + h, we will have in general, under thiscondition,

f (x + h) − f (x)

h= f ′(x + θh);(7)

then, by writing Δx instead of h, we will derive

f (x + Δx) − f (x) = f ′(x + θΔx)Δx .(8)

It is essential to observe that, in equations (7) and (8), θ always denotes an unknownnumber, but less than unity.

Examples. – By applying formula (7) to the functions xa, lx,15 we find16

(x + h)a − xa

h= a(x + θh)a−1,

l(x + h) − lxh

= 1

x + θh.

14Without much fanfare, hidden in a corollary, Cauchy has just proven the Mean Value Theoremfor Derivatives, one of the foundational theorems of calculus. His version of the theorem is not quitethe one we use today; modern versions generally do not require the derivative to be continuous overthe interval. Although not established at the time Cauchy’s text was written, Darboux’s Theorem(named after Jean-Gaston Darboux (1842–1917)

)demonstrates this requirement is not necessary,

as the derivative of a function has the intermediate value property regardless of whether or not it iscontinuous.15The original 1823 edition has Lx here and in the subsequent formula but are both corrected inits ERRATA.16The 1899 text reads, (x+h)a−xa

h = a(x − θh)a−1.

EIGHTH LECTURE.

DIFFERENTIALS OF FUNCTIONS OF SEVERALVARIABLES. PARTIAL DERIVATIVESAND PARTIAL DIFFERENTIALS.

Letu = f (x, y, z, . . . )be a function of several independent variables x, y, z, . . . .Wedenote by i an infinitely small quantity, and by

ϕ(x, y, z, . . . ),

χ(x, y, z, . . . ),

ψ(x, y, z, . . . ),

. . . . . . . . . . . .

the limits toward which the ratios

f (x + i, y, z, . . . ) − f (x, y, z, . . . )

i,

f (x, y + i, z, . . . ) − f (x, y, z, . . . )

i,

f (x, y, z + i, . . . ) − f (x, y, z, . . . )

i,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

converge, while i indefinitely approaches zero. ϕ(x, y, z, . . . ) will be the derivativethatwededuce fromthe functionu = f (x, y, z, . . . )byconsidering x as theonlyvari-able, or what we call the partial derivative of u with respect to x. In the same way,χ(x, y, z, . . . ), ψ(x, y, z, . . . ), . . . will be the partial derivatives of u with respect tothe variables y, z, . . . .

This granted, consider that we attribute to the variables x, y, z, . . . any incrementsΔx,Δy, Δz, . . . , and let Δu be the corresponding increment of the function u, sothat we have

Δu = f(x + Δx, y + Δy, z + Δz, . . .

) − f(x, y, z, . . .

).(1)


37


https://doi.org/10.1007/978-3-030-11036-9_8


If we assign to Δx,Δy, Δz, . . . the finite values h, k, l, . . . , the value of Δu, givenby equation (1), will become what we call the finite difference of the function u, andwill usually be a finite quantity. If, on the other hand, α denoting an infinitely smallratio,1 we suppose

Δx = α h, Δy = α k, Δz = α l, . . . ,(2)

the value of Δu, namely

f(x + α h, y + α k, z + α l, . . .

) − f(x, y, z, . . .

)

will usually be an infinitely small quantity;2 but, by then dividing this value by α,

we will obtain the fraction

Δu

α= f

(x + α h, y + α k, z + α l, . . .

) − f(x, y, z, . . .

)

α,(3)

which will converge, in general, toward a finite limit different from zero. This limitis what we call the total differential, or simply the differential of the function u. Weindicate it, with the help of the letter d, by writing

du or d f (x, y, z, . . . ).

Thus, regardless of the number of independent variables that are included in thefunction u, its differential is found defined by the formula

du = limΔu

α.(4)

If we successively make u = x, u = y, u = z, . . . , we will deduce from equations(2) and (4)

dx = h, dy = k, dz = l, . . . .(5)

By consequence, the differentials of the independent variables x, y, z, . . . are noth-ing other than the finite constants h, k, l, . . . .

We easily determine the total differential of the function

f (x, y, z, . . . )

when we know its partial derivatives. In fact, if, in this function we increase, oneafter the other, the variables x, y, z, . . . by any quantities Δx, Δy, Δz, . . . ,we will

1Recall Cauchy’s definition of an infinitely small quantity as a variable whose limit is zero but isitself not necessarily zero.2Cauchy is clearly assuming a well-behaved function whose partial derivatives are all continuous.Perhaps, this is why he intuitively includes the word “usually.”

EIGHTH LECTURE. 39

derive from formula (8) of the preceding lecture a sequence of equations of the form

f(x + Δx, y, z, . . .

) − f(x, y, z, . . .

)

= Δx ϕ(x + θ1Δx, y, z, . . .

),

f(x + Δx, y + Δy, z, . . .

) − f(x + Δx, y, z, . . .

)

= Δy χ(x + Δx, y + θ2Δy, z, . . .

),

f(x + Δx, y + Δy, z + Δz, . . .

) − f(x + Δx, y + Δy, z, . . .

)

= Δz ψ(x + Δx, y + Δy, z + θ3Δz, . . .

),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

θ1, θ2, θ3, . . . denoting unknown numbers, but all contained between zero and unity.By adding these same equations member to member, we will find

⎧⎪⎨

⎪⎩

f(x + Δx, y + Δy, z + Δz, . . .

) − f(x, y, z, . . .

)

= Δx ϕ(x + θ1Δx, y, z, . . .

) + Δy χ(x + Δx, y + θ2Δy, z, . . .

)

+ Δz ψ(x + Δx, y + Δy, z + θ3Δz, . . .

) + · · · .

(6)

If, in this last formula, whose first member can be replaced byΔu,we setΔx = α h,

Δy = α k, Δz = α l, and if, in addition, we divide the two members by α, we willobtain the following

⎧⎨

⎩

Δu

α= h ϕ

(x + θ1αh, y, z, . . .

) + k χ(x + αh, y + θ2αk, z, . . .

)

+l ψ(x + αh, y + αk, z + θ3αl, . . .

) + · · · ,

(7)

from which we will deduce, by passing to the limits, then writing dx, dy, dz . . .

instead of h, k, l, . . . ,

du = ϕ(x, y, z, . . .

)dx + χ

(x, y, z, . . .

)dy + ψ

(x, y, z, . . .

)dz + · · · .(8)

Examples. –

d(x + y + z + · · · ) = dx + dy + dz + · · · , d(x − y) = dx − dy,

d(ax + by + cz + · · · ) = a dx + b dy + c dz + · · · ,

d(xyz · · · ) = yz · · · dx + xz · · · dy + xy · · · dz + · · · ,

d(xa yb · · · ) = xa yb · · ·(

adx

x+ b

dy

y+ · · ·

), d

(x

y

)= y dx − x dy

y2,

d x y = yx y−1 dx + x y lx dy, . . . .


It is essential to observe that, in the value of du given by equation (8), the termϕ(x, y, z, . . . ) dx is precisely the differential that we would obtain for the function

u = f (x, y, z, . . . ),

by considering in this function x as the only variable quantity, and y, z, . . . as con-stant quantities. It is for this reason that the term in question is called the partialdifferential of the function u with respect to x. Similarly,3

χ(x, y, z, . . . ) dy, ψ(x, y, z, . . . ) dz, . . . ,

are the partial differentials of u with respect to y, with respect to z, . . . . If weindicate these partial differentials by placing at the base of the letter d the variablesto which they relate, as we see here,

dx u, dyu, dzu, . . . ,

we will have

ϕ(x, y, z, . . . ) = dx u

dx, χ(x, y, z, . . . ) = dyu

dy, ψ(x, y, z, . . . ) = dzu

dz, . . . ,(9)

and equation (8) can be presented under one or the other of the two forms

du = dx u + dyu + dzu + · · · ,(10)

du = dx u

dxdx + dyu

dydy + dzu

dzdz + · · · .(11)

For brevity, we usually omit the letters that we have placed at the base of the char-acter d in the equations in (9), and we simply represent the partial derivatives of u,

taken relative to x, y, z, . . . , by the notations

du

dx,

du

dy,

du

dz, . . . .(12)

Then dudx is not the quotient of du by dx; and, to express the partial differential of

u taken relative to x, the notation dudx dx must be employed, which is not susceptible

for simplification, unless we reestablish the letter x at the base of the character d.When we accept these conventions, formula (11) is reduced to

du = du

dxdx + du

dydy + du

dzdz + · · · .(13)

But, as it is no longer possible to remove in this last expression the differentialsdx, dy, dz, . . . , nothing can replace formula (10).

3The final ellipsis is omitted in the 1823 and 1899 editions but is clearly implied.

EIGHTH LECTURE. 41

The definitions and the formulas established above extend without any difficultyto the case where the function u becomes imaginary. Thus, for example, if we setu = x + y

√−1, the partial derivatives of u and its total differential will be given,respectively, by the equations

du

dx= 1,

du

dy= √−1, du = dx + √−1 dy.

We will indicate, in finishing, a very simple way of reducing the calculation oftotal differentials to those of derived functions. If, in the expression f (x + αh, y +αk, z + αl, . . . ), we consider α as the only variable, and if we set, by consequence,

f (x + αh, y + αk, z + αl, . . . ) = F(α),(14)

we will have, not only

u = F(0),(15)

but alsoΔu = F(α) − F(0),

and as a result,

du = limF(α) − F(0)

α= F ′(0).(16)

Thus, to form the total differential du, it will suffice to calculate the particular valuethat the derived function, F ′(α), receives for α = 0.4

4The reader will notice Cauchy has clearly relaxed and loosened his level of rigor while developinghis multiple variable results. A similar statement can be argued for his complex presentation. Thismay be an indication he felt the main goal of his course should be the rigorous development ofsingle-variable real calculus.

NINTH LECTURE.

USE OF PARTIAL DERIVATIVES IN THEDIFFERENTIATION OF COMPOSED FUNCTIONS.DIFFERENTIALS OF IMPLICIT FUNCTIONS.

Let s = F(u, v,w,. . . ) be any function of the variable quantities u, v,w,. . . that wesuppose to be themselves functions of the independent variables x, y, z, . . . . s willbe a composed function of these latter variables; and, if we designate by Δx,Δy,Δz,. . . the arbitrary simultaneous increments attributed to x, y, z, . . . , the correspondingincrementsΔu,Δv,Δw, . . ., Δs of the functions u, v, w, . . ., s will be related amongthemselves by the formula

Δs = F(u + Δu, v + Δv, w + Δw, . . . ) − F(u, v, w, . . . ).(1)

Moreover, let

Φ(u, v, w, . . . ), X (u, v, w, . . . ), Ψ (u, v, w, . . . ), . . .

be the partial derivatives of the function F(u, v, w, . . . ) taken successively with re-spect to u, v, w, . . .. Since equation (6) of the preceding lecture holds for any valuesof the variables x, y, z, . . . and of their incrementsΔx, Δy, Δz, . . .,wewill deduce,in replacing x, y, z, . . . by u, v, w, . . . , and the function f by the function F,

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

F(u + Δu, v + Δv, w + Δw, . . . ) − F(u, v, w, . . . )

= Δu Φ(u + θ1Δu, v, w, . . . ) + Δv X (u + Δu, v + θ2Δv, w, . . . )

+ Δw Ψ (u + Δu, v + Δv, w + θ3Δw, . . . )

+ · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · .

(2)

In this last equation, θ1, θ2, θ3, . . . always denote unknown numbers, but less thanunity.1 If we now set

Δx = αh = α dx, Δy = αk = α dy, Δz = αl = α dz, . . . ,

1Having been seen several times already, Cauchy uses this word frequently to denote the value ofone.


43


https://doi.org/10.1007/978-3-030-11036-9_9


and divide the two members of equation (2) by α, we will derive

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

Δs

α= Φ(u + θ1αh, v, w, . . . )

Δu

α

+ X (u + αh, v + θ2αk, w, . . . )Δv

α

+ Ψ (u + αh, v + αk, w + θ3αl, . . . )Δw

α+ · · · ;

(3)

then, by letting α converge toward the limit zero,

ds = Φ(u, v, w, . . . )du + X (u, v, w, . . . )dv + Ψ (u, v, w, . . . )dw + · · · .(4)

The value of ds produced by equation (4) is similar to the value of du producedby equation (8) of the preceding lecture. The main difference consists in that thedifferentials dx, dy, dz, . . . included in the value of du are arbitrary constants, whilethe differentials du, dv, dw, . . . included in the value of ds are new functions of theindependent variables x, y, z, . . . combined in a certain manner with the arbitraryconstants dx, dy, dz, . . . .

By applying formula (4) to particular cases, we will find2

d(u + v + w + · · · ) = du + dv + dw + · · · ,

d(u − v) = du − dv,

d(au + bv + cw + · · · ) = a du + b dv + c dw + · · · ,

d(uvw · · · ) = vw · · · du + uw · · · dv + uv · · · dw + · · · ,

d(u

v

)= v du − u dv

v2,

duv = vuv−1 du + uvlu dv,

etc. We have already obtained these equations (see the fifth lecture) by supposingu, v, w, . . . functions of a single independent variable x; but, we see that they remainvalid regardless of the number of independent variables.

In the particular case where we assume u a function of the single variable x, v afunction of the single variable y, w a function of the single variable z, . . . , we candirectly achieve equation (4) by starting from formula (10) of the preceding lecture.In fact, by virtue of this formula, we will generally have

ds = dx s + dys + dzs + · · · .(5)

Moreover, since among the quantities u, v, w, . . ., the first is, by hypothesis, the onlyone which contains the variable x, by considering s as a function of a function of

2Several relationships within this grouping have errors in both the original 1823 and the 1899reprint editionswhich have all been corrected here. The 1899 edition includes d(u−dv) = du−dv,d v

u = u dv−v duu2

, and duv = vuv−1 du + vvlu dv + · · · . The original 1823 edition contains the

expression d( u

v

) = u dv−v duu2

.

NINTH LECTURE. 45

this variable and having regard to formula (10) of the fourth lecture, we will find

dx s = dx F(u, v, w, . . . ) = Φ(u, v, w, . . . ) dx u = Φ(u, v, w, . . . ) du.

We will also find

dys = X (u, v, w, . . . ) dv, dzs = Ψ (u, v, w, . . . ) dw, . . . .

If we substitute these values of dx s, dys, dzs, . . . into formula (5), it will obviouslycoincide with equation (4).

Now, let r be a second function of the independent variables x, y, z, . . . . If wehave identically, that is to say, for any values of these variables,

s = r,(6)

we will infer

ds = dr.(7)

In the particular case where the function r is reduced either to zero or to a constantc, we find

dr = 0,

and as a result, the equation

s = 0 or s = c(8)

leads to the following

ds = 0.(9)

Equations (7) and (9) are among those that we call differential equations. The secondcan be presented under the form

Φ(u, v, w, . . . ) du + X (u, v, w, . . . ) dv + Ψ (u, v, w, . . . ) dw + · · · = 0,(10)

and remains valid even in the case where some of the quantities u, v, w, . . . arereduced to a few of the independent variables x, y, z, . . . . Thus, for example, wewill find, by supposing F(x, v) = 0,

Φ(x, v) dx + X (x, v) dv = 0;by supposing F(x, y, w) = 0,

Φ(x, y, w) dx + X (x, y, w) dy + Ψ (x, y, w) dw = 0;etc. In these last equations, v is obviously an implicit function of the variable x; wis an implicit function of the variables x, y; . . . .


Similarly, if we accept that the variables x, y, z, . . . , ceasing to be independent,are related among themselves by an equation of the form

f (x, y, z, . . . ) = 0,(11)

then, by making use of the notations adopted in the preceding lecture, we will obtainthe differential equation

ϕ(x, y, z, . . . ) dx + χ(x, y, z, . . . ) dy + ψ(x, y, z, . . . ) dz + · · · = 0,(12)

by means of which we can determine the differential of one of the variables, con-sidered as an implicit function of all the others. Thus, for example, we will find, bysupposing x2 + y2 = a2,

x dx + y dy = 0, dy = − x

ydx;

by supposing y2 − x2 = a2,

y dy − x dx = 0, dy = x

ydx;

by supposing x2 + y2 + z2 = a2,

x dx + y dy + z dz = 0, dz = − x

zdx − y

zdy.

Moreover, since we will have, in the first case,

y = ±√

a2 − x2,

and in the second,

y = ±√

a2 + x2,

we will deduce from the preceding formulas3

d√

a2 − x2 = − x dx√a2 − x2

, d√

a2 + x2 = x dx√a2 + x2

,(13)

which is easy to verify directly.When we designate by u the function f (x, y, z, . . . ), equations (11) and (12) can

be simply written, as it follows

u = 0, du = 0.

3Equation (13) has been corrected. Both the 1823 and the 1899 editions read,

(13) d√

a2 − x2 = − x dx√a2 − x2

, d√

a2 + x2 = x dx√a + x

.

NINTH LECTURE. 47

If the variables x, y, z, . . . , instead of being subject to a single equation of theform u = 0, were related between them by two equations of this type, such as

u = 0, v = 0,(14)

then we would have at the same time, the two differential equations

du = 0, dv = 0,(15)

with the help of which we could determine the differentials of two variables, con-sidered as implicit functions of all the others.

In general, if n variables x, y, z, . . . are related among themselves bym equations,such as

u = 0, v = 0, w = 0, . . . ,(16)

then we will have at the same time, the m differential equations

du = 0, dv = 0, dw = 0, . . . ,(17)

with the help of which we can determine the differentials of m variables, consideredas implicit functions of all the others.

TENTH LECTURE.

THEOREM OF HOMOGENEOUS FUNCTIONS.MAXIMA ANDMINIMA OF FUNCTIONSOF SEVERAL VARIABLES.

We say that a function of several variables is homogeneous, when, by letting all ofthe variables grow or decline in a given ratio, we obtain for a result the originalvalue of the function multiplied by a power of this ratio. The exponent of this poweris the degree of the homogeneous function. By consequence, f (x, y, z, . . . )will be ahomogeneous function of x, y, z, . . . and of degree a, if, t denoting a new variable,we have regardless of t,1

f (t x, t y, t z, . . . ) = ta f (x, y, z, . . . ).(1)

This granted, the theorem of homogeneous functions can be stated as follows.

THEOREM. – If we multiply the partial derivatives of a homogeneous func-tion of degree a by the variables to which they relate, the sum of the products thusformed will be equivalent to that which we would obtain in multiplying the functionitself by a.

Proof. – Let u = f (x, y, z, . . . ) be the given function, and let ϕ(x, y, z, . . . ),χ(x, y, z, . . . ), ψ(x, y, z, . . . ), . . . be its partial derivatives with respect to x, to y,

to z,, etc. If we differentiate the two members of equation (1) by considering t as theonly variable, we will find

ϕ(t x, t y, t z, . . . ) x dt + χ(t x, t y, t z, . . . ) y dt + ψ(t x, t y, t z, . . . ) z dt + · · ·= ata−1 f (x, y, z, . . . ) dt;

then, in dividing by dt, and setting t = 1,

xϕ(x, y, z, . . . )+yχ(x, y, z, . . . )+z ψ(x, y, z, . . . )+ · · · = a f (x, y, x, . . . ),(2)

1The rightmost ellipsis is omitted in equation (1) in the 1899 reprint but is present in the original1823 edition. Except for the lack of specificity as to whether Cauchy has a finite or infinite numberof variables in mind (the term “several” is used), Cauchy’s definition is very similar to a modern-day definition of a real homogeneous function.


49


https://doi.org/10.1007/978-3-030-11036-9_10



xdu

dx+ y

du

dy+ z

du

dz+ · · · = au.(3)

Corollary. – For a homogeneous function of degree null, we will have

xdu

dx+ y

du

dy+ z

du

dz+ · · · = 0.(4)

Examples.2 – Apply the theorem to the functions Ax2 + Bxy + Cy2, and L(

xy

).

When a real function of several independent variables x, y, z, . . . attains a par-ticular value which exceeds all neighboring values, that is to say, all those that wewould obtain by varying x, y, z, . . . more or less by very small quantities, this par-ticular value of the function is what we call a maximum.

When a particular value of a real function of x, y, z, . . . is less than all the neigh-boring values, it takes the name of minimum.

The study of maxima and minima of functions of several variables is easily re-duced to the study of maxima and minima of functions of a single variable. Suppose,in fact, that

u = f (x, y, z, . . . )

becomes a maximum for certain particular values attributed to x, y, z, . . . . We willhave, for these particular values and for very small numerical values of α,

f (x + αh, y + αk, z + αl, . . . ) < f (x, y, z, . . . ),(5)

regardless, moreover, of the finite constants h, k, l, . . . , provided that they have beenselected in a manner such that the first member of formula (5) remains real. Now,for brevity, if we let

f (x + αh, y + αk, z + αl, . . . ) = F(α),(6)

formula (5) will be found reduced to the following

F(α) < F(0).(7)

This relationship above remains valid regardless of the sign of α, and it results that, ifonly α varies, F(α), considered as a function of this single variable, will alwaysbecome a maximum for α = 0.

2The original 1823 edition prints the y in the L(

xy

)expression of the second example in a slightly

different font. However, this misprint is corrected in its ERRATA.

TENTH LECTURE. 51

We will recognize as well that, if f (x, y, z, . . . ) becomes a minimum for certainparticular values attributed to x, y, z, . . . , the corresponding value of F(α) will al-ways be a minimum for α = 0.

Observe now that, if the two functions F(α), F ′(α) are, one and the other, con-tinuous with respect to α in the vicinity of the particular value α = 0, this value canonly produce a maximum or a minimum of the first function, so long as it will causethe second to vanish (see the sixth lecture), that is to say, such that we will have

F ′(0) = 0.(8)

Moreover, when we write dx, dy, dz, . . . instead of h, k, l, . . . , equation (8) takesthe form

du = 0(9)

(see the eighth lecture). In addition, since the functions F(α) and F ′(α) are thosethat become u and du when we replace x by x + αh,3 y by y + αk, z by z + αl,. . . , it is clear that, if these two functions are discontinuous with respect to α in thevicinity of the particular value α = 0, the two expressions u and du, consideredas functions of the variables x, y, z, . . . , will be discontinuous with respect to thesevariables in the vicinity of the particular values which are attributed to them. Bycomparing these remarks to those which have been said above, we must concludethat the only values of x, y, z, . . . that work to produce maxima or minima of thefunction u are those which render the functions u and du discontinuous, as well asthose which satisfy equation (9), regardless of the finite constants dx, dy, dz, . . . .These principles being accepted, it will be easy to resolve the following question.

PROBLEM. – Find the maxima and minima of a function of several variables.

Solution. – Let u = f (x, y, z, . . . ) be the proposed function.Wewill first look forthe values of x, y, z, . . . that render the function u or du discontinuous, amongwhichwe must count those that we deduce from the formula

du = ±∞.(10)

In the second place, we will look for the values of x, y, z, . . . which satisfy equation(9), regardless of the finite constants dx, dy, dz, . . . . This equation, which can beplaced in the form

du

dxdx + du

dydy + du

dzdz + · · · = 0,(11)

3There is a typographical error in the 1899 reprint that is not present in the original 1823 edition.The 1899 reprint reads, “…when we replace x by x = αh.”


obviously leads to the following

du

dx= 0,

du

dy= 0,

du

dz= 0, . . . ,(12)

of which we obtain the first by setting dx = 1, dy = 0, dz = 0, . . . ; the second bysetting dx = 0, dy = 1, dz = 0, . . . ; etc. We remark in passing that, the number ofequations in (12) being equal to those of the unknowns x, y, z, . . . , we will usuallydeduce only a limited number of values for these unknowns.

Let us now conceive that we consider in particular one of the systems of val-ues that the previous study provides for the variables x, y, z, . . . , and denote byx0, y0, z0, . . . the values of which it is composed. The corresponding value of thefunction f (x, y, z, . . . ), namely f (x0, y0, z0, . . . ), will be a maximum, if for verysmall numerical values of α and for any values of h, k, l, . . . , the difference

f (x0 + αh, y0 + αk, z0 + αl, . . . ) − f (x0, y0, z0, . . . )(13)

is constantly negative. On the other hand, f (x0, y0, z0, . . . ) will be a minimum, ifthis difference is constantly positive. Finally, if this difference passes from posi-tive to negative while we change either the sign of α or the values of h, k, l, . . . ,f (x0, y0, z0, . . . ) will no longer be a maximum nor a minimum.

Example. – The function Ax2+ Bxy +Cy2+dx + Ey + F admits a maximum ora minimum when we have B2 − 4AC < 0, and no longer admits one when we haveB2 − 4AC > 0.

Note. – The nature of the function u can be such that an infinite number of dif-ferent systems of values attributed to x, y, z, . . . corresponds to values of u equal be-tween them, but greater than or less than all neighboring values, and each of whichare, by consequence, a type of maximum or minimum. When this circumstance oc-curs for systems in the vicinity for which the functions u and du remain continuous,these systems certainly satisfy the equations in (12). These equations can thereforesometimes admit an infinite number of solutions. This is what always happens whenthey are deduced in part one from the others.

Example. – If we take u = (cy −bz + l)2 + (az −cx +m)2 + (bx −ay +n)2, theequations in (12) will only produce

cy − bz + l

a= az − cx + m

b= bx − ay + n

c,

and the function u will admit an infinite number of values equal to

(al + bm + cn)2

a2 + b2 + c2,

each of which can be considered as a minimum.

ELEVENTH LECTURE.

USE OF INDETERMINATE FACTORS IN THESTUDY OF MAXIMA ANDMINIMA.

Let

u = f (x, y, z, . . . )(1)

be a function of the n variables x, y, z, . . . . But, let us conceive that these variables,instead of being independent of each other, as we have supposed in the tenth lecture,are related among themselves by m equations of the form

v = 0, w = 0, . . . .(2)

To deduce the maxima and the minima of the function u from the method that wehave indicated, we should begin by eliminating from this function m different vari-ables with the help of the formulas in (2). After this elimination, the variables whichremain, to number n − m, should be considered as independent; and, the systems ofvalues of these variables should be sought which render the function u or the func-tion du discontinuous, as well as those which satisfy, whatever the differentials ofthese same variables, the equation

du = 0.(3)

Now, the study of maxima and minima which corresponds to equation (3) can besimplified by the following considerations.

Ifwedifferentiate the functionu, by retaining all the givenvariables x, y, z, . . . ,

equation (3) will be presented under the form

du

dxdx + du

dydy + du

dzdz + · · · = 0,(4)

and will contain the n differentials dx, dy, dz, . . . . But, it is important to observethat, among these differentials, the only ones which we can arbitrarily dispose willbe those of the n − m variables regarded as independent. The other differentials willbe found determined according to the first and to the variables themselves by the


53


https://doi.org/10.1007/978-3-030-11036-9_11


formulas dv = 0, dw = 0, . . . , which, when expanded,1 respectively become

(5)

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

dv

dxdx + dv

dydy + dv

dzdz + · · · = 0,

dw

dxdx + dw

dydy + dw

dzdz + · · · = 0,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

This granted, since equation (4) must be satisfied, regardless of the differentialsof the independent variables, it is clear that, if we eliminate from this equation anumber m of the differentials with the help of the formulas in (5), the coefficients ofthe remaining n − m differentials must be separately equal to zero. Now, to performthe elimination, it is sufficient to add to equation (4) the formulas in (5) multiplied bythe indeterminate factors, −λ, −μ, − . . . , and to choose these factors in a mannerso as to eliminate the coefficients of the m successive differentials in the resultingequation. Moreover, as the resultant equation will be of the form

(6)

⎧⎪⎪⎨

⎪⎪⎩

(du

dx− λ

dv

dx− μ

dw

dx− · · ·

)

dx

+(

du

dy− λ

dv

dy− μ

dw

dy− · · ·

)

dy + · · · = 0,

and after having eliminated the coefficients of m differentials, the coefficients ofthose differentials remaining will still be equal to zero, it is possible to deduce thatthe values of λ, μ, ν, . . . , derived from a few of the formulas

(7)

⎧⎪⎪⎨

⎪⎪⎩

du

dx− λ

dv

dx− μ

dw

dx− · · · = 0,

du

dy− λ

dv

dy− μ

dw

dy− · · · = 0, . . . ,

must satisfy all the others. By consequence, the values of x, y, z, . . . that work toconfirm the formulas in (4) and (5) must satisfy the condition equations that providethe elimination of the indeterminate factors λ, μ, ν, . . . among the formulas in (7).The number of these condition equations will be n − m. By combining with theformulas in (2), we will obtain a total of n equations, from which we will deduce forthe given variables x, y, z, . . . several systems of values, among which are nec-essarily found those which, without rendering discontinuous one of the functions uand du, will provide for the first a maxima or a minima.

It is good to note that the condition equations produced by the elimination ofλ, μ, ν, . . . among the formulas in (2) would not be altered in any manner, if

1Cauchy uses the term développe here which is literally translated as developed. However, theterm also conveys the idea of growth or expansion. This latter meaning has been used here andthroughout the text to significantly improve the readability of his work and to more accuratelyconvey Cauchy’s true message.

ELEVENTH LECTURE. 55

we were to exchange in these formulas the function u with one of the functionsv, w, . . . . As a result, we would always achieve the same condition equations, if,instead of looking for the maxima and minima of the function u by supposingv = 0, w = 0, . . . , we were looking for the maxima and minima of the functionv by supposing u = 0, w = 0, . . . , as well as that of the function w by supposingu = 0, v = 0, . . . . Without altering the condition equations, we could even replacethe functions u, v, w, . . . by the following u − a, v − b, w − c, . . . , a, b, c, . . . den-oting arbitrary constants.

In the particular case where we want to obtain the maxima or the minima of thefunction u by supposing x, y, z, . . . subject to a single equation

v = 0,(8)

the formulas in (7) become

du

dx− λ

dv

dx= 0,

du

dy− λ

dv

dy= 0,

du

dz− λ

dv

dz= 0, . . . ,(9)

and we conclude, by the elimination of λ,

(dudx

)

(dvdx

) =(

dudy

)

(dvdy

) =(

dudz

)

(dvdz

) = · · · .(10)

This last formula is equivalent to n − 1 distinct equations, which combined withequation (8), will determine the values sought for x, y, z, . . . .

First Example. – Suppose that, a, b, c, . . . , r denoting constant quantities, andx, y, z, . . . variables subject to the equation

x2 + y2 + z2 + · · · = r2 or x2 + y2 + z2 + · · · − r2 = 0,

we request the maximum and the minimum of the function

u = ax + by + cz + · · · .

In this hypothesis, formula (10) will be found reduced to

a

x= b

y= c

z= · · · ,(11)

we will deduce (see Algebraic Analysis, Note II)

ax + by + cz + · · ·x2 + y2 + z2 + · · · = ±

√a2 + b2 + c2 + · · ·

√x2 + y2 + z2 + · · ·

oru

r2= ±

√a2 + b2 + c2 + · · ·

r,


and by consequence,

u = ± r√

a2 + b2 + c2 + · · · .(12)

To ensure that the two values of u given by equation (12) are a maximum and aminimum, it is sufficient to observe that we will always have

(13)

{(ax + by + cz + · · · )2 + (bx − ay)2 + (cx − az)2 + · · · + (cy − bz)2 + · · ·

= (a2 + b2 + c2 + · · · )(x2 + y2 + z2 + · · · ),

and as a result,u2 < (a2 + b2 + c2 + · · · )r2,

unless the values of x, y, z, . . . do not satisfy formula (11).

SecondExample. –Suppose that, a, b, c, . . . , k denotingconstantquantities, andx, y, z, . . . variables subject to the equation

ax + by + cz + · · · = k,

weseek theminimumof the function u = x2 + y2 + z2 + · · · . In this hypothesis,wewill again obtain formula (11), fromwhich we will deduce

k

u= ±

√a2 + b2 + c2 + · · ·√

u,

and as a result

u = k2

a2 + b2 + c2 + · · · .(14)

If the variables x, y, z, . . . are reduced to three and denote rectangular coordinates,the value of

√u, given by equation (14), will obviously represent the shortest dis-

tance from the origin to a fixed plane.

Third Example. – Conceive that we are looking for the semi-axes of an ellipse orof a hyperbola relative to its center and represented by the equation

Ax2 + 2Bxy + Cy2 = K .

Each of these semi-axes will be a maximum or a minimum of the radius vector r,leading from the origin to the curve, and determined by the formula

r2 = x2 + y2.

ELEVENTH LECTURE. 57

This granted, since we will have

dr = 1

r(x dx + y dy),

we can only cause dr to vanish by supposing

r = ∞ or x dx + y dy = 0.

The first hypothesis cannot be admitted for a hyperbola. By accepting the second,we will derive from formula (10)

x

Ax + By= y

Cy + Bx= x2 + y2

x(Ax + By) + y(Cy + Bx)= r2

K,

K

r2− A = B

y

x,

K

r2− C = B

x

y,

(K

r2− A

)(K

r2− C

)

= B2.(15)

Observe now that real values of r will always correspond to positive valuesof r2, and that equation (15) will produce two positive values for r2 if we haveAK > 0, AC − B2 > 0, and only one if we have AC − B2 < 0. Indeed, the curve,being an ellipse in the first case, will have two real axes; while in the second case, itwill change into a hyperbola, and will only have one real axis.2

2Cauchy describes the situation well enough to avoid the use of a diagram here. Following atradition set out by Lagrange, Cauchy purposefully omits diagrams of any type from both hisCalcul infinitésimal and Cours d’analyse books.

TWELFTH LECTURE.

DIFFERENTIALS AND DERIVATIVESOF VARIOUS ORDERS FOR FUNCTIONSOF A SINGLE VARIABLE. CHANGE OF THEINDEPENDENT VARIABLE.

As the functions of a single variable x ordinarily have for derivatives other functionsof this variable, it is clear that from a given function, y = f (x), we can deduce ingeneral a multitude of new functions each of which will be the derivative of theprevious one. These new functions are what we name the derivatives of variousorders of y or f (x), and we indicate them with the help of the notations

y′, y′′, y′′′, yIV , yV , . . . , y(n)

orf ′(x), f ′′(x), f ′′′(x), f IV (x), f V (x), . . . , f (n)(x).

This granted, y′ or f ′(x) will be the derivative of first order of the proposed functiony = f (x); y′′ or f ′′(x)will be the derivative of secondorder of y, and at the same timethe derivative of first order of y′, . . . ; finally, y(n) or f (n)(x) (n denoting any integernumber) will be the derivative of order n of y, and at the same time the derivative offirst order of y(n−1).

Now, let dx = h be the differential of the assumed independent variable x . Wewill have, after what we just said,

y′ = dy

dx, y′′ = dy′

dx, y′′′ = dy′′

dx, . . . , y(n) = dy(n−1)

dx,(1)

or to what amounts to the same thing,1

dy = y′h, dy′ = y′′h, dy′′ = y′′′h, . . . , dy(n−1) = y(n)h.(2)

Moreover, as the differential of a function of the variable x is another function ofthis variable, nothing prevents differentiating y several times in sequence. We will

1There is a small mistake in Cauchy’s original 1823 text and the 1899 reprint which both havedy(n) = y(n−1)h for the rightmost relation on this line. This error has been corrected.


59


https://doi.org/10.1007/978-3-030-11036-9_12


obtain by this manner, the differentials of various orders of the function y, namely

dy = y′h = y′ dx,

ddy = h dy′ = y′′h2 = y′′ dx2,

dddy = h2dy′′ = y′′′h3 = y′′′ dx3,

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

For brevity, we simply write d2y instead of ddy, d3y instead of dddy, . . . , so thatthe differential of first order is represented by dy, the differential of second order byd2y, that of third order by d3y, . . . , and generally, the differential of order n by dn y.

These conventions being accepted, we will obviously have

(3)

{dy = y′ dx, d2y = y′′ dx2, d3y = y′′′ dx3,

d4y = yIV dx4, . . . . . . . . . . . . , dn y = y(n) dxn,

and as a result,

(4)

⎧⎪⎪⎨⎪⎪⎩

y′ = dy

dx, y′′ = d2y

dx2, y′′′ = d3y

dx3,

yIV = d4y

dx4, . . . . . . . . . , y(n) = dn y

dxn.

It follows from the last of the formulas in (3) that the derivative of order n, namelyy(n), is precisely the coefficient by which it must multiply the nth power of theconstant h = dx to obtain the differential of order n. It is for this reason that y(n) issometimes called the differential coefficient of order n.2

The methods by which we determine the differentials and the derivatives of firstorder for the functions of a single variable serve equally to calculate their differ-entials and their derivatives of higher orders. The calculations of this type are per-formed very easily, as we shall show by examples.

First, let y = sin x . Since, in denoting by a a constant quantity, we have generally

d sin (x + a) = cos (x + a) d(x + a)

= sin(x + a + 1

2π)

dx,

we will infer

d sin x = sin(x + 1

2π)

dx,

d sin(x + 1

2π) = sin

(x + π

)dx,

d sin(x + π

) = sin(x + 3

2π)

dx,

. . . . . . . . . . . . . . . . . . . . . . . . ,

2Cauchy’s differential coefficient terminology never caught on. Instead, we now generally refer toy(n) as the nth derivative of y.

TWELFTH LECTURE. 61

and as a result, we will find, for y = sin x,

y′ = sin(x + 1

2π),

y′′ = sin(x + π

),

y′′′ = sin(x + 3

2π),

. . . . . . . . . . . . . . . . . . ,

y(n) = sin(x + n

2π).

By operating the same, we will also find, for y = cos x,

y′ = cos(x + 1

2π),

y′′ = cos(x + π

),

y′′′ = cos(x + 3

2π),

. . . . . . . . . . . . . . . . . . ,

y(n) = cos(x + n

2π);

for y = Ax ,

y′ = Ax lA,

y′′ = Ax (lA)2,

y′′′ = Ax (lA)3,

. . . . . . . . . . . . ,

y(n) = Ax (lA)n;for y = xa,

y′ = axa−1,

y′′ = a(a − 1)xa−2,

. . . . . . . . . . . . . . . . . . . . . . . . ,

y(n) = a(a − 1)(a − 2) · · · (a − n + 1)xa−n .

It is essential to observe that each of the expressions sin(x + 1

2nπ), cos

(x + 1

2nπ),

admits only four distinct values that reproduce periodically and always in the sameorder. These four values, from which we obtain the first, the second, the third, or thefourth, depending on whether the integer number n divided by 4 gives for remain-der 0, 1, 2, or 3, are, respectively, sin x, cos x, − sin x, − cos x for the expressionsin

(x + 1

2nπ), and cos x, − sin x, − cos x, sin x for the expression cos

(x + 1

2nπ).


Moreover, if, in the functions Ax , xa, we replace the letter A by the number e, whichserves as the base of the Napierian logarithm,3 and the quantity a by the integernumber n, we will recognize that the successive derivatives of ex are all equal toex ,4 while that for the function xn, the derivative of order n is reduced to the con-stant quantity 1 · 2 · 3 · · · n, and the following to zero.

By substituting the differentials for derivatives, we will derive from the formulasthat we have just established

dn sin x = sin(x + 1

2nπ)

dxn,

dn cos x = cos(x + 1

2nπ)

dxn,

dn Ax = Ax (lA)n dxn,

dnex = ex dxn,

dn(xa) = a(a − 1) · · · (a − n + 1)xa−n dxn,

dn(xn) = 1 · 2 · 3 · · · n dxn,

dnlx = dx dn−1(x−1) = (−1)n−1 1 · 2 · 3 · · · (n − 1)

xndxn,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Consider again the two functions f (x + a) and f (ax). We will find, for y =f (x + a),

y′ = f ′(x + a), y′′ = f ′′(x + a), . . . , y(n) = f (n)(x + a),

dn y = f (n)(x + a) dxn;for y = f (ax),

y′ = a f ′(ax), y′′ = a2 f ′′(ax), . . . , y(n) = an f (n)(ax),

dn y = an f (n)(ax) dxn .

Examples. –

dn(x + a)n = 1 · 2 · 3 · · · n dxn, dneax = aneax dxn,

dn sin ax = · · · .

Now, let y = f (x) and z be two functions of x related by the equation

z = F(y).(5)

3Named after John Napier (1550–1617), the Napierian logarithm is often used to denote the naturallogarithm.4An incredible property of this centrally important function.

TWELFTH LECTURE. 63

By differentiating this equation several times in sequence, we will find

(6)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

dz = F ′(y) dy,

d2z = F ′′(y) dy2 + F ′(y) d2y,

d3z = F ′′′(y) dy3 + 3F ′′(y) dy d2y + F ′(y) d3y,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Examples.5 −dn(a + y) = dn y, dn(−y) = −dn y,

dn(ay) = a dn y, dn(axn) = 1 · 2 · 3 · · · n · a dxn,

d ey = ey dy, d2ey = ey(dy2 + d2y

),

d3ey = ey(dy3 + 3dy d2y + d3y

), . . . . . . . . .

If the variable x were to cease to be independent, the equation

y = f (x),(7)

being differentiated several times in sequence, would create new formulas perfectlysimilar to those of the equations in (6), namely6

(8)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

dy = f ′(x) dx,

d2y = f ′′(x) dx2 + f ′(x) d2x,

d3y = f ′′′(x) dx3 + 3 f ′′(x) dx d2x + f ′(x) d3x,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We derive from those above

(9)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

f ′(x) = dy

dx,

f ′′(x) = dx d2y − dy d2x

dx3= 1

dxd

dy

dx,

f ′′′(x) = dx (dx d3y − dy d3x) − 3d2x (dx d2y − dy d2x)

dx5

= 1

dxd

dx d2y − dy d2x

dx3,

. . . . . . . . . . . . . . . . . . . . . .

5The comma following the third example here is misprinted in the original 1823 version. Thissmall typographical error is corrected in its ERRATA.6The d2x in the third line of equation (8) is also misprinted in the original 1823 edition but iscorrected in its ERRATA as well.


To return to the case where x is the independent variable, it is sufficient to supposethe differential dx is constant, and as a result, d2x = 0, d3x = 0, . . . . Then, theformulas in (9) become

(10) f ′(x) = dy

dx, f ′′(x) = d2y

dx2, f ′′′(x) = d3y

dx3, . . . ,

that is to say, they are reduced to the equations in (4). From these latter equations,compared to the equations in (9), it follows that, if we express the successive deriva-tives of f (x)with the help of the differentials of the variables x and y = f (x) : 1◦ inthe case where the variable x is assumed independent; 2◦ in the case where it ceasesto be, the derivative of first order will be the only one whose expression remains thesame in the two hypotheses. Add that, to pass from the first case to the second, itwill be necessary to replace

d2y

dx2by

dx d2y − dy d2x

dx3,

d3y

dx3by

dx (dx d3y − dy d3x) − 3d2x (dx d2y − dy d2x)

dx5,

etc. It is by substitutions of this nature that we can perform a change of independentvariable.

Among the composed functions of a single variable, there are some whose suc-cessive differentials are presented under a very simple form. Consider, for example,that we denote by u, v, w, . . . various functions of x . By differentiating each of thecomposed functions

u + v, u − v, u + v√−1, au + bv + cw + · · ·

n times, we will find

(11)

⎧⎪⎨⎪⎩

dn(u + v) = dnu + dnv,

dn(u − v) = dnu − dnv,

dn(u + v

√−1) = dnu + dnv

√−1,

dn(au + bv + cw + · · · ) = a dnu + b dnv + c dnw + · · · .(12)

It follows from formula (12) that the differential dn y of the entire function7

y = axm + bxm−1 + cxm−2 + · · · + px2 + qx + r

is reduced, for n = m, to the constant quantity 1·2 · 3 · · · m · a · dxm, and for n > m,

to zero.

7Recall this is the phrase Cauchy uses to denote a polynomial with non-negative integer powers.

THIRTEENTH LECTURE.

DIFFERENTIALS OF VARIOUS ORDERS FORFUNCTIONS OF SEVERAL VARIABLES.

Let u = f (x, y, z, . . . ) be a function of several independent variables x, y, z, . . . .

If we differentiate this function several times in sequence, either with respect toall the variables or with respect to only one of them, we will obtain several newfunctions, each of which will be the total or partial derivative of the preceding one.We could also conceive that the successive differentiations are sometimes with re-spect to one variable, sometimes to another one. In all cases, the result of one, of two,of three, …differentiations, successively performed, is what we call a totalor partial differential of first, of second, of third, …order. Thus, for example, bydifferentiating several times in sequence with respect to all the variables, we willgenerate the total differentials du, ddu, dddu, . . . that we denote, for brevity, by thenotations du, d2u, d3u, . . . . On the other hand, by differentiating several times insequence with respect to the variable x, we will generate the partial differen-tials dx u, dx dx u, dx dx dx u, . . . that we denote by the notations dx u, d2

x u, d3x u, . . . .

In general, if n is any integer number, the total differential of order n will be rep-resented by dnu, and the differential of the same order relative to only one of thevariables x, y, z, . . . by dn

x u, dny u, dn

z u, . . . .1 If we differentiate twice or severaltimes in sequence with respect to two or to several variables, we would obtain thepartial differentials of second order or of higher orders, designated by the notationsdx dyu, dydx u, dx dzu, . . . , dx dydzu, . . . . Now, it is easy to see that the differentialsof this type retain the same values when we reverse the order in which the differ-entiations relative to the various variables must be performed. We will have, forexample,

dx dyu = dy dx u.(1)

It is actually this that we can demonstrate as follows.Let us conceive that we indicate by the letter x, placed at the base of the char-

acter Δ, the increment that a function of x, y, z, . . . receives when only x grows an

1A small typographical error in the 1899 reprint exists here. It reads dnx u, dn

y , dnz u, . . . . This error

does not occur in the 1823 version.


65


https://doi.org/10.1007/978-3-030-11036-9_13


infinitely small quantity α dx . We will find

Δx u = f (x + α dx, y, z, . . . ) − f (x, y, z, . . . ), dx u = limΔx u

α,(2)

Δx dyu = dy(u + Δx u) − dyu = dyΔx u,(3)

and as a result,Δx dyu

α= dy Δx u

α= dy

Δx u

α;

then, by letting α converge toward zero, and having regard to the second of the for-mulas in (2), we will obtain equation (1). We would establish by the same manner,the identical equations dx dzu = dz dx u, dy dzu = dz dyu, . . . .

Example. – If we set u = arctang(

xy

), we will find

dx u = y

x2 + y2dx, dyu = −x

x2 + y2dy, dy dx u = dx dyu = x2 − y2

(x2 + y2)2dx dy.

Equation (1) being demonstrated once, it follows that in an expression of the formdx dy dz · · · u, it is always allowed to exchange between them, the variables whichrelate two consecutive differentiations. Now, it is clear that with the help of one orof several exchanges of this type, we can interchange in all possible manners theorder of the differentiations. Thus, for example, to deduce the differential dz dy dx ufrom the differential dx dy dzu, it will suffice to first bring the letter x to the spot ofthe letter z by two consecutive exchanges, then to exchange in sequence the lettersy and z to finally return the letter y to the second spot. We can therefore assert thata differential of the form dx dy dz · · · u has a value independent of the order in whichthe differentiations relative to the various variables are performed. This propositionremains valid even in the case where several differentiations are with respect to oneof the variables, as happens for the differentials dx dy dx u, dx dy dx dx u, . . . . Whenthis circumstance occurs, and two or more consecutive differentiations are relativeto the variable x,wewrite for brevity, d2

x instead of dx dx , d3x instead of dx dx dx , . . . .

This granted, we will have2

d2x dy u = dx dy dx u = dy d2

x u,

d3x dy dz u = dx dy dx dz dx u = dy d3

x dz u = · · · ,

d2x d3

y u = d3y d2

x u,

dx d2y d3

z u = dx d3z d2

y u = d2y dx d3

z u = · · · ,

2In the second expression, the 1899 reprint has a typographical error that is not present in theoriginal 1823 edition. The 1899 edition reads, · · · = dy d3

x dz u = · · · .

THIRTEENTH LECTURE. 67

and generally, l, m, n, . . . being any integer numbers,

dlx dm

y dnz · · · u = dl

x dnz dm

y · · · u = dmy dl

x dnz · · · u = · · · .(4)

Since, by differentiating a function of the independent variables x, y, z, . . . withrespect to one of them, we obtain for a result a new function of these variablesmultiplied by the finite constant dx, or dy, or dz, . . . , and that, in the differentiationof a product, the constant factors always pass out of the character d, it is clear that,if we perform, one after the other, l differentiations relative to x, m differentiationsrelative to y, n differentiations relative to z, . . . , on the function u = f (x, y, z, . . . ),the differential that will result from these various operations, namely dl

x dmy dn

z · · · u,

will be the product of a new function of x, y, z, . . . by the factors dx, dy, dz, . . . ,the first raised to the power l, the second raised to the power m, the third raised tothe power n, . . . . The new function in question here is that which we call a partialderivative of u of order l + m + n + · · · . If we denote it by�(x, y, z, . . . ),wewillhave

dlx dm

y dnz · · · u = �(x, y, z, . . . ) dxl dym dzn · · · ,(5)

and as a result,

�(x, y, z, . . . ) = dlx dm

y dnz · · · u

dxl dym dzn · · · .(6)

It is easy to express the total differentials d2u, d3u, . . . with the help of the partialdifferentials of the function u, or of its partial derivatives. In fact, we derive fromformula (10) (eighth lecture)

d2u = d du = dx du + dy du + dz du + · · ·= dx (dx u + dyu + dzu + · · · ) + dy(dx u + dyu + dzu + · · · )

+ dz(dx u + dyu + dzu + · · · ),and it follows

d2u = d2x u + d2

y u + d2z u + · · · + 2 dx dyu + 2 dx dzu + · · · + 2 dy dzu + · · · ,(7)

or to what amounts to the same thing

⎧⎪⎪⎨

⎪⎪⎩

d2u = d2x u

dx2dx2 + d2

y u

dy2dy2 + d2

z u

dz2dz2 + · · ·

+ 2dx dyu

dx dydx dy + 2

dx dzu

dx dzdx dz + · · · + 2

dy dzu

dy dzdy dz + · · · .

(8)

We would obtain with the same ease the values of d3u, d4u, . . . .


Examples. –

d2(xyz) = 2(x dy dz + y dz dx + z dx dy),

d3(xyz) = 6 dx dy dz,

d2(x2 + y2 + z2 + · · · ) = 2(dx2 + dy2 + dz2 + · · · ),d3(x3 + y3 + z3 + · · · ) = 6(dx3 + dy3 + dz3 + · · · ),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

For brevity, we ordinarily omit, in equations (6), (8), etc., the letters that we havewritten at the base of the character d, and we replace the second member of formula(6) by the notation

dl+m+n+···udxl dym dzn · · · .(9)

Then, the partial derivatives of second order are found represented by

d2u

dx2,

d2u

dy2,

d2u

dz2, . . . ,

d2u

dx dy,

d2u

dx dz, . . . ,

d2u

dy dz, . . . ;

the partial derivatives of third order by

d3u

dx3,

d3u

dx2 dy,

d3u

dx dy2, . . . ;

and the value of d2u is reduced to

(10)

⎧⎪⎪⎨

⎪⎪⎩

d2u = d2u

dx2dx2 + d2u

dy2dy2 + d2u

dz2dz2 + · · ·

+ 2d2u

dx dydx dy + 2

d2u

dx dzdx dz + · · · + 2

d2u

dy dzdy dz + · · · .

But, it is no longer allowed to remove, in this value, the differentials dx, dy, dz, . . . ,

since d2udx2 ,

d2udx dy , . . . do not denote the quotients that we would obtain in dividing

d2u by dx2, or by dx dy, . . . .

If, instead of the function u = f (x, y, z, . . . ), we were to consider the following

s = F(u, v, w, . . . ),(11)

the quantities u, v, w, . . . being themselves any functions of the independent vari-ables x, y, z, . . . , the values of d2s, d3s, . . . would be deduced without difficulty

THIRTEENTH LECTURE. 69

from the principles established in the ninth lecture. Indeed, by differentiating for-mula (11) several times, we would find3

(12)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

ds = d F(u, v, w, . . . )

dudu + d F(u, v, w, . . . )

dvdv

+ d F(u, v, w, . . . )

dwdw + · · · ,

d2s = d2F(u, v, w, . . . )

du2du2 + · · · + 2

d2F(u, v, w, . . . )

du dvdu dv

+ · · · + d F(u, v, w, . . . )

dud2u + · · · ,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Examples. –

dn(u + v) = dnu + dnv,

dn(u − v) = dnu − dnv,

dn(u + v√−1 ) = dnu + √−1 dnv,

dn(au + bv + cw + · · · ) = a dnu + b dnv + c dnw + · · · .

We would again obtain with the greatest ease the differentials of implicit func-tions of several independent variables. It would be sufficient to differentiate theequations which determine these same functions, once or several times in sequence,by considering the differentials of the independent variables as constants, and theother differentials as new functions of these variables.

3The 1823 edition has a typographical error that is corrected in the 1899 reprint. The originalversion has ds to begin the second equation instead of d2s.

FOURTEENTH LECTURE.

METHODS THAT WORK TO SIMPLIFY THESTUDY OF TOTAL DIFFERENTIALS FORFUNCTIONS OF SEVERAL INDEPENDENTVARIABLES. SYMBOLIC VALUES OF THESEDIFFERENTIALS.

Let u = f (x, y, z, . . . ) always be a function of several independent variables x, y,

z, . . . ; and, denote by

ϕ(x, y, z, . . . ), χ(x, y, z, . . . ), ψ(x, y, z, . . . ), . . .

its first order partial derivatives relative to x, to y, to z, . . . . If we make, as in theeighth lecture,

F(α) = f (x + α dx, y + α dy, z + α dz, . . . ),(1)

then, differentiate the two members of equation (1) with respect to the variable α,

we will find

(2)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

F ′(α) = ϕ(x + α dx, y + α dy, z + α dz, . . . ) dx

+ χ(x + α dx, y + α dy, z + α dz, . . . ) dy

+ ψ(x + α dx, y + α dy, z + α dz, . . . ) dz

+ · · · .

If, in this last formula, we set α = 0, we will obtain the following

(3)

{F ′(0) = ϕ(x, y, z, . . . ) dx + χ(x, y, z, . . . ) dy

+ ψ(x, y, z, . . . ) dz + · · · = du,

which is in accordance with equation (16) of the eighth lecture. Moreover, it obvi-ously results from the comparison of equations (1) and (2) that by differentiating afunction of the variable quantities

x + α dx, y + α dy, z + α dz, . . . ,(4)

with respect to α, we obtain for the derivative another function of these quantitiescombined in a certain manner with the constants dx, dy, dz, . . . . As further differ-entiations relative to the variable α must produce new functions of the same type, we


71


https://doi.org/10.1007/978-3-030-11036-9_14


are right to conclude that the expressions in (4) will be the only variable quantitiescontained, not only in F(α) and F ′(α), but also in F ′′(α), F ′′′(α), . . . , and generally,in F (n)(α), n denoting any integer number. As a result, the differences

F(α) − F(0), F ′(α) − F ′(0),

F ′′(α) − F ′′(0), . . . , F (n)(α) − F (n)(0),

will be precisely equal to the increments that the functions of x, y, z, . . . , representedby

F(0), F ′(0), F ′′(0), . . . , F (n)(0),

receive when we attribute to the independent variables the infinitely smallincrements α dx, α dy, α dz, . . . . This granted, since we have F(0) = u, we willfind successively, by letting α converge toward the limit zero,1

F ′(0) = limF(α) − F(0)

α= lim

�u

α= du,

F ′′(0) = limF ′(α) − F ′(0)

α= lim

�du

α= d du = d2u,

F ′′′(0) = limF ′′(α) − F ′′(0)

α= lim

�d2u

α= d d2u = d3u,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

F (n)(0) = limF (n−1)(α) − F (n−1)(0)

α= lim

�dn−1u

α= d dn−1u = dnu.

In summary, we will have

(5)

⎧⎪⎨

⎪⎩

u = F(0), du = F ′(0),

d2u = F ′′(0), d3u = F ′′′(0),

. . . . . . . . . , dnu = F (n)(0).

Thus, to generate the total differentials

du, d2u, . . . , dnu,

it will suffice to calculate the particular values that the derived functions

F ′(α), F ′′(α), . . . , F (n)(α)

receive in the case where the variable α vanishes.

1Both the original 1823 and the 1899 reprint texts share a typographical error in the last expressionof this group of equations. The texts read

F (n)(0) = limF (n−1)(α) − F (n−1)(0)

α= lim

�dn−1u

α= d dn−1 = dnu,

but this error has been corrected here for clarity.

FOURTEENTH LECTURE. 73

Among the methods that work to simplify the study of total differentials, weshould also distinguish those which are based on the consideration of symbolic val-ues of these differentials.

In analysis, we call a symbolic expression or symbol, any combination of alge-braic signs which does not mean anything by itself or to which we attribute a dif-ferent value from that which it should naturally have. We also call symbolic equa-tions, any of those which, taken literally and interpreted after generally establishedconventions, are inaccurate or do not make sense, but from which we can deduceaccurate results by modifying or altering, according to settled rules, either theseequations themselves or the symbols that they contain. In the number of symbolicequations that it is useful to know, we should understand the imaginary equations(see Algebraic Analysis, Chapter VII)2 and those that we will now establish.

If we denote by a, b, c, . . . constant quantities, and by l, m, n, . . . , p, q, r, . . .integer numbers, the total differential of the expression

a dlx dm

y dnz · · · u + b d p

x dqy dr

z · · · u + · · ·(6)

will be given by the formula

(7)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

d(a dl

x dmy dn

z · · · u + b d px dq

y drz · · · u + · · · )

= dx(a dl

x dmy dn

z · · · u + b d px dq

y drz · · · u + · · · )

+ dy(a dl

x dmy dn

z · · · u + b d px dq

y drz · · · u + · · · )

+ dz(a dl

x dmy dn

z · · · u + b d px dq

y drz · · · u + · · · ) + · · ·

= a dl+1x dm

y dnz · · · u + a dl

x dm+1y dn

z · · · u

+ a dlx dm

y dn+1z · · · u + · · · + b d p+1

x dqy dr

z · · · u + · · · .

From this formula, combined with equation (7) of the thirteenth lecture,we immediately deduce the following proposition.

THEOREM. – To obtain the total differential of expression (6), it is sufficient tomultiply by d the product of the two factors

a dlx dm

y dnz · · · + b d p

x dqy dr

z · · · + · · ·

and u, by supposing d = dx + dy + dz + · · · , and proceeding as if the notationsd, dx , dy, dz, . . . represent actual quantities, distinct from each other, to expand thenew product, by writing, in the different terms, the factors a, b, c, . . . in the first posi-tion and the letter u in the last; then, to consider that, in each term, the notationsdx , dy, dz, . . . cease to represent quantities and to resume their original meaning.

2Cauchy investigates complex (or “imaginary” as he refers to them) expressions and functions atlength in this and subsequent chapters within his 1821 Cours d’analyse.


Examples. – By determining, with the help of this theorem, the total differentialof the expression

dx u + dyu + dzu + · · · ,(8)

we will precisely obtain the value of ddu, or d2u, that generates equation (7) ofthe previous lecture. By applying the theorem further to this value of d2u, we willobtain those of d3u, and so on.

Note. – When we only indicate the multiplications, with the help of which wecan, according to the theorem, calculate the total differential of expression (6), weobtain, instead of equation (7), the symbolic formula

(9)

{d(a dl

x dmy dn

z · · · u + b d px dq

y drz · · · u + · · · )

= (a dl

x dmy dn

z · · · + b d px dq

y drz · · · + · · · )(dx + dy + dz + · · · )u.

Since, in formula (9), the notations dx , dy, dz, . . . are employed to represent differ-entials, this formula, taken literally, is nonsense; but, it becomes accurate as soonas we expand its second member with the help of the ordinary rules of algebraicmultiplication, and by proceeding as if dx , dy, dz, . . . were actual quantities.

When to expression (6) we substitute expression (8), and differentiate this lastexpression several times in sequence, we obtain by the same procedure the symbolicvalues of the total differentials d2u, d3u, . . . , namely,

(dx + dy + dz + · · · )(dx + dy + dz + · · · )u,

(dx + dy + dz + · · · )(dx + dy + dz + · · · )(dx + dy + dz + · · · )u,

etc. By joining with these symbolic values those of du, then writing for brevity,(dx + dy + dz + · · · )2

instead of (dx + dy + dz + · · · )(dx + dy + dz + · · · ),

and (dx + dy + dz + · · · )3

instead of(dx + dy + dz + · · · )(dx + dy + dz + · · · )(dx + dy + dz + · · · ),

etc., we will generate the symbolic equations

(10)

{du = (

dx + dy + dz + · · · )u, d2u = (dx + dy + dz + · · · )2u,

d3u = (dx + dy + dz + · · · )3u, . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

FOURTEENTH LECTURE. 75

and we will generally have, n denoting any integer number,

dnu = (dx + dy + dz + · · · )n

u.(11)

Now, let

s = F(u, v, w, . . . ),(12)

u, v, w, . . . being functions of the independent variables x, y, z, . . . . We will againfind

dns = (dx + dy + dz + · · · )n

s.(13)

It is very easy to expand the second member of this last equation in the particularcase where we assume u a function of x only, v a function of y only, w a functionof z only, etc. Moreover, to pass from this particular case to the general case, it willobviously suffice to replacedx u, d2

x u, d3x u, . . . bydu, d2u, d3u, . . . ;dyv, d2

y v, . . . bydv, d2v, . . . ; . . . ; that is to say, to remove the letters x, y, z, . . . placed at the base ofthe character d. Therefore, it will be easy, in all cases, to derive the value of dns fromformula (13). Take, to settle the idea, s = uv. By proceeding as we have just said, wewill find successively3

(14)

⎧⎪⎪⎨

⎪⎪⎩

dn(uv) = u dny v + n

1dx u dn−1

y v + n(n − 1)

1 · 2 d2x u dn−2

y v + · · ·

+ n

1dyv dn−1

x u + v dnx u

and4

(15)

⎧⎪⎪⎨

⎪⎪⎩

dn(uv) = u dnv + n

1du dn−1v + n(n − 1)

1 · 2 d2u dn−2v + · · ·

+ n

1dv dn−1u + v dnu.

The last formula remains valid, regardless of the values of u, v in x, y, and even inthe case where u, v are reduced to two functions of x .5

Example. –

dn eax

x= aneax

x

[

1 − n

ax+ n(n − 1)

a2x2− n(n − 1)(n − 2)

a3x3+ · · · ± n(n − 1) · · · 3 · 2 · 1

an xn

]

dxn .

3An error in equation (14) from the original 1823 edition is corrected in its ERRATA. The originalhas dn

x u instead of v dnx u as its final term on the right.

4A typographical error in equation (15) occurs in both the original 1823 and the 1899 reprinteditions. Both texts have · · · + n

1 dv dn−2u + v dnu as the final two terms in the equation. This errorhas been corrected here to avoid confusion.5The equation Cauchy is referring to, equation (15), plays a prominent role in his subsequentLecture Nineteen work.

FIFTEENTH LECTURE.

RELATIONSHIPS WHICH EXIST BETWEENFUNCTIONS OF A SINGLE VARIABLE ANDTHEIR DERIVATIVES OR DIFFERENTIALS OFVARIOUS ORDERS. USE OF THESEDIFFERENTIALS IN THE STUDY OF MAXIMAANDMINIMA.

Suppose that the function f (x) vanishes for the particular value x0 of the variablex . Consider, moreover, that this same function and its successive derivatives, up tothat of order n, are continuous in the vicinity of the particular value in question, andthat the continuity remains valid for each of them between the two limits x = x0, x =x0 + h. Equation (6) of the seventh lecture will become

f (x0 + h) = f (x0) + h f ′(x0 + θh) = h f ′(x0 + θh),

θ denoting a number less than unity; or in other words,

f (x0 + h) = h f ′(x0 + h1),(1)

h1 denoting a quantity of the same sign as h, but of a lesser numerical value. If thederived functions f ′(x), f ′′(x), . . . , f (n−1)(x) vanish in their turn for x = x0, wewill also find

(2)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f ′(x0 + h1) = h1 f ′′(x0 + h2),

f ′′(x0 + h2) = h2 f ′′′(x0 + h3),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

f (n−1)(x0 + hn−1) = hn−1 f (n)(x0 + hn),

h1, h2, h3, . . . , hn representing quantities which will all be of the same sign, butwhose numerical values will decrease more and more. By combining the equationsin (2) with equation (1), we will deduce the following without difficulty

f (x0 + h) = hh1h2 · · · hn−1 f (n)(x0 + hn),(3)

in which hn will be a quantity of the same sign as h, and the product hh1h2 · · · hn−1 aquantity of the same sign as hn . Add that the two ratios1 hn

h ,hh1h2···hn−1

hn will obviouslybe numbers contained between the limits 0 and 1, such that in designating by θ and

1The ratio hh1h2···hnhn is used in both of Cauchy’s 1823 text and its 1899 reprint.


77


https://doi.org/10.1007/978-3-030-11036-9_15


� two numbers of this type, we can present equation (3) under the form

f (x0 + h) = �hn f (n)(x0 + θh).(4)

If we imagine that the quantity h becomes infinitely small, formula (4) will alwaysremain valid, and we will find, by writing i instead of h,

f (x0 + i) = �i n f (n)(x0 + θ i).(5)

Moreover, since, for very small numerical values of i, the expression f (n)(x0 + θ i)will differ very little from f (n)(x0), we will immediately deduce from equation (5)the proposition that I will now state.2

THEOREM I. – Suppose that the function f (x) and its successive derivatives,up to that of ordern, being continuous with respect to x in the vicinity of the par-ticular value x = x0, all vanish, with the exception of f (n)(x), for this same value.Then, in designating by i a quantity which differs very little from zero, and settingx = x0 + i, we will obtain for f (x) a quantity affected by the same sign as the prod-uct in f (n)(x0).

It is easy to verify this theorem on the function

f (x) = (x − x0)nϕ(x).

When the function f (x) ceases to vanish for x = x0, Theorem I can be replacedby the following.

THEOREM II. – Suppose that the functions

f (x), f ′(x), f ′′(x), . . . , f (n)(x),

being continuous with respect to x in the vicinity of the particular value x = x0,all vanish for this same value, with the exception of the first, f (x), and of the last,f (n)(x). In designating by i a quantity which differs very little from zero, we willobtain for the infinitely small difference, f (x0 + i) − f (x0), a value affected by thesame sign as the product in f (n)(x0).

Proof. – To deduce Theorem II from Theorem I, it is sufficient to substitute forthe function f (x), the function f (x) − f (x0), which has the same derivatives as thefirst, and which, moreover, vanishes for x = x0. By virtue of the same substitution,equation (5) will be found replaced by the following

f (x0 + i) − f (x0) = � i n f (n)(x0 + θ i).(6)

2Equations (4) and (5) are both referenced in Cauchy’s further development of his version ofl’Hôpital’s Rule when he revisits this topic in the ADDITION at the end of his text.

FIFTEENTH LECTURE. 79

If we now write x instead of x0, and if we set

f (x) = y, Δx = i = αh,

equation (6) will take the form

Δy = �αn(dn y + β),(7)

β denoting, as well as α, an infinitely small quantity. However, it is essential toobserve that formula (7) will remain valid only for the particular value x = x0.

Corollary. – The same conditions being admitted as in Theorem II, suppose thatafter having assigned to the variable x the value x0,we attribute to this same variablean infinitely small increment. The corresponding increment of the function f (x)willbe, if n denotes an even number, a quantity constantly affected by the same sign asthe value of f (n)(x) or of dn y, corresponding to x = x0. If, on the other hand, nrepresents an odd number, the increment of the function will change sign with thatof the variable.

We have shown in the sixth lecture that the values of x, which, without render-ing discontinuous one of the functions, f (x), f ′(x), gave for the first a maxima orminima, were necessarily roots of the equation

f ′(x) = 0.(8)

Now, with the help of the foregoing, we can decide in general, if a root of equation(8) produces a maximum or a minimum of f (x). In fact, let x0 be this root and letf (n)(x) be the first of the derivatives of f (x) which does not vanish with f ′(x), forthe particular value x = x0. Suppose, in addition that, in the vicinity of this particularvalue, the functions

f (x), f ′(x), . . . , f (n)(x)

are all continuous with respect to x . It obviously follows from Theorem II that f (x0)will be a maximum, if, n being an even number, f (n)(x0) has a negative value anda minimum, if, n being an even number, f (n)(x0)3 has a positive value. If n werean odd number, the increment of the function changing sign with that of the vari-able, f (x0)4 would neither be a maximum nor a minimum. Moreover, by observingthat the differentials d f (x), d2 f (x), . . . always vanish with the derived functionsf ′(x), f ′′(x), . . . , and that, for even values of n, dn f (x) = f (n)(x) dxn has thesame sign as f (n)(x), we will be naturally led to the following proposition.

THEOREM III. – Let y = f (x) be a given function of the variable x . To decideif a root of the equation dy = 0 produces a maximum or a minimum of the pro-posed function, it will ordinarily suffice to calculate the values of d2y, d3y, d4y, . . .

3The 1899 reprint has f (n)(x).4The 1899 reprint has f (x).


corresponding to this root. If the value of d2y is positive or negative, the value of ywill be a minimum in the first case, a maximum in the second. If the value of d2y isreduced to zero, we should seek among the differentials d3y, d4y, . . . , the first whichwill not vanish. Denote this differential by dn y. If n is an odd number, the valueof y will be neither a maximum nor a minimum. If, on the other hand, n is an evennumber, the value of y will be a minimum whenever the differential dn y is positiveand a maximum whenever the differential dn y is negative.

Note. – It must be admitted for Theorem III, as for the first two, that the functiony and its successive derivatives, up to that of order n, are continuous in the vicinityof the particular value attributed to the variable x .

If, instead of taking for y an explicit function of the variable x,wewere to assumethe value of y in x given by an equation of the form u = 0, Theorem III would still beapplicable. Only, in this hypothesis, the successive values of dy, d2y, d3y, . . . wouldbe deduced from the differential equations

du = 0, d2u = 0, d3u = 0, . . . .

Example. – Lety = xae−x ,

a denoting a positive quantity. We will have

ly = a lx − x .

By differentiating the latter equation two times in sequence, we will find

dy

y=

(a

x− 1

)dx,

d2y

y−

(dy

y

)2

= −a

(dx

x

)2

;

then, by setting dy = 0, and ignoring the zero value that y cannot receive,5

0 = a

x− 1,

d2y

y= −a

(dx

x

)2

.(9)

The value of d2y given by the second of the formulas in (9) being negative, it resultsthat the value of x given by the first generates a maximum of y.

5A variation between the two editions occurs here. The original 1823 edition uses the phrasefaisant abstraction de la valeur zéro que y peut recevoir (ignoring the zero value that y can receive),whereas the 1899 edition has faisant abstraction de la valeur zéro que y ne peut recevoir (ignoringthe zero value that y cannot receive). The 1899 modification conveys the warning regarding adivision by zero error more clearly.

SIXTEENTH LECTURE.

USE OF DIFFERENTIALS OF VARIOUS ORDERSIN THE STUDY OF MAXIMA ANDMINIMA OFFUNCTIONS OF SEVERAL VARIABLES.

Let u= f (x, y, z, . . . ) be a function of the independent variables x, y, z, . . . , and set,as in the tenth lecture,

f (x + α dx, y + α dy, z + α dz, . . . ) = F(α).(1)

So that the value of u relative to certain particular values of x, y, z, . . . is either amaximum or a minimum, it will be necessary and sufficient that the correspond-ing value of F(α) always becomes a maximum or a minimum, by virtue of theassumption α = 0. We conclude (see the tenth lecture) that the systems of values ofx, y, z, . . . , which, without rendering discontinuous one of the two functions, u anddu, generates for the first, a maxima or aminima, and necessarily satisfies, regardlessof dx, dy, dz, . . . , the equation

du = 0,(2)

and by consequence, the following

du

dx= 0,

du

dy= 0,

du

dz= 0, . . . .(3)

Let x0, y0, z0, . . . be the particular values of x, y, z, . . . which compose one of thesesystems. The corresponding value of F(α) will become a maximum or a minimumforα = 0, regardless of the differentials dx, dy, dz, . . . , if, for all the possible valuesof these differentials, the first of the quantities F ′(0), F ′′(0), F ′′′(0), . . . whichwill not be null, corresponds to an even index and always maintains the samesign (see the fifteenth lecture). Add that F(0) will be a maximum, if the quantity inquestion is always negative, and a minimum, if it is always positive. When those ofthe quantities F ′(0), F ′′(0), F ′′′(0), . . . which cease to vanish first, correspond toan odd index for all the possible values of dx, dy, dz, . . . , or only for particular val-ues of these same differentials, as well as when this quantity is sometimes positive,sometimes negative, then F(0) can no longer be a maximum nor a minimum. If we


81


https://doi.org/10.1007/978-3-030-11036-9_16


now have regard to the equations in (5) of the fourteenth lecture, namely,

F(0) = u, F ′(0) = du, F ′′(0) = d2u, . . . ,

we will deduce from the remarks that we have just made, the following proposition.

THEOREM.–Let u = f (x, y, z, . . . )beagiven functionof the independentvari-ables x, y, z, . . . . To decide if a system of values of x, y, z, . . . works to satisfythe formulas in (3) and produce a maximum or a minimum of the function u, wewill calculate the values of d2u, d3u, d4u, . . . that correspond to this system, andthat will obviously be polynomials in which they will no longer have the arbitrarydifferentials dx = h, dy = k, dz = l, . . . . Let

dnu = dnu

dxnhn + dnu

dynkn + · · · + n

1

dnu

dxn−1 dyhn−1k + · · ·(4)

be the first of these polynomials which will not vanish, n designating an integernumber which can depend on the values attributed to the differentials h, k, l, . . . .If, for all possible values of these differentials, n is an even number and dnu is apositive quantity, the proposed value of u will be a minimum. It will be a maximum,if, n always being even, dnu always remains negative. Finally, if the number n issometimes odd, or if the differential dnu is sometimes positive, sometimes negative,the calculated value of u will be neither a maximum nor a minimum.

Note. – The foregoing theorem remains valid, by virtue of the principles estab-lished in the fifteenth lecture, whenever the functions F(α), F ′(α), . . . , F (n)(α)

are continuous with respect to α in the vicinity of the particular value α = 0, or towhat amounts to the same thing, whenever u, du, d2u, . . . , dnu are continuous withrespect to the variables x, y, z, . . . in the vicinity of the particular values attributedto these same variables.

Corollary I. – Consider that, to apply the theorem, we first find the value of theexpression

d2u = d2u

dx2h2 + d2u

dy2k2 + · · · + 2

d2u

dx dyhk + · · · ,(5)

by substituting the values of x, y, z, . . . drawn from the formulas in (3) into thederived functions

d2u

dx2,

d2u

dy2, . . . ,

d2u

dx dy, . . . .

We will find zero for a result if all these derivatives vanish. In the contrary hypothe-sis, d2u will be a homogeneous function of the arbitrary quantities h, k, l, . . . ; and,if we then vary these quantities, one of three things will happen. Either the differ-ential d2u will constantly maintain the same sign without ever vanishing; or, it willvanish for certain values of h, k, l, . . . and will return to the same sign whenever

SIXTEENTH LECTURE. 83

it ceases to be null; or, it will sometimes be positive and sometimes negative. Theproposed value of u will always be a maximum or a minimum in the first case,sometimes in the second, never in the third. Add that we will obtain, in the secondcase, a maximum or a minimum, if, for each of the systems of values of h, k, l, . . .that work to satisfy the equation d2u = 0, the first of the differentials d3u, d4u, . . .

which does not vanish is always of an even order and affected by the same sign asthose of the values of d2u which differ from zero.

Corollary II. – If the substitution of the values attributed to x, y, z, . . . were toreduce all of the derivatives of second order to zero, then, d2u being identically null,the proposed value of u could be neither a maximum nor a minimum, unless thesame substitution also caused d3u to vanish by reducing all of the derivatives of thirdorder to zero.

Corollary III. – If the substitution of the values attributed to x, y, z, . . . were tocause all of the derivatives of second order and of third order to vanish, we wouldhave identically d2u = 0, d3u = 0, and would need to resort to the first of the differ-entials d4u, d5u, . . . that would not be identically null. If this differential were of anodd order, the proposed value of u would have neither a maximum nor a minimum.If it were of an even order, or of the form

d2mu = d2mu

dx2mh2m + d2mu

dy2mk2m + · · · + 2m

1

d2mu

dx2m−1 dyh2m−1k + · · · ,(6)

one of three things could happen. Either the differential in question would constantlyretain the same sign while we vary h, k, l, . . . , without ever vanishing; or rather, itwould vanish for certain values of h, k, l, . . . , and would return to the same signwhenever it would cease to be null; or, it would sometimes be positive, sometimesnegative. The proposed value of u would always be a maximum or a minimum in thefirst case, sometimes in the second, never in the third. In addition, to finally decide,in the second case, whether there is a maximum or minimum, it would be needed,for each system of values of h, k, l, . . . that works to satisfy the equation d2mu = 0,to seek among the differentials of an order greater than 2m, the first of these thatceases to vanish, and see if this differential is always of even order and affected bythe same sign as the values of d2mu that differ from zero.

It is essential to observe that the value of d2mu given by formula (6),1 beingan entire function, and by consequence, continuous for the quantities h, k, l, . . . ,does not pass from positive to negative, while these quantities are varied, withoutbecoming null in the interval. We remark in addition that, if the quantity u were animplicit function of the variables x, y, z, . . . , or if some of these variables were tobecome implicit functions of all the others, each of the quantities du, d2u, d3u, . . .

would be found determined by means of one or of several differential equations,based on the differentials of the independent variables.

1The 1899 reprint has d2u.


Example. – Suppose that, a, b, c, . . . , k, p, q, r, . . . denoting positive constantsand x, y, z, . . . variables subject to the equation ax + by + cz + · · · = k, we seekthe maximum of the function u = x p yq zr · · · ; we will find

du

u= p

dx

x+ q

dy

y+ r

dz

z+ · · · ,

d2u

u−

(du

u

)2

= −p

(dx

x

)2

− q

(dy

y

)2

− r

(dz

z

)2

− · · · ,

and as a result, we will derive from formula (10) (eleventh lecture)

p

ax= q

by= r

cz= · · · = p + q + r + · · ·

k,

x = p

a

k

p + q + r + · · · , y = q

b

k

p + q + r + · · · , z = · · · .

Since the preceding values of x, y, z, . . . will render du constantly null and d2uconstantly negative, they will generate a maximum of the function u.

SEVENTEENTH LECTURE.

CONDITIONS WHICH MUST BE FULFILLED FORA TOTAL DIFFERENTIAL TO NOT CHANGESIGNWHILE WE CHANGE THE VALUESATTRIBUTED TO THE DIFFERENTIALS OF THEINDEPENDENT VARIABLES.

After what we have seen in the preceding lectures, if we denote by u a function ofthe independent variables x, y, z, . . . , and if we disregard the values of these vari-ables which render one of the functions u, du, d2u, . . . discontinuous, the functionu can only become a maximum or a minimum in the case where one of the totaldifferentials d2u, d4u, d6u, . . . , namely, the first of these that will not be constantlynull, will maintain the same sign for all possible values of the arbitrary quantitiesdx = h, dy = k, dz = l, . . . , or at least for the values of these quantities which willnot reduce it to zero. Add that, in the latter assumption, each of the systems of val-ues of h, k, l, . . . that work to make the total differential in question vanish, mustchange another total differential of even order into a quantity affected by the signthat maintains the first differential, as long as it does not vanish. Moreover, the differ-entials d2u, d4u, d6u, . . . are reduced, for the given values of x, y, z, . . . , to entireand homogeneous functions of the arbitrary quantities h, k, l, . . . . In addition, if wecall r, s, t, . . . the ratios of the first, of the second, of the third of these quantities,. . . to the last among them, the differential

d2mu = d2mu

dx2mh2m + d2mu

dy2mk2m + d2mu

dz2ml2m + · · · + 2m

1

d2mu

dx2m−1 dyh2m−1k + · · ·(1)

will obviously be affected by the same sign as the entire function of r, s, t, . . . towhich we achieve in dividing d2mu by the power 2m of the last of the quantitiesh, k, l, . . . , that is to say, of the same sign as the polynomial

d2mu

dx2mr2m + d2mu

dy2ms2m + d2mu

dz2mt2m + · · · + 2m

1

d2mu

dx2m−1 dyr2m−1s + · · · .(2)

By substituting a polynomial of this type to each differential of even order, we willrecognize that the study of maxima and minima requires the solution of the follow-ing questions.


85


https://doi.org/10.1007/978-3-030-11036-9_17


PROBLEM I. – Find the conditions which must be fulfilled for an entire functionof the quantities r, s, t, . . . to not change sign while these quantities vary.

Solution. –Let F(r, s, t, . . . ) be the given function, and first suppose the quantitiesr, s, t, . . . are reduced to a single r. So that the function F(r) does not ever changesign, it will be necessary and sufficient that the equation

F(r) = 0(3)

does not have a single real root nor an odd number of real reals. In fact, if, r0 denot-ing a real root of equation (3), m an integer number, and R a polynomial not divisibleby r − r0, we would have

F(r) = (r − r0)R

orF(r) = (r − r0)

2m+1R,

it is clear that, for two values of r differing very little from r0, but one larger andthe other smaller, the function F(r) would obtain two values of opposite signs. Inaddition, since a continuous function of r would not change sign while r variesbetween two given limits without becoming null in the interval, it is possible toassert that, if equation (3) does not have real roots, its first member will alwaysmaintain the same sign without ever vanishing, and that it will sometimes vanishwithout ever changing sign, if it is the product of several factors of the form

(r − r0)2m

by a polynomial which cannot be reduced to zero for any real value of r.Return now to the case where the quantities r, s, t, . . . are any number. Then, so

that the function F(r, s, t, . . . ) cannot change sign, it will be necessary and sufficientthat the equation

F(r, s, t, . . . ) = 0,(4)

resolved with respect to r, does not ever supply single real roots nor real roots equalto an odd number, regardless, moreover, of any s, t, . . . .

Corollary I. – The function F(r) or F(r, s, t, . . . ), constantly maintains the samesign when equation (3) or (4) does not have real roots. (See, for the determination ofthe number of real roots in algebraic equations, the XVIIth Notebook of the Journalof the Polytechnic School, page 457.)

Corollary II. – Let u = f (x, y). The total differential

d2u = d2u

dx2h2 + d2u

dy2k2 + 2

d2u

dx dyhk(5)

SEVENTEENTH LECTURE. 87

will constantly maintain the same sign, if the equation

d2u

dx2r2 + 2

d2u

dx dyr + d2u

dy2= 0(6)

does not have real roots, that is to say, if we have

d2u

dx2

d2u

dy2−

(d2u

dx dy

)2

> 0.(7)

The same differential could vanish without ever changing sign, if the first memberof formula (7) were to reduce to zero; and, it would admit values of opposite signs,if this first member were to become negative.

Corollary III. – Let u = f (x, y, z). The total differential1

d2u = d2u

dx2h2 + d2u

dy2k2 + d2u

dz2l2 + 2

d2u

dx dyhk + 2

d2u

dx dzhl + 2

d2u

dy dzkl(8)

will constantly maintain the same sign, if the equation

d2u

dx2r2 + 2

(d2u

dx dys + d2u

dx dz

)r + d2u

dy2s2 + 2

d2u

dy dzs + d2u

dz2= 0,(9)

resolved with respect to r, does not ever have real roots, that is to say, if we have,regardless of s,

(10)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

[d2u

dx2

d2u

d2y−

(d2u

dx dy

)2]

s2

+ 2

(d2u

dx2

d2u

dy dz− d2u

dx dy

d2u

dx dz

)s + d2u

dx2

d2u

dz2−

(d2u

dx dz

)2

> 0.

This latter condition will itself be satisfied when we will have

(11)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

d2u

dx2

d2u

d2y−

(d2u

dx dy

)2

> 0,

and[d2u

dx2

d2u

dy2−

(d2u

dx dy

)2] [

d2u

dx2

d2u

dz2−

(d2u

dx dz

)2]

−(

d2u

dx2

d2u

dy dz− d2u

dx dy

d2u

dx dz

)2

> 0.

1The original 1823 edition has d2u = · · · + 2 d2udx dy hk + 2 d2u

dx dy hl + 2 d2udy dz kl for equation (8). This

error is corrected in the 1899 reprint.


Scholium. – Let u = f (x, y, z, . . . ) be a function of the n independent variablesx, y, z, . . . , and set

(12)

⎧⎪⎪⎨⎪⎪⎩

F(r, s, t, . . . ) = d2u

dx2r2 + d2u

dy2s2 + d2u

dz2t2 + · · ·

+ 2d2u

dx dyrs + 2

d2u

dx dzrt + 2

d2u

dy dzst + · · · .

The differential d2u and the function F(r, s, t, . . . ) will always be affected by thesame sign as the quantity d2u

dx2 , if the product d2udx2 F(r, s, t, . . . ) is always positive.

Now, it is obviously this that will occur, if each of the products2

d2u

dx2F(r),

d2u

dx2F(r, s),

d2u

dx2F(r, s, t), . . .(13)

obtain a positive quantity for a minimum value. Moreover, if we let

D1 = d2u

dx2, D2 = d2u

dx2

d2u

dy2−

(d2u

dx dy

)2

, . . . ,

and if we generally designate by Dn the common denominator of the values ofh, k, l, . . . derived from the equations (see Algebraic Analysis, page 80)3

(14)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

d2u

dx2h + d2u

dx dyk + d2u

dx dzl + · · · = 1,

d2u

dx dyh + d2u

dy2k + d2u

dy dzl + · · · = 1,

d2u

dx dzh + d2u

dy dzk + d2u

dz2l + · · · = 1,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

we will prove without difficulty that the maxima or minima values of the functions4

F(r), F(r, s), F(r, s, t), . . . are, respectively,

D2

D1,

D3

D2,

D4

D3, . . . ,

Dn

Dn−1.(15)

2The ERRATA notes an error in equation (13). The original version reads, d2udx2

F(r, 0, 0, 0, . . . ),d2udx2

F(r, s, 0, 0, . . . ), d2udx2

F(r, s, t, 0, . . . ), . . . .3The second equation within (14) in the original 1823 edition has · · · + d2u

dy2k + d2u

dx dz l + · · · = 1.This error is corrected in the 1899 reprint.4The original 1823 edition again uses the improper notation F(r, 0, 0, 0, . . . ), F(r, s, 0, 0, . . . ),F(r, s, t, 0, . . . ), . . . here. Each of these is corrected in the ERRATA.

SEVENTEENTH LECTURE. 89

Therefore, the differential d2u will constantly maintain the same sign, if the fractionsin (15), multiplied by D1, result in positive products, or to what amounts to the samething, if D2, D3, D4, . . . , Dn are affected by the same signs as D2

1, D31, D4

1, . . . , Dn1 .

When we simply suppose u a function of the three variables x, y, z, the condi-tions that we just stated are reduced to the following two, D2 > 0, D1D3 > 0, andcoincide with those that generate the formulas in (11).

PROBLEM II. – Being given two entire functions of the variables r, s, t, . . . ,find the conditions which must be fulfilled so that the second function retains a defi-nite sign whenever the first vanishes.

Solution. – LetF(r, s, t, . . . )

be the first function, and letR = F(r, s, t, . . . )

be the second. We will eliminate r between the two equations

F(r, s, t, . . . ) = 0 and R = F(r, s, t, . . . ).

The resultant equation, being resolved with respect to R,must provide for this quan-tity a value affected by the agreed sign whenever we will attribute to the variabless, t, . . . real values which correspond to a real value of the variable r.

EIGHTEENTH LECTURE.

DIFFERENTIALS OF ANY FUNCTION OFSEVERAL VARIABLES EACH OF WHICH IS INITS TURN A LINEAR FUNCTION OF OTHERSUPPOSED INDEPENDENT VARIABLES.DECOMPOSITION OF ENTIRE FUNCTIONS INTOREAL FACTORS OF FIRST OR OF SECONDDEGREE.

Let a, b, c, . . . , k be constant quantities, and let

u = ax + by + cz + · · · + k(1)

be a linear function of the independent variables x, y, z, . . . . The differential

du = a dx + b dy + c dz + · · ·(2)

will itself be a constant quantity, and as a result, the differentials d2u, d3u, . . .

will all be reduced to zero. We immediately conclude from this remark that thesuccessive differentials of the functions

f (u), f (u, v), f (u, v, w, . . . ), . . .

retain the same form in the case where the variables u, v, w, . . . are considered as in-dependent, and in the case where u, v, w, . . . are linear functions of the independentvariables x, y, z, . . . . Thus, we will find in both cases, for s = f (u),

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

ds = f ′(u) du,

d2s = f ′′(u) du2,

d3s = f ′′′(u) du3,

. . . . . . . . . . . . . . . ,

dns = f (n)(u) dun;

(3)


91


https://doi.org/10.1007/978-3-030-11036-9_18


for s = f (u, v),

⎧⎪⎨

⎪⎩

dns = dn f (u, v)

dundun + n

1

dn f (u, v)

dun−1 dvdun−1 dv + · · ·

+ n

1

dn f (u, v)

du dvn−1du dvn−1 + dn f (u, v)

dvndvn;

(4)

for s = f (u) f (v),

⎧⎪⎨

⎪⎩

dns = f (n)(u) f (v) dun + n

1f (n−1)(u) f ′(v) dun−1 dv + · · ·

+ n

1f ′(u) f (n−1)(v) du dvn−1 + f (u) f (n)(v) dvn;

(5)

etc.It is easy to ensure that, if we represent by f (u), f (v), f (u, v), . . . entire func-

tions of the variables u, v, w, . . . , the formulas in (3), (4), (5), …will remain valideven when, u, v, w, . . . being linear functions of x, y, z, . . . , the constants a, b, c,. . . , k, . . . included in u, v, w, . . . will become imaginary.Wewill have, for example,for s = f

(x + y

√−1),

⎧⎪⎨

⎪⎩

ds = f ′(x + y√−1

)(dx + √−1 dy

),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

dns = f (n)(x + y

√−1)(

dx + √−1 dy)n;

(6)

for s = f(x − y

√−1),

⎧⎪⎨

⎪⎩

ds = f ′(x − y√−1

)(dx − √−1 dy

),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

dns = f (n)(x − y

√−1)(

dx − √−1 dy)n;

(7)

for s = f(x + y

√−1)

f(x − y

√−1),

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

dns = f (n)(x + y

√−1)

f(x − y

√−1)(

dx + √−1 dy)n

+ n

1f (n−1)

(x + y

√−1)

f ′(x − y√−1

)(dx + √−1 dy

)n−1(dx − √−1 dy

)

+ · · ·+ n

1f ′(x + y

√−1)

f (n−1)(x − y

√−1)(

dx + √−1 dy)(

dx − √−1 dy)n−1

+ f(x + y

√−1)

f (n)(x − y

√−1)(

dx − √−1 dy)n

.

(8)

From this last formula we will deduce the following proposition without difficulty.

EIGHTEENTH LECTURE. 93

THEOREM I. – Let f (x) be a real and entire function of x . If we set

s = f(x + y

√−1)

f(x − y

√−1),(9)

we can always satisfy, by real values of the variables x and y, the equation

s = 0.(10)

Proof. – Let n be the degree of the function f (x), so that we would have

f (x) = a0xn + a1xn−1 + · · · + an−1x + an,(11)

a0, a1, . . . , an−1, an denoting constants, the first of which, a0, does not vanish. Letus conceive in addition that, the variables x, y being supposed real, we represent byr, ρ, R, R1, R2, . . . the moduli of the imaginary expressions

x + y√−1, dx + dy

√−1,

f(x + y

√−1), f ′(x + y

√−1), f ′′(x + y

√−1), . . . ,

and we have, by consequence,

{x + y

√−1 = r(cos t + √−1 sin t

),

dx + dy√−1 = ρ

(cos τ + √−1 sin τ

),

(12)

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

f(x + y

√−1) = R

(cos T + √−1 sin T

),

f ′(x + y√−1

) = R1(cos T1 + √−1 sin T1

),

f ′′(x + y√−1

) = R2(cos T2 + √−1 sin T2

),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

f (n)(x + y

√−1) = Rn

(cos Tn + √−1 sin Tn

).

(13)

r, ρ, R, R1, R2, . . . , Rn will be positive quantities; t, τ, T, T1, T2, . . . , Tn will be realarcs; and, we will have

r =√

x2 + y2(14)


and1

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

s = R2 =[a0r

n cos nt + a1rn−1 cos (n − 1)t + · · · + an−1r cos t + an

]2

+[a0r

n sin nt + a1rn−1 sin (n − 1)t + · · · + an−1r sin t

]2

= r2n

(

a20 + 2a0a1 cos t

r+ a2

1 + 2a0a2 cos t

r2+ · · ·

)

.

(15)

It follows from these latter formulas that the quantity s, which represents an en-tire function, and by consequence, a continuous function of the variables x, y, willalways remain positive and will grow indefinitely if we attribute to these two vari-ables, or only to one among them, and as a result, to the modulus r, larger and largernumerical values. We must conclude that the function s will admit one or severalminima corresponding to one or to several systems of finite values of the variables xand y. We consider one of these systems in particular and calculate the correspond-ing values of the expressions

{f ′(x + y

√−1), f ′′(x + y

√−1),

. . . . . . . . . . . . , f (n)(x + y

√−1).

(16)

Some of these values can vanish; but, they will not ever be null all at once, sincethe expression f (n)

(x + y

√−1)is reducing, along with f (n)(x), to the product 1 ·

2 · 3 · · · n · a0, a constant value and different from zero. This granted, let f (m)(x +

y√−1

)be the first of the expressions in (16) whose value does not vanish. If the

expression f(x + y

√−1)itself obtains a value different from zero, dms will be, by

virtue of formula (8), the first of the differentials of s which will cease to vanish. Onthe other hand, if we have

f(x + y

√−1) = 0,(17)

the differential dms will become null itself. Now, I say that this latter case is the onlyone admissible. Because, in the first case, we would derive from formula (8)

⎧⎪⎨

⎪⎩

dms = f (m)(x + y

√−1)

f(x − y

√−1)(

dx + √−1 dy)m

+ f(x + y

√−1)

f (m)(x − y

√−1)(

dx − √−1 dy)m

= 2R Rmρm cos (Tm − T + mτ),

(18)

and as a result, the differential dms, changing sign when we replace τ by τ + πm ,

would not always remain positive regardless of the quantities dx and dy, since thismust necessarily happen every time the function s becomes a minimum. So, all the

1The 1899 edition incorrectly prints equation (15) witha21+2a0a2 cos 2t

r2for its final displayed term.

EIGHTEENTH LECTURE. 95

systems of values of x and y that work to provide minima of the function s willsatisfy equation (17), which we can also put in the form

R(cos T + √−1 sin T

) = 0,

and from which we derive

R = 0, s = R2 = 0.

Therefore, the function s will become null for the real and finite values of the vari-ables x and y whenever it attains a minima, which we have demonstrated aboveexists.

Corollary. – The real function s = R2 can only vanish along with the modulus R,

and the imaginary functions

f(x + y

√−1) = R

(cos T + √−1 sin T

),

f(x − y

√−1) = R

(cos T − √−1 sin T

)

will always vanish at the same time as it. By consequence, all real values of x and ywhich work to satisfy equation (10) will also satisfy equation (17) and the following

f(x − y

√−1) = 0.(19)

To these values of x and y will correspond real values of r and t that work to satisfythe two equations

f(r cos t + r sin t

√−1) = 0, f

(r cos t − r sin t

√−1) = 0.(20)

In the particular case where the value of y vanishes, the formulas in (17), (19), and(20) coincide with the single equation

f (x) = 0,(21)

which is then found satisfied by a real value of x . From these remarks we immedi-ately deduce the proposition that I will now state.

THEOREM II. – f (x) denoting a real and entire function of the variable x, wecan always satisfy equation (21), either by real values of this variable or by pairs ofimaginary conjugate values of the form

x = r(cos t + √−1 sin t

), x = r

(cos t − √−1 sin t

).(22)

Scholium. – If we call x0 a real or imaginary root of equation (20), the polynomialf (x) will be divisible by the linear factor x − x0. Therefore, to the two imaginary


conjugate roots of the form

r(cos t + √−1 sin t

), r

(cos t − √−1 sin t

),

will correspond the two linear factors

x − r cos t − r sin t√−1, x − r cos t + r sin t

√−1,

which will still be conjugates, one to another, and will give for a product a real factorof second degree, namely,

(x − r cos t)2 + r2 sin2 t = x2 − 2r x cos t + r2.

This granted, it results from Theorem II that any real and entire function of the vari-able x is divisible by a real factor of first or of second degree. The division beingperformed, we will obtain for a quotient another real and entire function which willitself be divisible by a new factor. By continuing in this manner, we will eventu-ally decompose the given function, that I called f (x), into real factors of first orof second degree. By equating these factors to zero, we will determine the real orimaginary roots of equation (21), which will be equal in number to the degree of thefunction. (See Algebraic Analysis, Chapter X.)2

2Cauchy’s Chapter X of his Cours d’analyse investigates real and imaginary roots of algebraicequations. It is here where Cauchy presents his Fundamental Theorem of Algebra.

NINETEENTH LECTURE.

USE OF DERIVATIVES AND DIFFERENTIALS OFVARIOUS ORDERS IN THE EXPANSION OFENTIRE FUNCTIONS.

It is easy to expand an entire function of x into a polynomial ordered accordingto the ascending powers of this variable, when we know the particular values ofthe function and of its successive derivatives, for x = 0. In fact, denote by F(x)

the given function, by n the degree of this function, and by a0, a1, a2, . . . , an theunknown coefficients of the various powers of x in the expansion we seek, so thatwe have

F(x) = a0 + a1x + a2x2 + · · · + an xn.(1)

By differentiating equation (1) n times in sequence, we will find⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

F ′(x) = 1 · a1 + 2a2x + · · · + nan xn−1,

F ′′(x) = 1 · 2 · a2 + · · · + (n − 1)nan xn−2,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

F (n)(x) = 1 · 2 · 3 · · · nan .

(2)

If we set, in these various formulas, x = 0, we will derive⎧⎪⎪⎨

⎪⎪⎩

a0 = F(0), a1 = 11 F ′(0),

a2 = 11·2 F ′′(0), . . . . . . . . . ,

an = 11·2·3···n F (n)(0),

(3)

and equation (1) will produce1

F(x) = F(0) + x

1F ′(0) + x2

1 · 2 F ′′(0) + · · · + xn

1 · 2 · 3 · · · nF (n)(0).(4)

1What has developed here so simply is a special case of a Maclaurin series, one in which thefunction is exactly represented by a finite polynomial.


97


https://doi.org/10.1007/978-3-030-11036-9_19


Example. – Let F(x) = (1 + x)n; we will obtain the known formula2

(1 + x)n = 1 + n

1x + n(n − 1)

1 · 2 x2 + n(n − 1)(n − 2)

1 · 2 · 3 x3 + · · · +n

1xn−1 + xn .(5)

Now, let u = f (x, y, z, . . . ) be an entire function of the variables x, y, z, . . . , andn the degree of this function, that is to say, the greatest sum that we can obtain byadding the exponents of the various variables taken in the same term. If we set

F(α) = f (x + α dx, y + α dy, z + α dz, . . . ),

F(α) will be an entire function of α of degree n, and we will have, by consequence,

F(α) = F(0) + α

1F ′(0) + α2

1 · 2 F ′′(0) + α3

1 · 2 · 3 F ′′′(0)

+ · · · + αn

1 · 2 · 3 · · · nF (n)(0).

This last formula, by virtue of the principles established in the fourteenth lecture,can be written as follows

{f (x +α dx, y + α dy, z + α dz, . . . )

= u + α1 du + α2

1 · 2d2u + α3

1 ·2 · 3d3u + · · · + αn

1 · 2 · 3 ··· n dnu.(6)

Additionally, this will remain valid for any values of α, either finite or infinitelysmall. If, for the simplest, we take α = 1, we will find

{f (x +dx, y + dy, z + dz, . . . )

= u + 11du + 1

1 · 2d2u + 11 · 2 · 3d3u + · · · + 1

1 · 2 · 3 ··· n dnu.(7)

In the particular case where the variables x, y, z, . . . are reduced to a single one, wehave

u = f (x),

du = f ′(x) dx,

d2u = f ′′(x) dx2,

. . . . . . . . . . . . . . . ,

dnu = f (n)(x) dxn,

2The known formula Cauchy is referring to is the binomial expansion for the case n a positiveinteger.

NINETEENTH LECTURE. 99

and we derive from formula (7) in replacing dx by h,

{f (x + h) = f (x) + h

1 f ′(x) + h2

1 · 2 f ′′(x)

+ h3

1 ·2 · 3 f ′′′(x) + · · · + hn

1 ·2 · 3 ··· n f (n)(x).(8)

Moreover, we could have directly deduced this last equation from formula (4).

Example. – If we suppose f (x) = xn, we will find{

(x + h)n = xn + n1 xn−1h + n(n−1)

1 · 2 xn−2h2 + · · ·+ n(n−1)

1 · 2 x2hn−2 + n1 xhn−1 + hn.

(9)

Note.3 – If f (x) is divisible by (x − a)m, or in other words, if we have

f (x) = (x − a)mϕ(x),(10)

ϕ(x) denoting an entire function of the variable x, the expansion of f (a + h), ac-cording to ascending powers of h, will obviously become divisible by hm . More-over, this expansion will be, by virtue of the preceding,

f (a) + h

1f ′(a) + h2

1 · 2 f ′′(a) + · · · + hm

1 · 2 · 3 · · · mf (m)(a) + · · · .

So, equation (10) being used, we shall deduce, not only f (a) = 0, but also f ′(a) =0, f ′′(a) = 0, . . . , f (m−1)(a) = 0. We would arrive at the same result by differen-tiating equation (10) several times in sequence, from which we would successivelyderive, with the help of formula (15) (fourteenth lecture),

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

f ′(x) = (x − a)mϕ′(x) + m(x − a)m−1ϕ(x),

f ′′(x) = (x − a)mϕ′′(x) + 2m(x − a)m−1ϕ′(x)

+m(m − 1)(x − a)m−2ϕ(x),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

f (m−1)(x) = (x − a)mϕ(m−1)(x) + · · · + m(m − 1) · · · 3 · 2 · (x − a)ϕ(x).

(11)

3This note turns out to be a fairly long aside. However, this tangent is a precursor to important workto follow. To provide motivation, recall Cauchy is writing this text in the early 1820s, a time periodin which many mathematicians were still following Lagrange’s earlier work on the calculus. TheTaylor series is central to Lagrange, as the derivative of the function f (x) in Lagrange’s calculus isdefined as the coefficient of the linear term in the Taylor series expansion of that function. Hence,the Taylor series was certainly a prominent topic for most of Cauchy’s contemporaries. Throughoutthis 1823 text, Cauchy fully develops his analysis of the Taylor and Maclaurin series, ending thebook with examples clearly demonstrating the failure of Lagrange’s theory. This current Note ispart of this pivotal complete development.


Thus, f (x) being an entire function of x, we can say that, if the equation

f (x) = 0(12)

admits m equal roots represented by a, each of the derivative equations⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

f ′(x) = 0,

f ′′(x) = 0,

f ′′′(x) = 0,

. . . . . . . . . ,

f (m−1)(x) = 0,

(13)

will be found verified by the assumption x = a. We should also remark that, f (x)being divisible by (x − a)m, f ′(x) will be divisible by (x − a)m−1, f ′′(x) by (x −a)m−2, . . . , and f (m−1)(x) only by x − a. As for the function f (m)(x), since it willbe determined by the equation

⎧⎪⎨

⎪⎩

f (m)(x) = (x − a)mϕ(m)(x) + m

1m(x − a)m−1ϕ(m−1)(x) + · · ·

+m

1m(m − 1) · · · 3 · 2 · (x − a)ϕ′(x) + m(m − 1) · · · 3 · 2 · 1 · ϕ(x),

(14)

it will reduce, for x = a, to

f (m)(a) = 1 · 2 · 3 · · · (m − 1)m · ϕ(a).(15)

All these remarks would remain valid even in the case where the value of f (x),

being given by equation (10), ϕ(x) would cease to be an entire function of the vari-able x . We know, moreover, the result that we can draw from these remarks for thedetermination of equal roots of algebraic equations.

Consider at present that y = F(x) and z = Υ (x) denote two entire functions ofx, divisible, one and the other, by (x − a)m . If the number m surpasses unity, thevalues of the fractions z

y and dzdy = z′

y′ , for x = a, are presented at the same timein an indeterminate form, and by consequence, we can no longer use the secondfraction to calculate the value of the first, as we have explained in the sixth lecture.4

However, the actual value of the fraction zy will not cease to be the limit toward

which the ratio ΔzΔy converges, while the differences Δy,Δz converge toward zero.

Moreover, by attributing to x the infinitely small increment Δx = α dx, we will

4In Lecture Six, Cauchy develops a technique to deal with this indeterminate situation in somecases. Within that lecture he shows in the limit, the ratio z

y approaches the same value as the ratioΔzΔy , or dz

dy , which has also been shown to equal z′y′ at the point of interest. However, in the present

case, withm > 1 as Cauchy has supposed, z′ and y′ are both still zero at x = a, and so, the techniquefails as this second fraction is still an indeterminate form of 0

0 .

NINETEENTH LECTURE. 101

derive from formula (6)5

Δy = α

1dy + α2

1 · 2 d2y + · · · + αm−1

1 · 2 · 3 · · · (m − 1)dm−1y

+ αm

1 · 2 · 3 · · · mdm y + αm+1

1 · 2 · 3 · · · (m + 1)dm+1y + · · · ,

Δz = α

1dz + α2

1 · 2 d2z + · · · + αm−1

1 · 2 · 3 · · · (m − 1)dm−1z

+ αm

1 · 2 · 3 · · · mdm z + αm+1

1 · 2 · 3 · · · (m + 1)dm+1z + · · · .

If we now assign to x the particular value a, since this value must cause the derivedfunctions y′, y′′, . . . , y(m−1), z′, z′′, . . . , z(m−1) to vanish, and by consequence, thedifferentials dy, d2y, . . . , d(m−1)y, dz, d2z, . . . , d(m−1)z, we will simply have

Δy = αm

1 · 2 · 3 · · · mdm y + αm+1

1 · 2 · 3 · · · (m + 1)dm+1y + · · ·

= αm

1 · 2 · 3 · · · m

(

dm y + α

m + 1dm+1y + · · ·

)

,

Δz = αm

1 · 2 · 3 · · · mdm z + αm+1

1 · 2 · 3 · · · (m + 1)dm+1z + · · ·

= αm

1 · 2 · 3 · · · m

(

dm z + α

m + 1dm+1z + · · ·

)

.

We will deduceΔz

Δy= dm z + α

m+1 dm+1z + · · ·dm y + α

m+1 dm+1y + · · · ;

then, by letting α converge toward the limit zero,

limΔz

Δy= dm z

dm y= z(m)

y(m).

Therefore, the value that the given fraction zy or Υ (x)

F(x)will receive, for x = a, will be

precisely equal to the corresponding value of the fraction

dm z

dm yor

Υ (m)(x)

F (m)(x).

5The expressions here have been corrected. In both of Cauchy’s texts, y and z terms occur at thebeginning of the right-hand side in each of these two equations, respectively, but they should notbe present.


Example. – ϕ(x) designating an entire function not divisible by x − a, and F(x)

another entire function divisible by (x − a)m, we will have, for x = a,

(x − a)mϕ(x)

F(x)= 1 · 2 · 3 · · · m ϕ(x) + 2 · 3 · · · m m (x − a)ϕ′(x) + · · ·

F (m)(x)(16)

= 1 · 2 · 3 · m ϕ(a)

F (m)(a).

TWENTIETH LECTURE.

DECOMPOSITION OF RATIONAL FRACTIONS.

Represent by Υ (x) and F(x) two entire functions of x, the first of degree m,the second of degree n.1 Υ (x)

F(x)will be what we call a rational fraction. Moreover, the

equation

F(x) = 0(1)

will admit n real or imaginary roots, equal or unequal; and if, by first supposingthem all unequal, we designate them by x0, x1, x2, . . . , xn−1, we will necessarilyhave2

F(x) = k(x − x0)(x − x1)(x − x2) · · · (x − xn−1),(2)

k being the coefficient of xn in F(x). This granted, let

ϕ(x) = k(x − x1)(x − x2) · · · (x − xn−1) andΥ (x0)

ϕ(x0)= A0.(3)

Equation (2) will take the form

F(x) = (x − x0)ϕ(x);(4)

and, since the difference

Υ (x)

ϕ(x)− A0 = Υ (x) − A0 ϕ(x)

ϕ(x)

will vanish for x = x0, it will be the same for the polynomial

Υ (x) − A0 ϕ(x).

1Cauchy is assuming m < n here.2The Fundamental Theorem of Algebra. Cauchy has developed this in his earlier 1821 Coursd’analyse Chapter X, §I, Theorem II.


103


https://doi.org/10.1007/978-3-030-11036-9_20


Therefore, this polynomial will be algebraically divisible by x − x0; so that we willhave

Υ (x) − A0 ϕ(x) = (x − x0)χ(x)

or

Υ (x) = A0 ϕ(x) + (x − x0)χ(x),(5)

χ(x) representing a new entire function of the variable x . If we now divide the twomembers of equation (5) by F(x), in having regard to formula (4), we will find

⎧⎪⎪⎨

⎪⎪⎩

Υ (x)

F(x)= A0

x − x0+ χ(x)

ϕ(x)

= A0

x − x0+ χ(x)

k(x − x1)(x − x2) · · · (x − xn−1).

(6)

We can therefore extract from the rational fraction Υ (x)

F(x)a simple fraction of the

form A0x−x0

, A0 designating a constant, in such a manner as to obtain for a remainderanother rational fraction whose denominator is that which becomes the polynomialF(x) when we omit in this polynomial the linear factor x − x0. Let us conceive that,as a result of similar operations, we successively extract from Υ (x)

F(x), then from χ(x)

ϕ(x),

. . . , a sequence of simple fractions of the form

A0

x − x0,

A1

x − x1,

A2

x − x2, . . . ,

An−1

x − xn−1,

in a manner so as to eliminate, one after the other, in the denominator of the re-maining fraction, all the linear factors of the polynomial F(x). The last of all theremainders will be a rational fraction whose denominator will be found reduced tothe constant k, that is to say, an entire function of the variable x . In denoting by Qthis entire function, we will have

Υ (x)

F(x)= Q + A0

x − x0+ A1

x − x1+ A2

x − x2+ · · · + An−1

x − xn−1.(7)

Since this latter formula leads to the following

{Υ (x) = QF(x) +A0

F(x)

x−x0+ A1

F(x)

x−x1

+A2F(x)

x−x2+ · · · + An−1

F(x)

x−xn−1,

(8)

in which all the terms which follow the product QF(x) are entire functions of xof a degree less than n,3 it is clear that the letter Q represents the quotient of thealgebraic division of Υ (x) by F(x). In addition, since all of these terms and that of

3Recall F(x) is degree n.

TWENTIETH LECTURE. 105

the product QF(x), with the exception of the first term, will be divisible by x − x0,we will obviously have, for x = x0,4

Υ (x) = A0F(x)

x − x0= A0

d F(x)

dx= A0 F ′(x).(9)

Therefore, to obtain the value of A0, it will suffice to set x = x0 in the fractionΥ (x)

F ′(x).

By similarly generating the values of A1, A2, . . . , we will find

A0 = Υ (x0)

F ′(x0), A1 = Υ (x1)

F ′(x1), A2 = Υ (x2)

F ′(x2), . . . , An−1 = Υ (xn−1)

F ′(xn−1).(10)

On inspection of these values, we recognize they are independent of the mode ofdecomposition adopted.5 Add that the value of A0, deduced from formula (9), canbe equivalently presented in one or the other of the two forms, Υ (x0)

F ′(x0)and Υ (x0)

ϕ(x0);

where it results that the first of the formulas in (10) is in accordance with the secondof the equations in (3).

When the two roots, x0, x1, are imaginary and conjugates, or of the form α +β√−1, α − β

√−1, then, in designating by A and B two real quantities that workto satisfy the equation

A − B√−1 = Υ (α + β

√−1 )

F ′(α + β√−1 )

,(11)

we find that the simple fractions corresponding to these roots are, respectively,

A − B√−1

x − α − β√−1

,A + B

√−1

x − α + β√−1

.(12)

By adding these two fractions, we obtain the following

2A(x − α) + 2Bβ

(x − α)2 + β2,(13)

which has for a numerator a real and linear function of x, and for a denominator asecond degree factor of the polynomial F(x).6

Examples. – Decompose the fractions 1x2−1 ,

xx2−1 ,

xm

xn±1 ,xn−1

xn±1 , . . . .

4Cauchy must mean a limit here, where he again appears to be using the l’Hôpital’s Rule typemaneuver he has employed earlier. Taking a limit as x tends to x0 in his equation (8) yields Υ (x0) =limx→x0

(A0

F(x)x−x0

), as all the other terms vanish. He is arguing the limit of A0

F(x)x−x0

is A0F ′(x).

5This technique for finding the coefficients in a partial fraction decomposition is inventive andvery cool. Cauchy’s method is far more interesting than the standard process of solving a systemof equations.6The modern irreducible quadratic factor.


We move on to the case where equation (1) has equal roots. Then, if we denoteby a, b, c, . . . the various roots, by p, q, r, . . . integer numbers, and by k a constantcoefficient, the polynomial F(x) will be of the form

F(x) = k(x − a)p(x − b)q(x − c)r · · · .(14)

If, in this new hypothesis, we let for brevity,

ϕ(x) = k(x − b)q(x − c)r · · · andΥ (a)

ϕ(a)= A,(15)

equation (14) will become

F(x) = (x − a)pϕ(x);(16)

and, since the two differences Υ (x)

ϕ(x)− A, Υ (x)− Aϕ(x)will vanish for x = a,wewill

necessarily have

Υ (x) = Aϕ(x) + (x − a)χ(x),(17)

χ(x) designating a new entire function of the variable x . This granted, we willderive from equations (14), (16), and (17)

⎧⎪⎪⎨

⎪⎪⎩

Υ (x)

F(x)= A

(x − a)p+ χ(x)

(x − a)p−1ϕ(x)

= A

(x − a)p+ χ(x)

k(x − a)p−1(x − b)q(x − c)r · · · .(18)

Thus, by extracting from the rational fraction Υ (x)

F(x)a simple fractionof the form A

(x−a)p ,

we obtain for a remainder another rational fraction whose denominator isthat which becomes the polynomial F(x) when we omit in this polynomial one ofthe factors equal to x − a. Let us conceive that with the help of several similardecompositions we successively remove from the denominator of the remainingfraction: 1◦ all the factors equal to x − a; 2◦ all the factors equal to x − b; 3◦ all thefactors equal to x − c; etc. The last of all the remainders will be a rational fractionof constant denominator, that is to say, an entire function of the variable x . So that,in denoting by Q this entire function, and by A, A1, A2, . . . , Ap−1, B, B1, B2, . . . ,

Bq−1, C, C1, C2, . . . , Cr−1, . . . the constant numerators of the various simple frac-tions, we will have

⎧⎪⎪⎨

⎪⎪⎩

Υ (x)

F(x)= Q + A

(x − a)p+ A1

(x − a)p−1+ · · ·

+ Ap−1

x − a+ B

(x − b)q+ · · · + C

(x − c)r+ · · · .

(19)

TWENTIETH LECTURE. 107

To prove: 1◦ that the polynomial Q is the quotient of the algebraic division ofΥ (x) by F(x); 2◦ that the values of the constants A, A1, . . . , Ap−1, B, . . . areindependent of the mode of decomposition adopted, it will suffice to observe thatformula (19) leads to the following

⎧⎪⎪⎨

⎪⎪⎩

Υ (x) = QF(x) +AF(x)

(x − a)p+ A1

F(x)

(x − a)p−1+ · · ·

+Ap−1F(x)

x − a+ B

F(x)

(x − b)q+ · · · ,

(20)

in which all the terms which follow the product QF(x) are entire functions of xof a degree less than that of the function F(x); and moreover, that, if in formula(20) we set x = a + z, the comparison of the constant terms and of the coefficientswhich will affect the similar powers of z in the two expanded members, accordingto the ascending powers of this variable (see the nineteenth lecture), will generatethe equations

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

Υ (a) = AF (p)(a)

1 · 2 · 3 · · · p,

Υ ′(a) = AF (p+1)(a)

1 · 2 · 3 · · · (p + 1)+ A1

F (p)(a)

1 · 2 · 3 · · · p,

Υ ′′(a) = · · · ,

(21)

from which we will deduce for the constants A, A1, A2, . . . , a single system of val-ues, namely,

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

A = 1 · 2 · 3 · · · p Υ (a)

F (p)(a),

A1 = 1 · 2 · 3 · · · (p + 1) Υ ′(a) − AF (p+1)(a)

(p + 1)F (p)(a),

(22)

etc. We would obtain by the same manner, the values of B, B1, B2, . . . , C, C1,

C2, . . . . It is essential to observe that the first of the formulas in (22) gives for theconstant A a value equal to that which the fraction

(x − a)pΥ (x)

F(x)= Υ (x)

ϕ(x)

receives, when we assume x = a, and by consequence, is equal to Υ (a)

ϕ(a). (See, for

more details, Algebraic Analysis, Chapter XI.)7

7Chapter XI of Cauchy’s Cours d’analyse devotes itself to the decomposition of rational fractions.

Part IIINTEGRAL CALCULUS.

TWENTY-FIRST LECTURE.

DEFINITE INTEGRALS.

Suppose that, the function y = f (x), being continuous1 with respect to the variable xbetween two finite limits x = x0, x = X, we denote by x1, x2, . . . , xn−1 new valuesof x interposed between these limits, which always go on increasing or decreasingfrom the first limit up to the second.2 We can use these values to divide the differenceX − x0 into elements

x1 − x0, x2 − x1, x3 − x2, . . . , X − xn−1,(1)

which will all be of the same sign. This granted, consider that we multiply each ele-ment by the value of f (x) corresponding to the origin of this same element, namely,the element x1−x0 by f (x0),3 the element x2−x1 by f (x1), . . . , finally, the elementX − xn−1 by f (xn−1); and, let

S = (x1 − x0) f (x0) + (x2 − x1) f (x1) + · · · + (X − xn−1) f (xn−1)(2)

be the sum of the products thus obtained. The quantity S will obviously dependupon: 1◦ the number of elements n into which we will have divided the differenceX − x0; 2◦ the values of these same elements, and by consequence, on the modeof division4 adopted. Now, it is important to remark that, if the numerical valuesof the elements become very small and the number n very considerable, the modeof division will no longer have a perceptible influence on the value of S. It is this,actually, that we can demonstrate as follows.

1Within this lecture, Cauchy will define his definite integral in a manner very similar to the moremodern definition of Georg Friedrich Bernhard Riemann (1826–1866) later in the 19th century.Riemann will build upon Cauchy’s early work on the definite integral and will remove the restric-tion Cauchy imposes here for f (x) to be continuous.2Notice Cauchy specifically does not state f (x) is the derivative of some other function. He makesa point of distancing himself from differentiation in this entire lecture to show the integral is inde-pendent of the derivative.3The 1899 edition has f (x).4Today’s partition, i.e. {x0, x1, x2, . . . , xn−1, X}.© Springer Nature Switzerland AG 2019D. M. Cates, Cauchy’s Calcul Infinitésimal,https://doi.org/10.1007/978-3-030-11036-9_21

111


https://doi.org/10.1007/978-3-030-11036-9_21


If we were to suppose all the elements of the difference X − x0 reduce to a singleone, which would be this difference itself, we would simply have

S = (X − x0) f (x0).(3)

When, on the other hand, we take the expressions in (1) for elements of the differ-ence X − x0, the value of S, determined in this case by equation (2), is equal to thesum of the elements multiplied by an average5 between the coefficients

f (x0), f (x1), f (x2), . . . , f (xn−1)

(see, in the Preliminaries of Analysis Course, the Corollary of Theorem III

)6 In

addition, these coefficients being particular values of the expression

f[x0 + θ(X − x0)

]

which correspond to the values of θ contained between zero and unity, we willprove, by arguments similar to those that we have made use of in the seventh lecture,that the average in question is another value of the same expression, correspondingto a value of θ contained between the same limits. We can, therefore, in equation(2) substitute the following7

S = (X − x0) f[x0 + θ(X − x0)

],(4)

in which θ will be a number less than unity.To pass from the mode of division that we have just considered, to another in

which the numerical values of the elements of X − x0 are even smaller, it will sufficeto partition each of the expressions in (1) into new elements.8 Then, we shouldreplace, in the second member of equation (2), the product (x1 − x0) f (x0) by a sumof similar products, for which we can substitute an expression of the form

(x1 − x0) f[x0 + θ0(x1 − x0)

],

θ0 being a number less than unity. We expect there will be a similar relationshipbetween this sum and the product (x1 − x0) f (x0), such as that which exists between

5Recall Cauchy’s definition of an average from his Cours d’analyse Note II, “We call an averageamong several given quantities a new quantity contained between the smallest and the largest ofthose that we consider.” Cauchy’s definition allows for an infinite number of averages for the sameset of values.6Cauchy duplicates this theorem with a complete proof at the end of his Cours d’analyse in NoteII as Theorem XII. Theorem XII is one of his interesting average theorems and has been includedin its entirety in this book as part of Appendix B. He is using the Corollary III result of this theoremwith αi = �xi , ai = f (xi ), and bi = 1 here.7To justify this result, Cauchy needs to use the Intermediate Value Theorem. It is for this reasonhe requires the continuity of f (x).8Cauchy takes care in noting that the partition is becoming finer and finer in the sense that everysubinterval becomes successively smaller, not just that more subintervals are being created.

TWENTY-FIRST LECTURE. 113

the values of S supplied by equations (4) and (3). By the same reasoning, we shouldsubstitute for the product (x2 − x1) f (x1), a sum of terms which can be presentedunder the form

(x2 − x1) f[x1 + θ1(x2 − x1)

],

θ1 again denoting a number less than unity. By continuing in this manner, we willfinally deduce that, in the new mode of division, the value of S will be of the form

⎧⎪⎨

⎪⎩

S = (x1 − x0) f[x0 + θ0(x1 − x0)

]

+ (x2 − x1) f[x1 + θ1(x2 − x1)

] + · · ·+ (X − xn−1) f

[xn−1 + θn−1(X − xn−1)

].

(5)

If, in this last equation, we let

f[x0 + θ0(x1 − x0)

] = f (x0) ± ε0,

f[x1 + θ1(x2 − x1)

] = f (x1) ± ε1,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

f[xn−1 + θn−1(X − xn−1)

] = f (xn−1) ± εn−1,

we will derive{

S = (x1 − x0)[

f (x0) ± ε0] + (x2 − x1)

[f (x1) ± ε1

] + · · ·+ (X − xn−1)

[f (xn−1) ± εn−1

];(6)

then, by expanding the products,{

S = (x1 − x0) f (x0) + (x2 − x1) f (x1) + · · · + (X − xn−1) f (xn−1)

± ε0(x1 − x0) ± ε1(x2 − x1) ± · · · ± εn−1(X − xn−1).(7)

Add that, if the elements x1− x0, x2− x1, . . . , X − xn−1 have very small numericalvalues, eachof the quantities ±ε0, ±ε1, . . . , ±εn−1 will differ very little fromzero;9

and as a result, it will be the same for the sum

± ε0(x1 − x0) ± ε1(x2 − x1) ± · · · ± εn−1(X − xn−1),

which is equivalent to the product of X − x0 by an average between these variousquantities. This granted, it follows from equations (2) and (7), when compared to

9It is here where Cauchy makes what some regard as a slight mistake, or at least is not clear in hisdescription. He claims if the elements x1 − x0, x2 − x1, . . . , X − xn−1 are all sufficiently small,then each of the ε j ’s are very close to zero. In the limit of increasing n, this argument technicallyrequires f (x) to be uniformly continuous, and not just continuous (in the modern sense), as there isno guarantee all of the ε j ’s can be made small for a given n. Cauchy’s own view of continuity fromLecture Two seems to fall somewhere between the two modern versions of simple and uniformcontinuity, and so, this aspect of his argument is not entirely clear. Some argue his claims are fine,and indeed are consistent with his earlier definitions; others argue his claims are not justified. Thedebate goes on.


each other, that we will not significantly alter the calculated value of S for a mode ofdivision in which the elements of the difference X − x0 have very small numericalvalues, if we pass to a second mode in which each of these elements are foundsubdivided into several others.

Let us now conceive that we consider both modes of division of the differenceX − x0, in each of which the elements of this difference have very small numericalvalues. We can compare these two modes to a third, chosen so that each element,whether from the first or from the second mode, is found formed by the union ofthe various elements from the third. For this condition to be fulfilled, it will sufficethat all the values of x interposed in the first two modes between the limits x0, X areemployed in the third, and we will prove that we alter the value of S very little bypassing from the first or from the second mode to the third, and by consequence, inpassing from the first to the second. Therefore, when the elements of the differenceX − x0 become infinitely small, the mode of division will no longer have a percep-tible influence on the value of S;10 and, if we decrease indefinitely the numericalvalues of these elements, by increasing their number, the value of S will eventuallybe substantially constant, or in other words, it will finally attain a certain limit whichwill depend uniquely on the form of the function f (x), and the extreme values x0,X attributed to the variable x . This limit11 is what we call a definite integral.12

10It is unlikely that at the time Cauchy wrote this lecture he had worked out all the details ofthis analysis, especially given his possible misconceptions surrounding the concepts of uniformcontinuity and completeness, but one can tell he had the right idea in mind about the proof thatwould eventually develop over time. We know today the type of sequence he is constructing hereis called a Cauchy sequence (informally, we can say sequences which have the property in whichits terms become arbitrarily close together if one looks out far enough in the sequence are Cauchysequences) and that this type of sequence always converges to some value, which is precisely whatCauchy has claimed here.11Cauchy makes it very clear he is not defining the definite integral to be a sum. Instead, he isdefining the definite integral to be the limit of a sequence of sums.12Most of the advancements in calculus throughout the 18th century by pioneers such as JakobBernoulli (1654–1705), Johann Bernoulli (Jakob’s younger brother and lifelong rival), the prolificLeonhard Euler, and many, many others, were mainly limited to differential calculus due to themistaken belief integration was merely the inverse operation of differentiation, and so, did not war-rant any special consideration. By the year 1800, integration had certainly settled into a secondaryposition of relevance behind differentiation. However, in the early 1800s, Joseph Fourier (1768–1830) changed the integral landscape forever. His representation of an arbitrary function requiredthe calculation of coefficients defined as definite integrals,

an = 1

π

∫ π

−π

f (x) cos (nx) dx and bn = 1

π

∫ π

−π

f (x) sin (nx) dx,

to construct his now famous Fourier series,

f (x) = a02

+∞∑

n=1

(an cos (nx) + bn sin (nx)

).

An integral theory independent of differential calculus was clearly needed. Cauchy is one of thefirst mathematicians to take on this challenge and seriously consider the question of the existenceof the definite integral. Here, he has demonstrated the definite integral does exist without everinvoking the concept of the derivative. Cauchy has shown the definite integral is important on itsown, and his proof in this lecture is one of the highlights of his 1823 text.

TWENTY-FIRST LECTURE. 115

Now observe that, if we denote by �x = h = dx a finite increment attributed tothe variable x, the different terms which compose the value S, such that the products

(x1 − x0) f (x0), (x2 − x1) f (x1), . . .

will all be included in the general formula

h f (x) = f (x) dx,(8)

fromwhichwewill deduce, one after the other, byfirst setting x = x0 and h = x1−x0,then x = x1 and h = x2−x1, etc.We can, therefore, state that the quantity S is a sumof products similar to expression (8), which we sometimes express with the help ofthe character

∑, by writing

S =∑

h f (x) =∑

f (x)�x .(9)

As to the definite integral that converges toward the quantity S, while the elementsof the difference X − x0 become infinitely small, we agree to represent this by thenotation

∫h f (x) or

∫f (x) dx, in which the letter

∫, substituted for the letter

∑, no

longer indicates a sum of products similar to expression (8), but the limit of a sum ofthis type.13 In addition, as the value of the definite integral that we consider dependson the extreme values x0, X attributed to the variable x, we agree to place these twovalues, the first below, the second above the letter

∫, or to write to the side of the

integral, that we designate as a consequence, by one of the notations

∫ X

x0

f (x) dx,

∫f (x) dx

[x0X

],

∫f (x) dx

[x = x0x = X

].(10)

The first of these notations, devised by Mr. Fourier, is the simplest. In the particularcase where the function f (x) is replaced by a constant quantity, a, we find, whateverthe mode of division of the difference X − x0,

S = a(X − x0),

and we conclude∫ X

x0

a dx = a(X − x0).(11)

If, in this last formula, we set a = 1, we will derive∫ X

x0

dx = X − x0.(12)

13Leibniz, in the late 1600s, had defined the integral as the sum of infinitely many infinitesimalsummands (the sum itself). This is why he chose the symbol of a large S to denote his sum. Hislarge S gradually evolved into the

∫symbol for integration we use today.

TWENTY-SECOND LECTURE.

FORMULAS FOR THE DETERMINATION OFEXACT OR APPROXIMATE VALUES OFDEFINITE INTEGRALS.

After what has been said in the last lecture, if we divide X − x0 into infinitely smallelements x1 − x0, x2 − x1, . . . , X − xn−1, the sum1

S = (x1 − x0) f (x0) + (x2 − x1) f (x1) + · · · + (X − xn−1) f (xn−1)(1)

will converge toward a limit represented by the definite integral

∫ X

x0

f (x) dx .(2)

From principles upon which we have founded this proposition, it follows that wewould again reach the same limit, if the value of S, instead of being determined byequation (1), were deduced from formulas similar to equations (5) and (6) (twenty-first lecture), that is to say, if we would assume

{S = (x1 − x0) f

[x0 + θ0(x1 − x0)

] + (x2 − x1) f[x1 + θ1(x2 − x1)

] + · · ·+ (X − xn−1) f

[xn−1 + θn−1(X − xn−1)

],

(3)

θ0, θ1, . . . , θn−1 designating some numbers less than unity, or just as well,{

S = (x1 − x0)[

f (x0) ± ε0] + (x2 − x1)

[f (x1) ± ε1

] + · · ·+ (X − xn−1)

[f (xn−1) ± εn−1

],

(4)

ε0, ε1, . . . , εn−1 designating numbers subject to vanish along with the elements ofthe difference X − x0. The first of the two preceding formulas is reduced to equation(1) when we take

θ0 = θ1 = · · · = θn−1 = 0.

1Cauchy’s 1823 and 1899 editions both read,

(1) S = (x1 − x0) f (x0) + (x1 − x2) f (x1) + · · · + (X − xn−1) f (xn−1),

a clear misprint which has been corrected here.


117


https://doi.org/10.1007/978-3-030-11036-9_22


If we make, on the other hand,2

θ0 = θ1 = · · · = θn−1 = 1,

we will find

S = (x1 − x0) f (x1) + (x2 − x1) f (x2) + · · · + (X − xn−1) f (X).(5)

When, in this last formula, we exchange between them the two quantities x0, X insuch a way that all the terms are placed at equal distances from the two extremes inthe sequence x0, x1, . . . , xn−1, X,we obtain a new value for S equal, but of oppositesign, to that generated by equation (1). The limit toward which this new value of Smust converge will therefore be equal, but of opposite sign, to the integral (2), fromwhich we will deduce by the mutual exchange of the two quantities x0, X. We willgenerally have, therefore,

∫ x0

Xf (x) dx = −

∫ X

x0

f (x) dx .(6)

We frequently employ formulas (1) and (5) in the study of the approximatevalues of definite integrals. For simplicity, we usually assume that the quantitiesx0, x1, . . . , xn−1, X included in these formulas are in an arithmetic progression.Then, the elements of the difference X − x0 all become equal to the fraction X−x0

n ;and, in denoting this fraction by i, we find that equations (1) and (5) are reduced tothe following two

S = i[

f (x0) + f (x0 + i) + f (x0 + 2i) + · · · + f (X − 2i) + f (X − i)],(7)

S = i[

f (x0 + i) + f (x0 + 2i) + · · · + f (X − 2i) + f (X − i) + f (X)].(8)

We could also suppose that the quantities x0, x1, . . . , xn−1, X form a geometric pro-gression whose ratio differs very little from unity. By adopting this hypothesis, andletting (

X

x0

) 1n

= 1 + α,

2Cauchy is clearly allowing θ = 0 and θ = 1 in this instance—describing the closed interval [0, 1].His wording usually leaves us in doubt as to whether he is describing an open or closed interval,as he generally uses imprecise language throughout the text to convey his intent. Phrases such as“contained between the limits …,” “less than …,” “included between the limits …,” or simply“between the limits …,” are used in nearly every part of Calcul infinitésimal—including someof his important theorems. As an example of his inexplicit language, in Lecture Four Cauchywrites, “…contained between the limits −π/2,+π/2,” to seemingly describe the domain of boththe sine function

(whose domain is actually

[ − π/2,+π/2])

and the tangent function(with an

actual domain of (−π/2,+π/2))within the same sentence!

TWENTY-SECOND LECTURE. 119

we will derive two new values of S from formulas (1) and (5), the first of which willbe

S = α

{x0 f (x0) + x0(1 + α) f

[x0(1 + α)

] + · · · + X

1 + αf

(X

1 + α

)}.(9)

It is essential to observe that, in several cases, we can deduce from equations (7)and (9) not only the approximate values of the integral (2), but also its exact value,or lim S. We will find, for example,3

∫ X

x0

x dx = lim(X − x0)(X + x0 − i)

2= lim

X2 − x20

2 + α= X2 − x2

0

2,(10)

∫ X

x0

Ax dx = limi(AX − Ax0)

Ai − 1= AX − Ax0

lA,

∫ X

x0

ex dx = eX − ex0 ,(11)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ X

x0

xa dx = limα(Xa+1 − xa+1

0 )

(1 + α)a+1 − 1= Xa+1 − xa+1

0

a + 1,

∫ X

x0

dx

x= lim nα = l

( X

x0

),

(12)

the last equation above being restricted to the case where the quantities x0, X havethe same sign. Add that, it is often easy to reduce the determination of a definiteintegral to that of another integral of the same type. In this way, for example, wewill derive from formula (1)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

x0

aϕ(x) dx = lim a[(x1 − x0)ϕ(x0) + · · · + (X − xn−1)ϕ(xn−1)

]

= a∫ X

x0

ϕ(x) dx,

(13)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

x0

f (x + a)dx = lim[(x1−x0) f (x0 + a)+ · · · + (X−xn−1) f (xn−1+a)

]

=∫ X+a

x0+af (x)dx,

(14)

∫ X

x0

f (x − a) dx=∫ X−a

x0−af (x) dx,

∫ X

x0

dx

x − a=

∫ X−a

x0−a

dx

x= l

( X − a

x0 − a

),(15)

3Cauchy often leaves very few clues as to his methods for solving given example problems inhis lectures. However, here we can use his intermediate results to determine whether he is usingequation (7) with an arithmetic progression or equation (9) with a geometric progression by his useof i for the former, or α for the latter.


the last equation above being restricted to the case where x0 − a and X − a arequantities affected by the same sign. Moreover, we will derive from formula (8),by setting x0 = 0, and replacing f (x) by f (X − x),

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

0f (X−x)dx = lim i

[f (X − i)+ f (X − 2i)+· · · + f (2i)+ f (i) + f (0)

]

=∫ X

0f (x)dx;

(16)

then, we will deduce, by having regard to equation (14),

∫ X−x0

0f (X − x) dx =

∫ X−x0

0f (x + x0) dx =

∫ X

x0

f (x) dx .(17)

Finally, if in formula (9) we set

f (x) = 1

x lxand l(1 + α) = β,

we will derive⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

x0

dx

x lx= lim β

(1

lx0+ 1

lx0 + β+ · · · + 1

lX − β

)eβ − 1

β

=∫ lX

lx0

dx

x= l

(lXlx0

),

(18)

the quantities x0, X above being positive, and both greater than or both less thanunity.

An important remark to make is that the forms under which the value of S ispresented in equations (4) and (5) of the preceding lecture apply equally to theintegral (2). In fact, these equations remaining valid, one and the other, while wesubdivide the difference X − x0 or the quantities x1− x0, x2− x1, . . . , X − xn−1 intoinfinitely small elements, will again be true in the limit, such that we will have4

∫ X

x0

f (x) dx = (X − x0) f[x0 + θ(X − x0)

](19)

and ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ X

x0

f (x) dx = (x1 − x0) f[x0 + θ0(x1 − x0)

]

+ (x2 − x1) f[x1 + θ1(x2 − x1)

] + · · ·+ (X − xn−1) f

[xn−1 + θn−1(X − xn−1)

],

(20)

4Cauchy has just proven the Mean Value Theorem for Definite Integrals. This is actually a specialcase of the General Mean Value Theorem for Definite Integrals which he will also prove in the nextlecture.

TWENTY-SECOND LECTURE. 121

θ, θ0, θ1, . . . , θn−1 designating unknown numbers, but all less than unity. If, forsimplicity, we assume the quantities x1−x0, x2−x1, . . . , X −xn−1 are equal to eachother, then by letting i = X−x0

n , we will find

∫ X

x0

f (x)dx = i[

f (x0+θ0i)+ f (x0 + i+θ1i) + · · · + f (X − i + θn−1i)].(21)

When the function f (x) is always increasing or always decreasing from x = x0 upto x = X, the second member of formula (21) obviously remains contained betweenthe two values of S generated by equations (7) and (8), values whose difference is±i

[f (X) − f (x0)

]. By consequence, in this hypothesis, by taking the half-sum of

these two values, or the expression

i[12 f (x0)+ f (x0 + i)+ f (x0 + 2i)+· · ·+ f (X − 2i)+ f (X − i)+ 1

2 f (X)],(22)

for the approximate value of the integral (21), we commit an error much smallerthan the half difference ±i

[12 f (X) − 1

2 f (x0)].

Example. – If we suppose

f (x) = 1

1 + x2, x0 = 0, X = 1, i = 1

4 ,

expression (22)will become 14

(12 + 16

17 + 45 + 16

25 + 14

) = 0.78 . . . . By consequence,

0.78 is the approximate value of the integral∫ 10

dx1+x2 . The error committed in this

case cannot exceed 14

(12 − 1

4

) = 116 . It will actually be below that of one-hundredth,5

as we will see later.

When the function f (x) is sometimes increasing and sometimes decreasing be-tween the limits x = x0, x = X, the error that we commit by taking one of the valuesof S generated by equations (7) and (8) for the approximate value of the integral (2)is obviously less than the product of ni = X − x0 by the largest numerical value thatcan be obtained by the difference

f (x + �x) − f (x) = �x f ′(x + θ�x),(23)

when we suppose x contained between the limits x0, X, and �x between the limits0, i. Therefore, if we call k the greatest numerical value that f ′(x)6 receives, whilex varies from x = x0 up to x = X, the error committed will certainly be includedbetween the limits

−ki(X − x0), +ki(X − x0).

5The 1899 reprint has 1100 .

6Both of Cauchy’s texts have f (x) here.

TWENTY-THIRD LECTURE.

DECOMPOSITION OF A DEFINITE INTEGRALINTO SEVERAL OTHERS. IMAGINARY DEFINITEINTEGRALS. GEOMETRIC REPRESENTATIONOF REAL DEFINITE INTEGRALS.DECOMPOSITION OF THE FUNCTION UNDERTHE

∫SIGN INTO TWO FACTORS IN WHICH

ONE ALWAYS MAINTAINS THE SAME SIGN.

To divide the definite integral

∫ X

x0

f (x) dx(1)

into several others of the same type, it suffices to decompose into several parts,either the function under the

∫sign or the difference X − x0. First, let us suppose

f (x) = ϕ(x) + χ(x) + ψ(x) + · · · .

We will infer

(x1 − x0) f (x0) + · · ·+(X − xn−1) f (xn−1)

= (x1 − x0)ϕ(x0) + · · · + (X − xn−1)ϕ(xn−1)

+ (x1 − x0)χ(x0) + · · · + (X − xn−1)χ(xn−1)

+ (x1 − x0)ψ(x0) + · · · + (X − xn−1)ψ(xn−1) + · · · ;


∫ X

x0

f (x) dx =∫ X

x0

ϕ(x) dx +∫ X

x0

χ(x) dx +∫ X

x0

ψ(x) dx + · · · .

From this last formula, combined with equation (13) (twenty-second lecture), wewill derive, in designating by u, v, w, . . . various functions of the variable x, and by

1If Cauchy is supposing a finite series, this step is fine. Otherwise, he needs to be more carefulwith this claim. Cauchy’s “passing to the limits” is assuming the limit of a sum is equivalent to thesum of the individual limits. This step actually requires an exchange of limits operation which isonly allowed under certain restrictions—the details of which were not worked out until later in the19th century.


123


https://doi.org/10.1007/978-3-030-11036-9_23


a, b, c, . . . constant quantities,∫ X

x0

(u + v + w + · · · ) dx =∫ X

x0

u dx +∫ X

x0

v dx +∫ X

x0

w dx + · · · ,(2)

∫ X

x0(u + v) dx =

∫ X

x0u dx +

∫ X

x0v dx,

∫ X

x0(u − v) dx =

∫ X

x0u dx −

∫ X

x0v dx,(3)

∫ X

x0(au + bv + cw + · · · ) dx = a

∫ X

x0u dx + b

∫ X

x0v dx + c

∫ X

x0w dx + · · · .(4)

When we extend the definition we have given for the integral (1) to the case wherethe function f (x) becomes imaginary, equation (4) remains valid for imaginary val-ues of the constants a, b, c, . . . . We have, as a result,

∫ X

x0

(u + v

√−1)

dx =∫ X

x0

u dx + √−1∫ X

x0

v dx .(5)

Supposenowthat,afterwehavedividedthedifferenceX − x0 intoafinitenumberofelements represented by x1 − x0, x2 − x1, . . . , X − xn−1, wepartition each of theseelements into several others whose numerical values are infinitely small, and thatwe modify, by consequence, the value of S supplied by equation (1) (twenty-secondlecture). The product (x1 − x0) f (x0) will be found replaced by a sum of similarproducts which will have for a limit the integral

∫ x1x0

f (x) dx . In the same way, theproducts (x2 − x1) f (x1), . . . , (X − xn−1) f (xn−1) willbereplacedbythesumswhichwillhaveforlimitsthedefiniteintegrals

∫ x2x1

f (x) dx, . . . ,∫ X

xn−1f (x) dx, respectively.

Moreover, by bringing together the different sums in question we will obtain as aresult, a total sum whose limit will be precisely the integral (1). Therefore, since thelimit of a sum of several quantities is always equivalent to the sum of their limits,we will have generally∫ X

x0

f (x) dx =∫ x1

x0

f (x) dx +∫ x2

x1

f (x) dx + · · · +∫ X

xn

f (x) dx .(6)

It is essential to recall that we must here attribute to the integer number n a finitevalue. When between the limits x0, X, we interpose a single value of x representedby ξ, equation (6) is reduced to∫ X

x0

f (x) dx =∫ ξ

x0

f (x) dx +∫ X

ξ

f (x) dx .(7)

It is easy to prove that equations (6) and (7) remain valid even in the case wheresome of the quantities x1, x2, . . . , xn−1, ξ would cease to be contained betweenthe limits x0, X, and in those where the differences x1−x0, x2−x1, . . . , X −xn−1,

ξ − x0, X − ξ would no longer be quantities of the same sign. Allow, for example,

TWENTY-THIRD LECTURE. 125

that the differences ξ −x0, X −ξ could be of opposite signs. Then, in addition to thatwe will suppose x0 is contained between ξ and X, or rather X is contained betweenx0 and ξ, and we will find

∫ X

ξ

f (x) dx =∫ x0

ξ

f (x) dx +∫ X

x0

f (x) dx,

or just as well,2 ∫ ξ

x0

f (x) dx =∫ X

x0

f (x) dx +∫ ξ

Xf (x) dx .

Now, formula (6) of the twenty-second lecture suffices to show how the two equa-tions that we have just obtained agree with equation (7). This last equation beingestablished in all the hypotheses, we can directly deduce equation (6), regardless ofx1, x2, . . . , xn−1.

We have seen, in the previous lecture, how easy it is to find, not only the approx-imate values of the integral (1), but also the limits of the errors committed when thefunction f (x) is always increasing or always decreasing from x = x0 up to x = X.

When this condition ceases to be satisfied, we can obviously, with the aid of formula(6), decompose the integral (1) into several others, for which the same condition isfulfilled for each.

Let us now conceive3 the limit X being greater than x0 and the function f (x) beingpositive from x = x0 up to x = X, with x, y denoting rectangular coordinates andA the surface contained, one part between the x-axis and the curve y = f (x), theother part between the ordinates f (x0), f (X). This surface, which has for its basethe length X − x0 measured on the x-axis, will be an average between the areas ofthe two rectangles constructed on the base X − x0, with heights, respectively, equalto the smallest and to the largest of the ordinates rising from the different points ofthis base. It will, therefore, be equivalent to a rectangle constructed on an averageordinate represented by an expression of the form f

[x0 + θ(X − x0)

]; so that, wewill have

A = (X − x0) f[x0 + θ(X − x0)

],(8)

θ designating a number less than unity. If we divide the base X − x0 into very smallelements, x1 − x0, x2 − x1, . . . , X − xn−1, the surface A will be found divided intocorresponding elements whose values will be given by equations similar to formula(8). We will, therefore, again have

(9)

{A = (x1 − x0) f

[x0 + θ0(x1 − x0)

] + (x2 − x1) f[x1 + θ1(x2 − x1)

]+ · · · + (X − xn−1) f

[xn−1 + θn−1(X − xn−1)

],

2The 1899 edition has X0 instead of X for the lower limit in its final integral to the right. This errordoes not occur in the original 1823 version.3Cauchy is going to geometrically connect his definite integral to the area under a curve, somethinghe has kept away from doing until now.


θ0, θ1, . . . , θn−1 designating numbers less than unity. If, in this last equation, wedecrease indefinitely the numerical values of the elements of X − x0, we will derive,by passing to the limits,

A =∫ X

x0

f (x) dx .(10)

Examples. – Apply formula (10) to the curves y = ax2, xy = 1, y = ex , . . . .

In closing this lecture, we will make known a remarkable property of real definiteintegrals. If we suppose f (x) = ϕ(x)χ(x), ϕ(x) and χ(x) being two new functionswhich remain, one and the other, continuous between the limits x = x0, x = X, andin which the second always retains the same sign between these limits, the value ofS given by equation (1) of the twenty-second lecture will become

(11)

{S = (x1 − x0)ϕ(x0)χ(x0)

+ (x2 − x1)ϕ(x1)χ(x1) + · · · + (X − xn−1)ϕ(xn−1)χ(xn−1),

and will be equivalent to the sum

(x1 − x0)χ(x0) + (x2 − x1)χ(x1) + · · · + (X − xn−1)χ(xn−1)

multiplied by an average between the coefficients ϕ(x0), ϕ(x1), . . . , ϕ(xn−1),4 or

to what amounts to the same thing, by a quantity of the form ϕ(ξ), ξ designating avalue of x contained between x0 and X. We will have, therefore,

S = [(x1 − x0)χ(x0) + (x2 − x1)χ(x1) + · · · + (X − xn−1)χ(xn−1)

]ϕ(ξ),(12)

and we will deduce, by finding the limit of S,

(13)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ X

x0

f (x) dx =∫ X

x0

ϕ(x)χ(x) dx

= ϕ(ξ)

∫ X

x0

χ(x) dx,

ξ always designating a value of x contained between x0 and X.5

4Cauchy is using one of the average theorem results found in his Cours d’analyse text. In this case,he employs Theorem I from his Preliminaries, restated with an accompanying proof as TheoremXII in Note II at the end of his 1821 text. Cauchy also uses this result in Lecture Seven to help provehis Mean Value Theorem for Derivatives. Because of its important role in these Calcul infinitésimallectures, a full translation of his Cours d’analyse Note II, Theorem XII—along with its proof—isincluded as Appendix B at the end of this book for reference.5This “remarkable” property of the definite integral is known today as the General Mean ValueTheorem for Definite Integrals. It is indeed remarkable.

TWENTY-THIRD LECTURE. 127

Examples. – If we successively take

χ(x) = 1, χ(x) = 1

x, χ(x) = 1

x − a,

we will obtain the formulas6

∫ X

x0

f (x) dx = f (ξ)

∫ X

x0

dx = (X − x0) f (ξ),(14)

∫ X

x0

f (x) dx = ξ f (ξ)

∫ X

x0

dx

x= ξ f (ξ)l

(X

x0

),(15)

∫ X

x0

f (x) dx = (ξ − a) f (ξ)

∫ X

x0

dx

x − a= (ξ − a) f (ξ)l

(X − a

x0 − a

),(16)

the first of which coincides with equation (19) of the twenty-second lecture.7 Let usadd that the ratio X

x0in the second formula and the ratio X−a

x0−a in the third must besupposed positive.

6There are two significant mistakes here. In the 1899 text, the last equation reads,

(16)∫ X

x0f (x) dx = (ξ − a) f (ξ − a)

∫ X

x0

dx

x − a= (ξ − a)l

(X − a

x0 − a

).

Interestingly, the original 1823 text is different, but is also incorrect. In this earlier edition, theequation reads,

(16)∫ X

x0f (x) dx = (ξ − a) f (ξ − a)

∫ X

x0

dx

x − a= (ξ − a) f (ξ − a)l

(X − a

x0 − a

).

A corrected version of (16) is given here.7Cauchy is referring to the Mean Value Theorem for Definite Integrals, a special case of theremarkable version derived earlier.

TWENTY-FOURTH LECTURE.

OF DEFINITE INTEGRALS WHOSE VALUES AREINFINITE OR INDETERMINATE. PRINCIPALVALUES OF INDETERMINATE INTEGRALS.

In the previous lectures, we have demonstrated several remarkable properties of thedefinite integral

∫ X

x0

f (x) dx,(1)

but by supposing: 1◦ that the limits x0, X were finite quantities; 2◦ that the functionf (x) would remain finite and continuous between these same limits. When thesetwo types of conditions are found fulfilled, then, in designating by x1, x2, . . . , xn−1

new values of x interposed between the extreme values x0, X, we have

∫ X

x0

f (x) dx =∫ x1

x0

f (x) dx +∫ x2

x1

f (x) dx + · · · +∫ X

xn−1

f (x) dx .(2)

When the interposed values are reduced to two, one differing very little from x0and represented by ξ0, the other differing very little from X and represented by ξ,

equation (2) becomes

∫ X

x0

f (x) dx =∫ ξ0

x0

f (x) dx +∫ ξ

ξ0

f (x) dx +∫ X

ξ

f (x) dx,

and can be rewritten as follows∫ X

x0

f (x) dx =(ξ0 − x0) f[x0 + θ0(ξ0 − x0)

]

+∫ ξ

ξ0

f (x) dx + (X − ξ) f[ξ + θ(X − ξ)

],

θ0, θ designating two numbers less than unity.1 If, in the last formula, we let ξ0converge toward the limit x0, and ξ toward the limit X, we will derive, by passing

1Cauchy has employed his Mean Value Theorem for Definite Integrals.


129


https://doi.org/10.1007/978-3-030-11036-9_24


to the limits,∫ X

x0

f (x) dx = lim∫ ξ

ξ0

f (x) dx .(3)

When the extreme values x0, X become infinite, or when the function f (x) doesnot remain finite and continuous from x = x0 up to x = X, we can no longer claimthat the quantity designated by S in the previous lectures has a fixed limit, and as aresult, we can no longer see what meaning we should attach to the notation in (1)which would serve to generally represent the limit of S. To lift any uncertainty andrender to the notation in (1), in every case, a clear and precise meaning, it sufficesto extend equations (2) and (3) by analogy, even in the case where they are not ableto be rigorously demonstrated. This is what we are going to see by some examples.

Consider, in the first place, the integral∫ +∞

−∞ex dx .(4)

If we denote by ξ0 and ξ two variable quantities, the first of which converges towardthe limit −∞, and the second toward the limit ∞, we will derive from formula (3)2

∫ +∞

−∞ex dx = lim

∫ ξ

ξ0

ex dx = lim(eξ − eξ0

) = e∞ − e−∞ = ∞.

Thus, the integral (4) has a positive infinite value.Consider, in the second place, the integral

∫ ∞

0

dx

x(5)

taken between two limits, one of which is infinite, while the other renders the func-tion under the

∫sign infinite, namely, 1

x . In designating by ξ0 and ξ two positivequantities, the first of which converges toward the limit zero, and the second towardthe limit ∞, we will derive from formula (3)

∫ ∞

0

dx

x= lim

∫ ξ

ξ0

dx

x= lim l

( ξ

ξ0

)= l

(∞0

)= ∞.

Thus, the integral (5) again has a positive infinite value.It is essential to observe that, if the variable x and the function f (x) remain finite,

one and the other, for one of the limits of the integral (1), we can reduce formula (3)to one of the following two

∫ X

x0

f (x) dx = lim∫ ξ

x0

f (x) dx,

∫ X

x0

f (x) dx = lim∫ X

ξ0

f (x) dx .(6)

2Cauchy is implying a double limit with his loose notation here. One as ξo approaches −∞, andthe other as ξ approaches ∞.

TWENTY-FOURTH LECTURE. 131

We will derive in particular from these last equations⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ 0

−∞ex dx = e0 − e−∞ = 1,

∫ ∞

0ex dx = e∞ − e0 = ∞,

∫ 0

−1

dx

x= l0 = −∞,

∫ 1

0

dx

x= l

(10

)= ∞.

(7)

Now consider the integral

∫ +1

−1

dx

x,(8)

in which the function under the∫sign, namely, 1

x , becomes infinite for the particularvalue x = 0, contained between the limits x = −1, x = +1. We will derive fromformula (2)

∫ +1

−1

dx

x=

∫ 0

−1

dx

x+

∫ +1

0

dx

x= −∞ + ∞.(9)

The value of the integral (8), therefore, appears indeterminate. To make sure thatit is actually so,3 it suffices to observe that, if we denote by ε, an infinitely smallnumber,4 and by μ, ν two positive, but arbitrary constants, we will have by virtueof the formulas in (6),

∫ 0

−1

dx

x= lim

∫ −εμ

−1

dx

x,

∫ 1

0

dx

x= lim

∫ 1

εν

dx

x.(10)

As a result, formula (9) will become

(11)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ +1

−1

dx

x= lim

(∫ −εμ

−1

dx

x+

∫ 1

εν

dx

x

)

= lim

(l(εμ) + l

( 1

εν

))= l

(μ

ν

),

and will generate for the integral (8), a completely indeterminate value, since thisvalue will be the Napierian logarithm of the arbitrary constant μ

ν.5

Let us now conceive that the function f (x) becomes infinite between the limitsx = x0, x = X, for the particular values of x represented by x1, x2, . . . , xm . Ifwe designate by ε an infinitely small number, and by μ1, ν1, μ2, ν2, . . . , μm,

3This is where Cauchy’s Lecture Twenty-Four starts to become particularly interesting. He under-stands this definite integral may still exist, but is simply not determinable by the methods developedthus far.4A positive variable whose limit is zero.5Clever. The resultant value is dependent upon one’s choice of μ and ν.


νm positive, but arbitrary, constants, we will derive from formulas (2) and (3)

(12)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

x0f (x) dx =

∫ x1

x0f (x) dx +

∫ x2

x1f (x) dx + · · · +

∫ X

xm

f (x) dx

= lim

[ ∫ x1−εμ1

x0f (x) dx +

∫ x2−εμ2

x1+εν1

f (x) dx + · · · +∫ X

xm+ενm

f (x) dx

].

If the limits x0, X are themselves found replaced by −∞ and +∞, we would have

(13)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ +∞

−∞f (x) dx = lim

[ ∫ x1−εμ1

− 1εμ

f (x) dx +∫ x2−εμ2

x1+εν1

f (x) dx

+ · · · +∫ 1

εν

xm+ενm

f (x) dx

],

μ, ν designating two new positive, but arbitrary constants. Add that, in the secondmember of formula (13), we must reestablish X in the place of 1

εν, or x0 in the place

of − 1εμ

, if, of the two quantities x0, X, only one becomes infinite. In all the cases,the values of the integrals

∫ X

x0

f (x) dx,

∫ +∞

−∞f (x) dx,(14)

deduced from equations (12) and (13) could be, dependent upon the nature of thefunction f (x), either infinite quantities, finite and determinate quantities, or inde-terminate quantities which will depend upon the values attributed to the arbitraryconstants μ, ν, μ1, ν1, . . . , μm, νm .

If, in formulas (12) and (13), we reduce the arbitrary constants μ, ν, μ1, ν1, . . . ,

μm, νm to unity, we will find

∫ X

x0f (x) dx = lim

[ ∫ x1−ε

x0f (x) dx +

∫ x2−ε

x1+εf (x) dx + · · · +

∫ X

xm+εf (x) dx

],(15)

∫ +∞

−∞f (x) dx = lim

[ ∫ x1−ε

− 1ε

f (x) dx +∫ x2−ε

x1+ε

f (x) dx + · · · +∫ 1

ε

xm+ε

f (x) dx

].(16)

Whenever the integrals in (14) become indeterminate, equations (15) and (16) sup-ply for each of them a single particular value to which we will give the name ofprincipal value.6 If we take, for example, the integral (8) whose general value isindeterminate, we will recognize that its principal value is reduced to zero.

6Today, the Cauchy Principal Value is defined along these same lines.

TWENTY-FIFTH LECTURE.

SINGULAR DEFINITE INTEGRALS.

Consider that an integral relative to x, and in which the function under the∫sign is

denoted by f (x), is taken between two limits infinitely close to a definite particularvalue a attributed to the variable1 x . If this value a is a finite quantity, and if thefunction f (x) remains finite and continuous in the neighborhood of x = a, then,by virtue of formula (19) (twenty-second lecture),2 the proposed integral will beessentially null. But, it can obtain a finite value different from zero or even an infinitevalue, if we have

a = ±∞ or else f (a) = ±∞.

In this latter case, the integral in question will become what we will call a singulardefinite integral. It will ordinarily be easy to calculate its value with the help offormulas (15) and (16) of the twenty-third lecture, as we shall see.

Let ε be an infinitely small number, and μ, ν two positive, but arbitrary, con-stants. If a is a finite quantity, but taken among the roots of the equation f (x) = ±∞,

and ifΥ designates the limit toward which the product (x−a) f (x) converges, whileits first factor converges toward zero, the values of the singular integrals

∫ a−εμ

a−ε

f (x) dx,∫ a+ε

a+εν

f (x) dx

will be very nearly(by virtue of formula (16), twenty-third lecture

)

∫ a−εμ

a−ε

f (x) dx = Υ lμ,(1)

∫ a+ε

a+εν

f (x) dx = Υ l(1

ν

).(2)

1The 1899 reprint mistakenly replaces Cauchy’s original word variable with the term valeur,which would change the translation to value. His original wording is used here.2The Mean Value Theorem for Definite Integrals.


133


https://doi.org/10.1007/978-3-030-11036-9_25


If we suppose, on the other hand, a = ±∞, by calling Υ the limit toward which theproduct x f (x) converges, while the variable x converges toward the limit ±∞, wewill essentially have

(twenty-third lecture, equation (15)

)

∫ − 1ε

− 1εμ

f (x) dx = Υ lμ,(3)

∫ 1εν

1ε

f (x) dx = Υ l(1

ν

).(4)

It is essential to observe that the limit of the product (x−a) f (x) or x f (x) sometimesdepends upon the sign of its first factor. In this way, for example, the product x(x2 +x4)− 1

2 converges toward the limit +1 or −1, depending on whether its first factor, inapproaching zero, remains positive or negative. It follows from this remark that thequantity designated by Υ sometimes changes its value in passing from equation (1)to equation (2), or from equation (3) to equation (4).

The consideration of singular definite integrals provides the means to calculatethe general value of an indeterminate integral when we know its principal value. Infact, let

∫ X

x0

f (x) dx(5)

be the integral in question, and consider that, by admitting the notations of the pre-vious lecture, we have

E =∫ x1−εμ1

x0

f (x) dx +∫ x2−εμ2

x1+εν1

f (x) dx + · · · +∫ X

xm+ενm

f (x) dx,(6)

F =∫ x1−ε

x0

f (x) dx +∫ x2−ε

x1+ε

f (x) dx + · · · +∫ X

xm+ε

f (x) dx .(7)

Further, let A = lim E be the general value, and B = lim F the principal value of theintegral (5). The difference,

A − B = lim (E − F),

will be equivalent to the sum of the singular integrals

∫ x1−εμ1

x1−εf (x) dx,

∫ x1+ε

x1+εν1

f (x) dx,∫ x2−εμ2

x2−εf (x) dx, . . . ,

∫ xm+ε

xm+ενm

f (x) dx,(8)

that is to say, to the limit which the sum of the integrals in (8) approaches, whileε decreases indefinitely. Moreover, if we designate by Υ1, Υ2, . . . , Υm the limits

TWENTY-FIFTH LECTURE. 135

toward which the products

(x − x1) f (x), (x − x2) f (x), . . . , (x − xm) f (x)

converge while their first factors converge toward zero, and if these limits are inde-pendent of the signs of these first factors, we will find that the sum of the integralsin (8) is essentially reduced to

Υ1l(μ1

ν1

)+ Υ2l

(μ2

ν2

)+ · · · + Υm l

(μm

νm

).(9)

Whenwe have x1 = x0 or xm = X, the difference A−B includes at least one singularintegral, namely the first or the last of the integrals in (8).

When we suppose x0 = −∞, X = +∞, equations (6) and (7) should be replacedby those which follow

E =∫ x1−εμ1

− 1εμ

f (x) dx +∫ x2−εμ2

x1+εν1

f (x) dx + · · · +∫ 1

εν

xm+ενm

f (x) dx,(10)

F =∫ x1−ε

− 1ε

f (x) dx +∫ x2−ε

x1+ε

f (x) dx + · · · +∫ 1

ε

xm+ε

f (x) dx .(11)

In the same hypothesis, it is necessary to add to the integrals in (8) the followingtwo3

∫ − 1ε

− 1εμ

f (x) dx,∫ 1

εν

1ε

f (x) dx,(12)

of which the sum will essentially be equivalent to the expression

Υ l(μ

ν

),(13)

if the product x f (x) converges toward the limit Υ, while the variable x convergestoward one of the two limits, −∞, +∞. If only one of the two quantities x0, Xbecomes infinite, it would be necessary to keep in the difference A− B only a singleone of the integrals in (12).

When, for the infinitely small values of ε, and for the finite or infinitely smallvalues of the arbitrary coefficients μ, ν, μ1, ν1, . . . , μm, νm, the singular integralsin (8) and (12), or at least some among them, obtain either infinite values, or finite

3Both the 1823 and the 1899 editions read,

(12)∫ − 1

ε

− 1εμ

f (x) dx,∫ 1

εμ

1ε

f (x) dx .

An obvious typographical error. A corrected version has been given here.


values,butdifferent fromzero, the integrals∫ Xx0

f (x) dx,∫ +∞−∞ f (x) dx areobviously

infinite or indeterminate. This iswhat occurswhenever the quantitiesΥ1, Υ2, . . . , Υm

are not simultaneously null. But, the converse is not true, and it could occur that,these quantities being null all at once, the integrals in (8) and (12), or at least someamong them, obtain finite values different from zero for infinitely small values ofthe coefficients μ, ν, μ1, ν1, . . . , μm, νm . Thus, for example, if we take f (x) = 1

x lx ,

the product x f (x) will vanish for x = 0; however, the singular integral4

∫ εν

ε

dx

x lx= l

(1 + lν

lε

)

will cease to vanish for the infinitely small values of ν.

When the singular integrals included in the difference A − B all vanish for theinfinitely small values of ε, regardless, moreover, of the values, finite or infinitelysmall, attributed to the coefficients μ, ν, μ1, ν1, . . . , μm, νm, we are assured thatthe general value of the integral (5) is reduced to a finite and determinate quantity.In fact, in this hypothesis, let δ be a very small number and suppose ε is chosen ina manner such that for values of μ, ν, μ1, ν1, . . . , μm, νm less than unity, each ofthe integrals in (8) and (12) has a numerical value less than 1

2(m+1) δ. The approx-imate value of B, represented by F, will be a finite quantity which will not containanything arbitrary; and, if we attribute to the coefficients μ, ν, μ1, ν1, . . . , μm, νminfinitely small values, E will indefinitely approach A by remaining contained be-tween the limits F − δ, F + δ. A will be, therefore, contained between the samelimits, and by consequence, we can find a finite quantity, F, which differs from A bya quantity less than a given number δ. We must deduce that the general value A ofthe integral (5) will be, in the admitted hypothesis, a finite and determinate quantity.

From principles that we have just established, we immediately deduce the fol-lowing proposition.

THEOREM. – For the general value of the integral (1) to be finite and determi-nate, it is necessary and sufficient that those singular integrals in (8) and (12) whichare found included in the difference A− B are reduced to zero for the infinitely smallvalues of ε, regardless, moreover, of the values, finite or infinitely small, attributedto the coefficients μ, ν, μ1, ν1, . . . , μm, νm .

Example. – Let Υ (x)F(x) be a rational fraction.

5 For the integral∫ +∞−∞

Υ (x)F(x) dx to retain

a finite and determinate value, it will be necessary and sufficient: 1◦ that the equa-tion F(x) = 0 does not have real roots; 2◦ that the degree of the denominator F(x)exceeds, by at least two units, the degree of the numerator, Υ (x).

4Let u = ln x, so that∫ du

u is easily evaluated. Cauchy determined this integral earlier in LectureTwenty-Two.5The 1899 edition here again mistakenly replaces the original fraction rationnelle, translated asrational fraction (Cauchy’s normal phrase in this case), with fonction rationnellemeaning rationalfunction.

TWENTY-SIXTH LECTURE.

INDEFINITE INTEGRALS.

If, in the definite integral∫ X

x0f (x) dx we vary one of the two limits, for example, the

quantity X, the integral itself will vary along with this quantity; and, if we replacethe limit X, to become the variable x, we will obtain as a result a new function of x,

which will be what we call an integral taken starting from the origin x = x0. Let

F(x) =∫ x

x0

f (x) dx(1)

be this new function. We will derive from formula (19) (twenty-second lecture)1

F(x) = (x − x0) f [x0 + θ(x − x0)], F(x0) = 0,(2)

θ being a number less than unity, and from formula (7) (twenty-third lecture)∫ x+α

x0

f (x) dx −∫ x

x0

f (x) dx =∫ x+α

xf (x) dx = α f (x + θα)

or

F(x + α) − F(x) = α f (x + θα).(3)

It follows from equations (2) and (3) that, if the function f (x) is finite and contin-uous in the neighborhood of a particular value attributed to the variable x, the newfunction F(x) will not only be finite, but also continuous in the neighborhood ofthis value, since an infinitely small increment of x will correspond to an infinitelysmall increment of F(x). So, if the function f (x) remains finite and continuous fromx = x0 up to x = X, it will be the same for the function F(x).2 Add that, if we divideboth members of formula (3) by α, we will deduce, in passing to the limits,

F′(x) = f (x).(4)

1Cauchy is again referring to his Mean Value Theorem for Definite Integrals here.2Cauchy has just argued his integrals are continuous functions, an important result of its own.


137


https://doi.org/10.1007/978-3-030-11036-9_26


Therefore, the integral (1), considered as a function of x, has for its derivative, thefunction f (x) included under the

∫sign in this integral. We would prove in the same

way, that the integral ∫ X

xf (x) dx = −

∫ x

Xf (x) dx,

considered as a function of x, has − f (x) for its derivative. We will have, therefore,3

d

dx

∫ x

x0

f (x) dx = f (x) andd

dx

∫ X

xf (x) dx = − f (x).(5)

If, to the various preceding formulas, we combine equation (6) of the seventh lec-ture, it will become easy to resolve the following questions.

PROBLEMI. –We require a function �(x) whose derivative � ′(x) is constantlynull. In other words, we propose to solve the equation

� ′(x) = 0.(6)

Solution. – If we desire that the function�(x) remains finite and continuous fromx = −∞ up to x = +∞, then, in designating by x0 a particular value of the variablex, we will derive from formula (6) (seventh lecture)4

�(x) − �(x0) = (x − x0)� ′[x0 + θ(x − x0)] = 0,

and as a result,5

�(x) = �(x0),(7)

or, if we designate by c the constant quantity �(x0),

�(x) = c.(8)

So, the function �(x) must then be reduced to a constant, and retains the samevalue c from x = −∞ up to x = ∞.We can add that this single value will be entirelyarbitrary, since formula (8) will satisfy equation (6), whatever c may be.

If we permit the function�(x) to continue to offer solutions of continuity6 corre-sponding to various values of x, and if we suppose that these values of x, arranged

3Cauchy has just proven the first version of the Fundamental Theorem of Calculus. One of themain reasons Newton and Leibniz are credited with inventing the calculus is their displayed under-standing of this inverse relationship in the 17th century.4The Mean Value Theorem for Derivatives.5The 1899 text has a typographical error in equation (7). It reads, � x = �(x0).6Recall “solutions of continuity” is the older mathematical phrase used by Cauchy to denote breaksin continuity. These are points in which continuity dissolves or disappears. This expression is nolonger in use, today referring to these as points of discontinuity.

TWENTY-SIXTH LECTURE. 139

in their order of magnitude, are represented by x1, x2, . . . , xm, then equation (7)must remainvalidonly from x = −∞up to x = x1, or from x = x1 up to x = x2, . . . ,or finally, from x = xm up to x = +∞, depending on whether the particular valueof x represented by x0 will be contained between the limits −∞ and x1, or else,between the limits of x1 and x2, . . . , or finally, between the limits xm and ∞. Byconsequence, it will no longer be necessary that the function �(x) keeps the samevalue from x = −∞ up to x = +∞, but only that it remains constant between twoconsecutive terms of the sequence

−∞, x1, x2, . . . , xm, +∞.

It is this that will happen, for example, if we take7

⎧⎪⎪⎨

⎪⎪⎩

�(x) = c0 + cm

2+ c1 − c0

2

x − x1√(x − x1)2

+ c2 − c12

x − x2√(x − x2)2

+ · · ·

+ cm − cm−1

2

x − xm√(x − xm)2

,

(9)

c0, c1, c2, . . . , cm designating constant, but arbitrary, quantities.8 In fact, in this case,the function�(x)will beconstantly equal toc0 between the limits x = −∞, x = x1;to c1 between the limitsx = x1, x = x2; . . . ; finally, to cm between the limits x =xm, x = ∞.

If we desire that �(x) be reduced to c0 for negative values, and to c1 for positivevalues of x, it will suffice to take

�(x) = c0 + c12

+ c1 − c02

x√x2

.(10)

PROBLEM II. – Find the general value of y that works to satisfy the equation

dy = f (x) dx .(11)

Solution. – If we denote by F(x) a particular value of the unknown y, and byF(x)+�(x) its general value, we will derive from formula (11), to which these twovalues must satisfy,

F ′(x) = f (x), F ′(x) + � ′(x) = f (x),

and as a result,� ′(x) = 0.

7Both of Cauchy’s texts print c in the second term of the right-hand side of equation (9), i.e.,c−c02

x−x1√(x−x1)2

. This term should be c1 and has been corrected here.

8This adds an interesting twist to Cauchy’s proof. �(x), his arbitrary constant, is not a constant atall. Instead, �(x) is a piecewise constant function.


Moreover, it results from the first of the equations in (5) that we would satisfy for-mula (11) by taking y = ∫ x

x0f (x) dx . Therefore, the general value of y will be

y =∫ x

x0

f (x) dx + �(x),(12)

�(x) designating a function that works to satisfy equation (6). This general valueof y, which includes, as a particular case, the integral (1), and which maintains thesame form, whatever the origin x0 of this integral, is represented in the calculusby the simple notation

∫f (x) dx, and receives the name of indefinite integral. This

granted, formula (11) always leads to the following

y =∫

f (x) dx,(13)

and vice versa, so that we have identically

d∫

f (x) dx = f (x) dx .(14)

If the function F(x) differs from the integral (1), the general value of y, or∫f (x) dx, can always be presented under the form

∫f (x) dx = F(x) + �(x),(15)

and must reduce to the integral (1), for a particular value of�(x),which will satisfy,at the same time, equation (6) and the following

F(x) =∫ x

x0

f (x) dx = F(x) + �(x).(16)

If, in addition, the functions f (x) and F(x) are, one and the other, continuous be-tween the limits x = x0, x = X, the function F(x) will itself be continuous, and as aresult, �(x) = F(x) − F(x) will constantly maintain the same value between theselimits, between which we will have

�(x) = �(x0),

F(x) − F(x) = F(x0) − F(x0) = −F(x0),

F(x) = F(x) − F(x0),

∫ x

x0

f (x) dx = F(x) − F(x0).(17)

TWENTY-SIXTH LECTURE. 141

Finally, if in equation (17), we set x = X, we will find9

∫ X

x0

f (x) dx = F(X) − F(x0).(18)

It results from equations (15), (17), and (18) that, being given a particular value F(x)

for y that works to satisfy formula (11), we can deduce: 1◦ the value of the indefi-niteintegral

∫f (x) dx; 2◦ thoseofthetwodefiniteintegrals

∫ xx0

f (x) dx,∫ X

x0f (x) dx,

in the case where the functions f (x), F(x) remain continuous between the limits ofthese two integrals.10

Example. – Since we verify the equation dy = dx1+x2 by taking y = arctang x, and

that the two functions 11+x2 , arctang x, remain finite and continuous between the

limits x = −∞, x = ∞, we will derive from formulas (15), (17), and (18)∫

dx

1 + x2= arctang x + �(x),

∫ x

0

dx

1 + x2= arctang x,

∫ 1

0

dx

1 + x2= π

4= 0.785 . . . .

Note. – When in equation (17) we wish to extend the value of x beyond a limitwhich renders the function f (x) discontinuous, it is usually required to add to thesecond member one or several singular integrals.

Example. – As we would satisfy the equation dy = dxx by taking y = 1

2 lx2, if we

denote by ε an infinitely small number, and byμ, ν two positive coefficients, we willfind, for x < 0, ∫ x

−1

dx

x= 1

2 lx2 − 1

2 l1 = 12 lx

2;

and, for x > 0,∫ x

−1

dx

x=

∫ −με

−1

dx

x+

∫ x

νε

dx

x= 1

2 lx2 − 1

2 l1 + lμ

ν

= 12 lx

2 +∫ −με

−ε

dx

x+

∫ ε

εν

dx

x.

9Cauchy has succeeded in proving the second version of the Fundamental Theorem of Calculus.10Cauchy has shown the limit process he developed in Lecture Twenty-One to evaluate a definiteintegral can be completely avoided by making use of this powerful relationship, if (in modernlanguage) a proper antiderivative can be identified. A tremendous shortcut.

TWENTY-SEVENTH LECTURE.

VARIOUS PROPERTIES OF INDEFINITEINTEGRALS. METHODS TO DETERMINETHE VALUES OF THESE SAME INTEGRALS.

From after what has been said in the previous lecture, the indefinite integral∫

f (x) dx(1)

is nothing other than the general value of the unknown y, subject to satisfy thedifferential equation

dy = f (x) dx .(2)

Moreover, being given a particular value F(x) of the same unknown, it will sufficeto obtain the general value, to add to F(x) a function �(x) that works to satisfy theequation � ′(x) = 0, or to what amounts to the same thing, an algebraic expressionwhich can only admit a finite number of constant values, each of which remainsvalid between certain limits assigned to the variable x . For brevity, we designate,henceforth, by the letter C, an expression of this nature, and we call it an arbitraryconstant, which will not mean that it must always retain the same value, regardlessof x .1 This granted, we will have

∫f (x) dx = F(x) + C.(3)

When we replace the function F(x) by the definite integral∫ x

x0f (x) dx, which is

itself a particular value of y, formula (3) is reduced to∫

f (x) dx =∫ x

x0

f (x) dx + C.(4)

By extending the definition that we have given of the integral (1) to the case wherethe function f (x) is supposed imaginary, we will easily recognize that, in this hy-pothesis, equations (3) and (4) still remain valid. Only, the arbitrary constant C then

1Cauchy continues to stress the fact his arbitrary constant is a piecewise constant function of x,

along with its inherent points of discontinuity.


143


https://doi.org/10.1007/978-3-030-11036-9_27


becomes imaginary at the same time that f (x) does; that is to say, it takes the form

C1 + C2

√−1,

C1 and C2 designating two arbitrary, but real, constants.Before going further, it is important to observe that by forming the sum or the

difference, or even any linear function of two or of several arbitrary constants, weobtain for a result a new arbitrary constant.

Several remarkable properties of indefinite integrals2 are easily deduced fromequation (4) combined with formulas (13) (twenty-second lecture) and (2), (3), (4),(5) (twenty-third lecture). In fact, if, after having replaced X by x in the twomembersof each of these formulas, we add to the integrals the arbitrary constants which theycontain, we will find, in denoting by a, b, c, . . . assumed known constants, and byu, v, w, . . . functions of the variable x,

∫au dx = a

∫u dx,(5)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫(u + v + w + · · · ) dx =

∫u dx +

∫v dx +

∫w dx + · · · ,

∫(u − v) dx =

∫u dx −

∫v dx,

∫(au + bv + cw + · · · ) dx = a

∫u dx + b

∫v dx + c

∫w dx + · · · ,

∫ (u + v

√−1)

dx =∫

u dx + √−1∫

v dx .

(6)

These equations remain valid even in the case where a, b, c, . . . , u, v, w, . . .

become imaginary.To integrate the differential formula f (x) dx, or in other words, to integrate equa-

tion (2), is to find the value of the indefinite integral∫

f (x) dx . The operation bywhich we achieve this is an indefinite integration. Definite integration consists offinding the value of a definite integral, such as

∫ Xx0

f (x) dx . We now make knownthe four main methods, with the help of which we can perform, in certain cases, thefirst of these two operations.

Immediate integration. –When in the formula f (x) dx we recognize the exact dif-ferential of a determinate function, F(x), the value of the indefinite integral

∫f (x)

is deduced immediately from equation (3). We extend the number of cases in whichthis type of integration is applicable by observing that the constant factors includedin f (x) can be placed at will within or outside the

∫sign

(see equation (5)

).

2What must be a typographical error occurs in the text of the 1899 edition. It replaces the 1823edition’s intégrales indéfinies with the phrase intégrales définies, which would change the transla-tion from an indefinite to a definite integral. Cauchy’s originally intended, and appropriate, 1823phrase is used here.

TWENTY-SEVENTH LECTURE 145

Examples. –∫

a dx = ax + C,

∫(a + 1)xa dx = xa+1 + C,

∫xa dx = xa+1

a + 1+ C,

∫x dx = 1

2 x2 + C,

∫dx

x2= −1

x+ C,

∫dx

xm= − 1

(m − 1)xm−1+ C,

∫dx√

x= 2

√x + C,

∫dx

x= 1

2 l(x2

) + C,

∫dx

1 + x2= arctang x + C,

∫dx√1 − x2

= arcsin x + C = C + 12π − arccos x,

∫ex dx = ex + C,

∫Ax lA dx = Ax + C,

∫Ax dx = Ax

lA+ C,

∫cos x dx = sin x + C,

∫sin x dx = − cos x + C,

∫dx

cos2 x= tang x + C,

∫dx

sin2 x= − cot x + C.

Integration by substitution. – Consider that for the variable x we substitute an-other variable z related to the first by an equation from which we derive z = ϕ(x)

and x = χ(z). Formula (2) will be found replaced by the following

dy = f[χ(z)

]χ ′(z) dz.(7)

If we make for brevity, f [χ(z)]χ ′(z) = Υ (z), the general value of y derived fromequation (7) will be represented by the indefinite integral

∫Υ (z) dz. Moreover, this

general value should coincide with the integral (1). Therefore, if, by virtue of theestablished relationship between x and z, we have identically

f (x) dx = Υ (z) dz,(8)

we will infer ∫f (x) dx =

∫Υ (z) dz.(9)

Now suppose that the value of∫

Υ (z) dz is given by an equation of the form∫

Υ (z) dz = F(z) + C.(10)

We will derive from this equation∫

f (x) dx = F[ϕ(x)

] + C.(11)


Examples. – By accepting formula (10) and successively setting

x ± a = z, ax = z,x

a= z, x2 + a2 = z,

lx = z, ex = z, sin x = z, cos x = z,

we will derive from formula (11), combined with equation (5),∫

Υ (x ± a) dx = F(x ± a) + C,

∫Υ (ax) dx = 1

aF(ax) + C,

∫Υ

( x

a

)dx = aF

( x

a

)+ C,

∫xΥ

(x2 + a2

)dx = 1

2F(x2 + a2

) + C,

∫xa−1Υ

(xa

)dx = 1

aF(xa

) + C,

∫Υ

(lx

) dx

x= F

(lx

) + C,

∫ex Υ

(ex

)dx = F

(ex

) + C,

∫cos x Υ (sin x) dx = F(sin x) + C,

∫sin x Υ (cos x) dx = −F(cos x) + C.

These last formulas being combined in their turn with those which result fromimmediate integration, we will find

∫dx

x − a= 1

2 l(x − a)2 + C,

∫dx

(x − a)m= − 1

(m − 1)(x − a)m−1+ C,

∫dx

1 + a2 x2= 1

aarctang (ax) + C,

∫dx

x2 + a2= 1

a2

∫dx

1 + (xa

)2 = 1

aarctang

( x

a

)+ C,

∫x dx

x2 + a2= 1

2 l(x2 + a2

) + C,

∫x dx√x2 + a2

=√

x2 + a2 + C,

∫eax dx = 1

aeax + C,

∫e−ax dx = −1

ae−ax + C,

∫cos ax dx = 1

asin ax + C,

∫sin ax dx = −1

acos ax + C,

∫lxx

dx = 12 (lx)2 + C,

∫dx

x lx= l(lx) + C,

∫dx

x(lx)m= −1

(m − 1)(lx)m−1+ C,

∫ex dx

e2x + 1= arctang (ex ) + C,

∫sin x dx

cos2 x= 1

cos x+ C = sec x + C,

∫cos x dx

sin2 x= − 1

sin x+ C.

TWENTY-SEVENTH LECTURE 147

Integration by decomposition. – This type of integration is performed with thehelp of the formulas in (6) when the function under the

∫sign can be decomposed

into several parts in such a manner that each part, multiplied by dx, gives for itsproduct an easily integrable expression. It applies particularly to the case where thefunction under the

∫sign is reduced, either to an entire function or to a rational

fraction.

Examples. –

∫dx

sin2 x cos2 x=

∫sin2 x + cos2 x

sin2 x cos2 xdx

=∫

dx

cos2 x+

∫dx

sin2 x

= tang x − cot x + C,

∫ (a + bx + cx2 + · · · ) dx = a

∫dx + b

∫x dx + c

∫x2 dx + · · ·

= ax + bx2

2+ c

x3

3+ · · · + C.

Integration by parts. – Let u and v be two different functions of x, and u′, v′ theirrespective derivatives. uv will be a particular value of y that works to satisfy thedifferential equation

dy = u dv + v du = uv′ dx + vu′ dx,

from which we will generally derive

y = uv + C =∫

uv′ dx +∫

vu′ dx =∫

u dv +∫

v du,

and as a result, ∫u dv = uv −

( ∫v du − C

),

or more simply3

∫u dv = uv −

∫v du,(12)

the arbitrary constant, −C, can be supposed contained in the integral∫

v du.

3Precisely in the form we commonly see integration by parts today.


Examples. –∫

lx dx = x lx −∫

xdx

x= x(lx − 1) + C,

∫x ex dx = ex (x − 1) + C,

∫x cos x dx = x sin x + cos x + C,

∫x sin x dx = −x cos x + sin x + C, . . . .

Note. – It is essential to observe that the arbitrary constants, which are supposedcontained in the indefinite integrals that include the two members of equation (12),can have very different numerical values. This remark suffices to explain the formula

∫dx

x lx= 1 +

∫dx

x lx,

to which we reach, upon setting in equation (12),4

u = 1

lxand v = lx .

4One might picture Cauchy proving 0 = 1 to his students with this idea, perhaps interjecting abit of a riddle and some humor into his highly intellectual and academic class – Cauchy himselfhoping for a rare chuckle from his students.

Throughout this text, Cauchy occasionally makes assumptions about characteristics of his func-tions which in general may not be true. During the latter part of the 19th century, several patholog-ical functions were generated which tested the new rigorous calculus theory. Although these func-tions may have taken Cauchy by surprise in 1823, the purpose was not to discredit the emergingtheory. Instead, their introduction was constructive and used to locate weaknesses or ambiguitiesin the foundation of the subject, so the theory could be strengthened. Among these are the Weier-strass Function, a real valued function designed by Karl Weierstrass that is continuous everywhere,but differentiable nowhere. This unusual function clearly demonstrates intuition cannot be trustedas this function shows continuity is not enough to guarantee differentiability, as Cauchy tends todo. This particular function would have been unimaginable to Cauchy in 1823. Riemann devel-oped a similarly bizarre function which is continuous everywhere, yet not differentiable almosteverywhere. Another classic pathological example is the Indicator Function, sometimes referredto as the Dirichlet Function, named after Lejeune Dirichlet (1805–1859). It is a strange functionthat is nowhere continuous. A final example is the Ruler Function, developed by Carl JohannesThomae (1840–1921). It contains an infinite number of discontinuities yet remains integrable and iscontinuous almost everywhere. These functions, along with many others, helped uncover and thenstrengthen the foundation of modern calculus. Pretty cool stuff.

TWENTY-EIGHTH LECTURE.

ON INDEFINITE INTEGRALS WHICH CONTAINALGEBRAIC FUNCTIONS.

We call algebraic functions those that we form by only employing the first opera-tions of the algebra, namely, addition, subtraction, multiplication, division, and theraising of variables to fixed powers. The algebraic functions of one variable are ra-tional when they contain only integer powers of this variable, that is to say, whenthey are reduced to entire functions or to rational fractions. They are irrational oth-erwise.

This granted, let us conceive that, f (x) denoting an algebraic function of x, weare looking for the value of the indefinite integral

∫f (x) dx . If the function f (x)

is rational, we will decompose the product f (x) dx into several terms which will bepresented under one of the forms

Axm dx,A dx

x − a,

A dx

(x − a)m,

(A ∓ B

√−1)dx

x − α ∓ β√−1

,

(A ∓ B

√−1)dx

(x − α ∓ β

√−1)m ,(1)

a, α, β, A, B designating real constants, and m an integer number; then, we willintegrate these different terms with the help of the equations

∫Axm dx = A

xm+1

m + 1+ C,

∫A dx

x − a= 1

2 Al(x − a)2 + C,

∫A dx

(x − a)m= − A

(m − 1)(x − a)m−1+ C,

∫ (A∓B

√−1)

dx

x − α∓β√−1

=(A∓B

√−1)∫

(x− α) dx

(x− α)2+β2+(

B± A√−1

)∫

β dx

(x− α)2+β2

= 12

(A∓B

√−1)l[(x − α)2+β2

]+(B± A

√−1)arctang

(x−αβ

)+C,

∫ (A ∓ B

√−1)

dx(x − α ∓ β

√−1)m = − A ∓ B

√−1

(m − 1)(x − α ∓ β

√−1)m−1 + C,


149


https://doi.org/10.1007/978-3-030-11036-9_28


the first of which are deduced from the principles established in the preceding lec-ture, and the last of which, derived by induction from the third, may be easily veri-fied after the fact.

Examples. –

∫ (A − B

√−1

x − α − β√−1

+ A + B√−1

x − α + β√−1

)

dx

= A l[(x − α)2 + β2

] + 2B arctangx − α

β+ C,

∫dx

x2 − 1=

∫1

2

(1

x − 1− 1

x + 1

)

dx = 1

4l( x − 1

x + 1

)2 + C,

∫x dx

x2 + 1= 1

2l(x2 + 1

) + C,

∫dx

x3 − 1=

∫1

3

(1

x − 1− x + 2

x2 + x + 1

)

dx

= 1

6l( (x − 1)2

x2 + x + 1

)− 1√

3arctang

2x + 1√3

+ C,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

When the function f (x), without ceasing to be algebraic, becomes irrational,there are no longer general rules by means of which we can calculate the value of∫

f (x) dx in finite terms. In truth, to achieve this it would be sufficient to substitutefor the variable x a second variable z chosen so that the expression f (x) dx is foundtransformed into another, Υ (z) dz, in which the function Υ (z) is rational. But, wehave no sure method to carry out such a transformation, except in a small numberof particular cases that we will now make known.

First, letΥ (x, z) be a rational function of x and z, z being an irrational function ofx, determined by an algebraic equation of any degree with respect to z, but of firstdegree with respect to x . To render the differential formula Υ (x, z) dx rational andintegrable, it will obviously suffice to substitute the variable z for the variable x . Weshould especially note the case where the value of z is provided, either by one of thebinomial equations

{zn − (ax + b) = 0,

(a0x + b0)zn − (a1x + b1) = 0,(2)

or by the second-degree equation

(a0x + b0)z2 − 2(a1x + b1)z − (a2x + b2) = 0,(3)

TWENTY-EIGHTH LECTURE. 151

a, b, a0, b0, a1, b1, a2, b2 being real constants, and n any integer number. Since wewould satisfy the equations in (2) by setting

z = (ax + b)1n or z =

(a1x + b1a0x + b0

) 1n

,

and equation (3) by setting

z = a1x + b1 + √(a1x + b1)2 + (a0x + b0)(a2x + b2)

a0x + b0,

it follows that we render the formulas

Υ[x, (ax + b)

1n

]dx or Υ

[

x,

(a1x + b1a0x + b0

) 1n

]

dx(4)

integrable by equating to z the radical that it contains, and the two formulas1

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

Υ

[

x,a1x + b1 + √

(a1x + b1)2 + (a0x + b0)(a2x + b2)

a0x + b0

]

dx,

Υ[x,

√(a1x + b1)2 + (a0x + b0)(a2x + b2)

]dx,

(5)

in substituting the value of x by z derived from equation (3), or to what amounts tothe same thing, from the following

√(a1x + b1)2 + (a0x + b0)(a2x + b2) = (a0x + b0)z − (a1x + b1).(6)

Let us now conceive how to render the expression

Υ(

x,√

Ax2 + Bx + C)

dx(7)

integrable, A, B, C being real constants. It will obviously suffice to employ equation(6), after having reduced the trinomial Ax2+ Bx +C to the form (a1x +b1)2+(a0x +b0)(a2x + b2). Now, we can perform this reduction in an infinite number of ways,

1The second expression in (5) from the 1899 edition reads,

Υ[x,

√(a1x + b1)2 + (a2x + b0)(a0x + b2)

]dx .

Additionally, in the original 1823 edition, this same expression reads,

Υ[x,

√(a1x + b1)2 + (a0x + b0)(a0x + b2)

]dx .

Both of these misprints have been corrected here.


by choosing a binomiala1x +b1 so that the difference Ax2 + Bx +C − (a1x +b1)2 isdecomposable into real factors of first degree, that is to say, so that we would have

Ab21 + Ca2

1 − Ba1b1 + 14 B2 − AC > 0.(8)

By seeking the simplest values of a1 and b1 which work to fulfill this latter condi-tion, we will find: 1◦ if 1

4 B2 − AC is positive,

a1 = 0, b1 = 0;2◦ if A is positive,

a1 = A12 , b1 = 0;

3◦ if C is positive,b1 = C

12 , a1 = 0.

Moreover, since we will have

Ax2 + Bx + C − (A

12 x

)2 = 1 × (Bx + C)

and

Ax2 + Bx + C − (C

12)2 = x(Ax + B),

we can take a0x + b0 = 1 in the second case, and a0x + b0 = x in the third. In sum-mary, if Ax2 + Bx +C is the product of two real factors a0x + b0, a2x + b2, we willrender formula (7) rational by setting

√(a0x + b0)(a2x + b2) = (a0x + b0)z or

a2x + b2a0x + b0

= z2.(9)

Otherwise, the radical√

Ax2 + Bx + C cannot be a real quantity, unless the twocoef-ficients A and C are positive. In all the cases, we will render expression (7)2 rationalby supposing

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

if A is positive,√

Ax2 + Bx + C = z − A12 x,

and

if C is positive,√

Ax2 + Bx + C = xz − C12 ,

or

√

A + B1

x+ C

1

x2= z − C

121

x.

(10)

It is easy to verify these various consequences of formula (6)3 after the fact.

2Both of Cauchy’s texts reference equation number (17), a clear misprint that has been correctedhere.3Cauchy references equation (16), another simple error which has been corrected.

TWENTY-EIGHTH LECTURE. 153

Examples. – We will derive from the first of the equations in (10)

∫dx√

Ax2 + Bx + C=

∫dz

A12 z + 1

2 B

=l(

Ax + 12 B + A

12

√Ax2 + Bx + C

)

A12

+ C,

∫dx√

x2 + 1= l

(x +

√x2 + 1

) + C,

∫dx√

x2 − 1= l

(x +

√x2 − 1

) + C,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

It is important to observe that, if we designate by Υ (u, v, w, . . . ) an entire func-tion of the variables u, v, w, . . . , and by p, q, r, . . . divisors of integer number n, thedifferential expressions

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

Υ[x, (ax + b)

1p , (ax + b)

1q , (ax + b)

1r , . . .

]dx,

Υ

[

x,

(a1x + b1a0x + b0

) 1p

,

(a1x + b1a0x + b0

) 1q

, . . .

]

dx(11)

will be of the same form as the expressions in (4) and can be integrated in the samemanner. Thus, we will find, by setting x = z6,

∫ (x

12 + x

23

)−1dx = 6

∫z2 dz

1 + z= 6

[12 z2 − z + 1

2 l(1 + z)2]

+ C.

Add that we will immediately reduce the differential expressions

Υ[xμ, (axμ + b)

1n

]xμ−1 dx, Υ

[

xμ,

(a1xμ + b1a0xμ + b0

) 1n

]

xμ−1 dx(12)

(μ denoting any constant) to the formulas in (4), and the expression

Υ[x, (a0x + b0)

12 , (a1x + b1)

12

]dx(13)

to formula (7), by setting xμ = y in the expressions in (12), and a0x + b0 = y2 inexpression (13).


Examples. – We integrate x2m+1√x2−1

dx, by setting x2 = y, y − 1 = z2, or simply

x2 − 1 = z2; and dx

(x−1)12 +(x+1)

12, by setting x − 1 = y2, then (y2 + 2)

12 = z − y, or

simply (x − 1)12 + (x + 1)

12 = z.

In finishing this lecture, we will remark that, in all the cases where we succeedto calculate the value of an indefinite integral which contains an algebraic function,this value is composed of several terms, each of which is presented under one of theforms4

Υ (x), Al(Υ (x)

), A arctang

(Υ (x)

),(14)

Υ (x) designating an algebraic function of x, and A a constant quantity. The expres-sions arcsin x = arctang x√

1−x2 , arccos x, and similar others, are obviously included

under the last of the three forms that we have just indicated.5

4Parentheses have been added in the third expression.5An incredible remark indeed.

TWENTY-NINTH LECTURE.

ON THE INTEGRATION AND THE REDUCTIONOF BINOMIAL DIFFERENTIALS AND OF ANYOTHER DIFFERENTIAL FORMULAS OF THESAME TYPE.

Let a, b, a1, b1, λ, μ, ν be real constants, y a variable quantity, and let us makeyλ = x . The expression (ayλ + b)μ dy, in which dx has for a coefficient a power ofthe binomial ayλ + b, will be what we call a binomial differential, and the indefiniteintegral

∫(ayλ + b)μ dy = 1

λ

∫(ax + b)μx

1λ−1 dx(1)

will be the product of 1λwith another integral included in the general formula

∫(ax + b)μ(a1x + b1)

ν dx,(2)

which we will now occupy ourselves.We easily determine integral (2) when the numerical values of the exponents

μ, ν and of their sum μ + ν are reduced to three rational numbers, one of which isan integer number. In fact, denote by l, m, n any integer numbers. To integrate thedifferential expressions1

(ax + b)±l(a1x + b1)± m

n dx,

(ax + b)±mn (a1x + b1)

±l dx,

(ax + b)±mn (a1x + b1)

±l± mn dx,

it will suffice to successively set (see the twenty-eighth lecture)

a1x + b1 = zn, ax + b = zn,ax + b

a1x + b1= zn.

The formula (ax + b)μ(a1x + b1)ν dx not always being integrable, it is good tosee how we can restore the determination of the integral (2) to those of several other

1In the 1823 edition the third expression reads, (ax + b)± mn (a1x + b1)±l∓ m

n dx .


155


https://doi.org/10.1007/978-3-030-11036-9_29


integrals of the same type, but inwhich the exponents of the binomialsax+b, a1x+b1are no longer the same. To achieve this in the most direct manner, we will haverecourse to equation (12) (twenty-seventh lecture) that we will present in the form

∫uv 1

2dlv2 = uv −∫

uv 12dlu2;(3)

then, we will suppose the functions u and v respectively proportional to certain pow-ers of two of the three quantities

ax + b, a1x + b1,ax + b

a1x + b1.(4)

Since these three quantities, combined in pairs, offer six different combinations, wesee that formula (3) will give rise to six distinct equations. We will simplify the cal-culation by operating as if u and v shall always remain positive, and by consequence,reducing formula (3) to this other

∫uv dlv = uv −

∫uv dlu;(5)

then, having regard to the equations

dl(ax + b) = a dx

ax + b,

dl(a1x + b1) = a1 dx

a1x + b1,

dl(

ax + b

a1x + b1

)= (ab1 − a1b) dx

(ax + b)(a1x + b1),

from which we will derive the value of dx to substitute into the integral (2). Let usconceive that, for brevity, we designate by A this same integral. We will find:

1◦ By supposing u to be proportional to a power of ax + b, and v to a power ofa1x + b1,

A =∫

(ax + b)μ(a1x + b1)ν+1

a1dl(a1x + b1)

=∫

(ax + b)μ

(ν + 1)a1(a1x + b1)

ν+1 dl(a1x + b1)ν+1

= (ax + b)μ(a1x + b1)ν+1

(ν + 1)a1−

∫(ax + b)μ(a1x + b1)ν+1

(ν + 1)a1dl(ax + b)μ,

⎧⎪⎪⎨⎪⎪⎩

∫(ax + b)μ(a1x + b1)

ν dx

= (ax + b)μ(a1x + b1)ν+1

(ν + 1)a1− μa

(ν + 1)a1

∫(ax + b)μ−1(a1x + b1)

ν+1 dx;(6)

TWENTY-NINTH LECTURE. 157

2◦ By supposing u to be proportional to a power of a1x + b1, and v to a power ofax + b,

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫(ax + b)μ(a1x + b1)

ν dx

= (ax + b)μ+1(a1x + b1)ν

(μ + 1)a

− νa1

(μ + 1)a

∫(ax + b)μ+1(a1x + b1)

ν−1 dx;

(7)

3◦ By supposing u to be proportional to a power of ax+ba1x+b1

, and v to a power ofa1x + b1,

A =∫

(ax + b)μ(a1x + b1)ν+1

a1dl(a1x + b1)

=∫ (

ax + b

a1x + b1

)μ(a1x + b1)μ+ν+1

(μ + ν + 1)a1dl(a1x + b1)

μ+ν+1

= (ax + b)μ(a1x + b1)ν+1

(μ + ν + 1)a1

−∫

(ax + b)μ(a1x + b1)ν+1

(μ + ν + 1)a1dl

(ax + b

a1x + b1

)μ

,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫(ax + b)μ(a1x + b1)

ν dx

= (ax + b)μ(a1x + b1)ν+1

(μ + ν + 1)a1

− μ(ab1 − a1b)

(μ + ν + 1)a1

∫(ax + b)μ−1(a1x + b1)

ν dx;

(8)

4◦ By supposing u to be proportional to a power of a1x+b1ax+b , and v to a power of

ax + b,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫(ax + b)μ(a1x + b1)

ν dx

= (ax + b)μ+1(a1x + b1)ν

(μ + ν + 1)a

− ν(a1b − ab1)

(μ + ν + 1)a

∫(ax + b)μ(a1x + b1)

ν−1 dx;

(9)


5◦ By supposing u to be proportional to a power of a1x + b1, and v to a power ofax+b

a1x+b1,

A =∫

(ax + b)μ+1(a1x + b1)ν+1

ab1 − a1bdl

(ax + b

a1x + b1

)

=∫

(a1x + b1)μ+ν+2

(μ + 1)(ab1 − a1b)

(ax + b

a1x + b1

)μ+1

dl(

ax + b

a1x + b1

)μ+1

= (ax + b)μ+1(a1x + b1)ν+1

(μ + 1)(ab1 − a1b)

−∫

(ax + b)μ+1(a1x + b1)ν+1

(μ + 1)(ab1 − a1b)dl(a1x + b1)

μ+ν+2,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫(ax + b)μ(a1x + b1)

ν dx

= (ax + b)μ+1(a1x + b1)ν+1

(μ + 1)(ab1 − a1b)

− (μ + ν + 2)a1

(μ + 1)(ab1 − a1b)

∫(ax + b)μ+1(a1x + b1)

ν dx;

(10)

6◦ By supposing u to be proportional to a power of ax + b, and v to a power ofax+b

a1x+b1,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫(ax + b)μ(a1x + b1)

ν dx

= (ax + b)μ+1(a1x + b1)ν+1

(ν + 1)(a1b − ab1)

− (μ + ν + 2)a

(ν + 1)(a1b − ab1)

∫(ax + b)μ(a1x + b1)

ν+1 dx .

(11)

With the help of formulas (6), (7), (8), (9), (10), (11), we can always replace theintegral (2) with another integral of the same type, but in which each of the binomialsax + b, a1x + b1 has an exponent contained between the limits 0 and −1. In fact, itwill suffice to achieve this, to employ formulas (8) and (9), or at least one betweenthem, once or several times in sequence,2 if the exponentsμ, ν are positive, or if, oneof them being positive, the other is already contained between the limits 0 and −1.On the other hand, we should employ formulas (10) and (11), or at least one between

2The 1899 text uses the phrase une ou deux de suite meaning once or twice in sequence, instead ofCauchy’s original une ou plusieurs fois de suite translated here.

TWENTY-NINTH LECTURE. 159

them, if the exponents μ, ν are both negative. Finally, if, one of the two exponentsbeing positive, the other is less than −1, we will have to use formula (6) or formula(7) for the simultaneous reduction of the numerical values of these two exponents,until one of them changes into a quantity contained between the limits 0 and −1.

When the exponents μ, ν have integer numerical values, then, by operating as wehave just said, we end by reducing both to one of the two quantities, 0 and −1. Thisreduction being performed, the integral (2) is necessarily found replaced by one ofthe following four

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫dx = x + C,

∫dx

ax + b= 1

2al(ax + b)2 + C,

∫dx

a1x + b1= 1

2a1l(a1x + b1)

2 + C,

∫dx

(ax + b)(a1x + b1)= 1

ab1 − a1b

∫dl

(ax + b

a1x + b1

)

= 1

2(ab1 − a1b)l(

ax + b

a1x + b1

)2

+ C.

(12)

In general, whenever the formula

(ax + b)μ(a1x + b1)ν dx

will be integrable, the reduction methods indicated above will allow substitution ofthe integral (2) with other simpler integrals whose values will be easy to obtain.

If we want to apply the same methods to the reduction of the integral (1), it willbe assumed in formula (5) that the quantities u and v are proportional to definitepowers, no longer of the quantities in (4), but of the following

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

ax + b = ay2 + b,

x = y2,

ax + b

x= ay2 + b

y2.

(13)

Example. – Consider how to reduce the integral∫

dy

(1 + y2)n=

∫ (1 + y2

)−ndy,


n denoting an integer number greater than unity. We will suppose u and v to beproportional to powers of y2 and of 1+y2

y2 ; and, since we will have

dl(1 + y2

y2

)= 2

(y

1 + y2− 1

y

)dy = − 2 dy

y(1 + y2),

we will derive from formula (5)⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫dy

(1 + y2)n=

∫ −y(1 + y2)−n+1

2dl

(1 + y2

y2

)

=∫

y−2n+3

2(n − 1)

(1 + y2

y2

)−n+1

dl(1 + y2

y2

)−n+1

= y(1 + y2)−n+1

2(n − 1)−

∫y(1 + y2)n−1

2(n − 1)dly−2n+3

= y

2(n − 1)(1 + y2)n−1+ 2n − 3

2n − 2

∫dy

(1 + y2)n−1.

(14)

THIRTIETH LECTURE.

ON INDEFINITE INTEGRALS WHICH CONTAINEXPONENTIAL, LOGARITHMIC, OR CIRCULARFUNCTIONS.

We call exponential functions or logarithmic functions those which contain vari-able exponents or logarithms, and trigonometric or circular functions those whichcontain trigonometric lines or arcs1 of a circle. It would be very useful to integratethe differential formulas which contain similar functions. But, we do not have suremethods to achieve this, except in a small number of particular cases that we willnow review.

First, if we denote by Υ a function such that the indefinite integral∫

Υ (z) dz hasa known value, we will deduce the values of

∫Υ (lx)

dx

x,

∫exΥ (ex ) dx,

∫cos x Υ (sin x) dx,

∫sin x Υ (cos x) dx,(1)

by successively setting, as in the twenty-seventh lecture,

lx = z, ex = z, sin x = z, cos x = z.

We would likewise determine the three integrals

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

∫Υ (arctang x)

dx

1 + x2,

∫Υ (arcsin x)

dx√1 − x2

,

∫Υ (arccos x)

dx√1 − x2

,

(2)

1Recall the phrase “trigonometric lines” is an old term used to denote the signed lengths of theline segments represented by the six trigonometric functions. Picturing a unit circle with a partic-ular signed angle along with its corresponding terminal side, the line segments referred to here arethose generated if one were to graphically construct the trigonometric function representations onthe corresponding right triangle about this circle. Today, we would simply refer to “trigonometriclines” as the values of the trigonometric functions of a particular angle. The term “arcs” is an-other older word referring to the values of the inverse trigonometric functions, or the signed anglementioned earlier. In the case of a unit circle, this coincides with the signed length of the arc.


161


https://doi.org/10.1007/978-3-030-11036-9_30


by setting, in the first, arctang x = z, and in the last two, arcsin x = z, or arccos x = z.Observe also that, if we denote by Υ (u), Υ (u, v), Υ (u, v, w, . . . ) algebraic func-

tions of the variables u, v, w, . . . , it will suffice to let ex = z to render algebraic, thedifferential expression included under the

∫sign in the integral

∫Υ (ex ) dx,(3)

and cos x = z, or sin x = z to produce the same effect on the two integrals⎧⎪⎪⎨

⎪⎪⎩

∫Υ (sin x, cos x) dx,

∫Υ (sin x, sin 2x, sin 3x, . . . , cos x, cos 2x, cos 3x, . . . ) dx,

(4)

the second of which has no more generality than the first, since we can replace thesines and cosines of the arcs 2x, 3x, 4x, . . . by their values in sin x and cos x, derivedfrom equations of the form2

cos nx + √−1 sin nx = (cos x + √−1 sin x

)n,

cos nx − √−1 sin nx = (cos x − √−1 sin x

)n.

Add that, if, in the first of the integrals in (4), we equate sin x, not to z, but to ±z12 ,

this integral will take the very simple form∫

Υ[±z

12 , (1 − z)

12

] ±dz

2z12 (1 − z)

12

.(5)

We will have, for example, in designating by μ, ν two constant quantities,∫

sinμ x cosν x dx = ±1

2

∫z

μ−12 (1 − z)

ν−12 dz.(6)

Finally, note that, by supposing known values of the integrals in (3) and (4), we willeasily deduce those of the following

∫Υ (eax ) dx,(7)

⎧⎪⎪⎨

⎪⎪⎩

∫Υ (sin bx, cos bx) dx,

∫Υ (sin bx, sin 2bx, sin 3bx, . . . , cos bx, cos 2bx, cos 3bx, . . . ) dx;

(8)

2These expressions represent de Moivre’s formula; named for Abraham de Moivre (1667–1754),a French mathematician of Newton’s time.

THIRTIETH LECTURE. 163

then, it will suffice to divide the functions obtained by a or by b, after having re-placed x by ax or by bx .

Now, let P, z be two functions of x, the first of which remains algebraic, and thesecond of which has an algebraic derivative, z′. If, by setting

∫P dx = Q,

∫Qz′ dx = R,

∫Rz′ dx = S,

. . . . . . . . . . . . ,

we obtain for Q, R, S, . . . , known functions of the variable x, we will determinewithout difficulty, with the help of several integration by parts,

∫Pzn dx,(9)

n being an integer number. In fact, we will find successively∫

Pzn dx = Qzn − n∫

Qz′zn−1 dx,

∫Qz′zn−1 dx = Rzn−1 − (n − 1)

∫Rz′zn−2 dx,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

and as a result,∫

Pzn dx = Qzn − n Rzn−1 + n(n − 1)Szn−2 − · · · + C.(10)

When the function z is reduced to a single term, it is necessarily presented underone of the two forms (see the twenty-eighth lecture)

Al[Υ (x)

], A arctangΥ (x),

A designating a constant quantity and Υ (x) an algebraic function of x .

Examples. – If we suppose the function P reduced to unity, and the function z toone of the following

lx, arcsin x, arccos x,

l(x +

√x2 + 1

), . . . ,


we will derive from formula (10)∫

(lx)n dx = x(lx)n[

1 − n

lx+ n(n − 1)

(lx)2− · · · ± n(n − 1) · · · 3 · 2 · 1

(lx)n

]

+ C,(11)

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

∫(arcsin x)n dx = (arcsin x)n

[

x + n√1 − x2

arcsin x

− n(n − 1)x

(arcsin x)2− n(n − 1)(n − 2)

√1 − x2

(arcsin x)3+ · · ·

]

+ C,

(12)

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

∫(arccos x)n dx = (arccos x)n

[

x − n√1 − x2

arccos x

− n(n − 1)x

(arccos x)2+ n(n − 1)(n − 2)

√1 − x2

(arccos x)3+ · · ·

]

+ C,

(13)

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

∫ [l(x +

√x2 + 1

)]ndx =

[l(x +

√x2 + 1

)]n{

x − n√

x2 + 1

l(x + √

x2 + 1)

+ n(n − 1)x[l(x + √

x2 + 1)]2 − n(n − 1)(n − 2)

√x2 + 1

[l(x + √

x2 + 1)]3 + · · ·

}

+ C,

(14)

etc. If we were to assume P = xa−1, and z = lx, we would find⎧⎪⎪⎪⎨

⎪⎪⎪⎩

∫xa−1(lx)n dx = xa

a(lx)n

[

1 − n

alx+ n(n − 1)

a2(lx)2− · · ·

± n(n − 1) · · · 3 · 2 · 1an(lx)n

]

+ C.

(15)

When we substitute z for x, the preceding formulas become∫

znez dz = znez

[

1 − n

z+ n(n − 1)

z2− · · · ± n(n − 1) · · · 3 · 2 · 1

zn

]

+ C,(16)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

∫zn cos z dz = zn

{

sin z

[

1 − n(n − 1)

z2+ · · ·

]

+ cos z

[n

z− n(n − 1)(n − 2)

z3+ · · ·

] }

+ C,

(17)

THIRTIETH LECTURE. 165

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

−∫

zn sin z dz = zn

{

cos z

[

1 − n(n − 1)

z2+ · · ·

]

− sin z

[n

z− n(n − 1)(n − 2)

z3+ · · ·

]}

+ C,

(18)

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

∫zn

(ez + e−z

2

)

dz = zn{

ez − e−z

2

[

1 + n(n − 1)

z2+ · · ·

]

− ez + e−z

2

[n

z+ n(n − 1)(n − 2)

z3+ · · ·

]}

+ C,

(19)

∫zneaz dz = zneaz

a

[

1 − n

az+ n(n − 1)

a2z2− · · · ± n(n − 1) · · · 3 · 2 · 1

anzn

]

+ C.(20)

We could directly establish these last formulas with the help of several integrationby parts that we would perform in a manner to continually diminish the exponentn, to make it finally disappear. Thus, for example, formula (20) is deduced from theequations

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

∫zneaz dz = zneaz

a− n

a

∫zn−1eaz dz,

∫zn−1eaz dz = zn−1eaz

a− n − 1

a

∫zn−2eaz dz,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(21)

A similar remark applies to all the integrals that we would deduce from the integral(10), assumed known, by substituting z for x .

Integration by parts can again be used to establish the values of∫

zneaz cos bz dz,∫

zneaz sin bz dz,(22)

a, b denoting constant quantities, and n an integer number. Thus, for example, wewill obtain the general values of the two integrals

∫eaz cos bz dz,

∫eaz sin bz dz by

adding arbitrary constants to the values of these same integrals, derived from theequations

∫eaz cos bz dz = eaz cos bz

a+ b

a

∫eaz sin bz dz,

∫eaz sin bz dz = eaz sin bz

a− b

a

∫eaz cos bz dz.


Moreover, the establishment of the integrals in (22) can be simplified by means ofthe following considerations.

Since we have (see the end of the fifth lecture)

d(cos x + √−1 sin x

) = (cos x + √−1 sin x

)dx

√−1,

we conclude

d[eaz( cos bz + √−1 sin bz

)] = (a + b

√−1)eaz( cos bz + √−1 sin bz

)dz,(23)

∫eaz

(cos bz + √−1 sin bz

)dz = eaz


)

a + b√−1

+ C,(24)

C admitting imaginary values. This granted, it is clear that the formulas in (21) and(20), which are a necessary result, will still remain valid if we replace the exponen-tial eaz by the product

eaz(cos bz + √−1 sin bz

),

and the divisor a bya + b

√−1.

We will have, therefore,⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫zneaz


)dz

= zneaz(cos bz + √−1 sin bz

)

a + b√−1

[

1 − n(a + b

√−1)z

+ · · ·

± n(n − 1) · · · 3 · 2 · 1(a + b

√−1)n

zn

]

+ C.

(25)

If we reduce the second member of this last equation to the form u + v√−1, u

and v denoting real quantities, these quantities will be precisely the values of theintegrals in (22). The two formulas which will determine these values will include,as a particular case, equations (16), (17), (18), and (20). Moreover, they will lead toequation (19) and will be reduced, if we suppose n = 0, to the following two

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

∫eaz cos bz dz = a cos bz + b sin bz

a2 + b2eaz + C,

∫eaz sin bz dz = a sin bz − b cos bz

a2 + b2eaz + C.

(26)

THIRTY-FIRST LECTURE.

ON THE DETERMINATIONAND THE REDUCTION OF INDEFINITEINTEGRALS IN WHICH THE FUNCTION UNDERTHE

∫SIGN IS THE PRODUCT OF TWO FACTORS

EQUAL TO CERTAIN POWERS OF SINESAND OF COSINES OF THE VARIABLE.

Let μ, ν be two constant quantities, and consider the integral∫

sinμ x cosν x dx .(1)

If we set sin2 x = z, or sin x = ±z12 , this integral will become

±1

2

∫z

μ−12 (1 − z)

ν−12 dz.(2)

Therefore, it can easily be determined (see the twenty-ninth lecture), when the nu-merical values of the two exponents μ−1

2 , ν−12 , and of their sum

μ + ν − 2

2,

are reduced to three rational numbers, of which one will be an integer number. Thisis what will necessarily happen whenever the quantities μ, ν have integer numericalvalues.

In all cases, we can at least reduce the establishment of the integral (1) or (2)to those of several other integrals of the same type, but in which the exponents ofsin x and cos x, or of z and 1− z, will no longer be the same. To achieve this, it willsuffice to once again employ formula (5) of the twenty-ninth lecture, namely,

∫uv dlv = uv −

∫uv dlu,(3)

by supposing the functions u, v proportional to certain powers of two of the threequantities z, 1−z, 1−z

z , or to what amounts to the same thing, of two of the followingthree

sin x, cos x,sin x

cos x= tang x = 1

cot x.(4)


167


https://doi.org/10.1007/978-3-030-11036-9_31


Consider, to settle the ideas, that we want to reduce the integral (1). We will beginby substituting into this integral the value of dx derived from one of the equations

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

dl(sin x) = cos x dx

sin x,

dl(cos x) = − sin x dx

cos x,

dl(tang x) = −dl(cot x) = dx

sin x cos x;

(5)

then, we will deduce from formula (3): 1◦ by supposing u proportional to a powerof sin x , and v to a power of cos x ,∫

sinμ x cosν x dx =∫

− sinμ−1 x cosν+1 x dl(cos x)

=∫ − sinμ−1 x

ν + 1cosν+1 x dl

(cosν+1 x

)

= − sinμ−1 x cosν+1 x

ν + 1+

∫sinμ−1 x cosν+1 x

ν + 1dl

(sinμ−1 x

),

∫sinμ x cosν x dx = − sinμ−1 x cosν+1 x

ν + 1+ μ − 1

ν + 1

∫sinμ−2 x cosν+2 x dx;(6)

2◦ by supposing u proportional to a power of cos x, and v to a power of sin x,

∫sinμ x cosν x dx = sinμ+1 x cosν−1 x

μ + 1+ ν − 1

μ + 1

∫sinμ+2 x cosν−2 x dx;(7)

3◦ by supposing u proportional to a power of tang x, and v to a power of cos x,

∫sinμ x cosν x dx =

∫− sinμ−1 x cosν+1 x dl(cos x)

=∫ −tangμ−1 x

μ + νcosμ+ν x dl

(cosμ+ν x

)

= − sinμ−1 x cosν+1 x

μ + ν+

∫sinμ−1 x cosν+1 x

μ + νdl

(tangμ−1 x

),

∫sinμ x cosν x dx = − sinμ−1 x cosν+1 x

μ + ν+ μ − 1

μ + ν

∫sinμ−2 x cosν x dx;(8)

THIRTY-FIRST LECTURE. 169

4◦ by supposing u proportional to a power of cot x, and v to a power of sin x,

∫sinμ x cosν xdx= sinμ+1 x cosν−1 x

μ+ ν+ν − 1

μ+ν

∫sinμ x cosν−2 xdx;(9)

5◦ by supposing u proportional to a power of cos x, and v to a power of tang x,

∫sinμ x cosν x dx =

∫sinμ+1 x cosν+1 x dl(tang x)

=∫

cosμ+ν+2 x

μ + 1tangμ+1 x dl

(tangμ+1 x

)

= sinμ+1 x cosν+1 x

μ + 1−

∫sinμ+1 x cosν+1 x

μ + 1dl

(cosμ+ν+2 x

),

∫sinμ x cosν xdx = sinμ+1 x cosν+1 x

μ + 1+μ+ν+2

μ + 1

∫sinμ+2 x cosν xdx;(10)

6◦ by supposing u proportional to a power of sin x, and v to a power of cot x,

∫sinμ x cosν xdx = − sinμ+1 x cosν+1 x

ν + 1+μ + ν+2

ν + 1

∫sinμ x cosν+2 xdx .(11)

With the help of formulas (6), (7), (8), (9), (10), (11), we can always transform theintegral (1) into another integral of the same type, but in which each of the quantitiessin x, cos x has an exponent contained between the limits −1,+1. In fact, to attainthis goal, it will suffice to employ formulas (8) and (9), or at least one betweenthem, once or several times in sequence, if the exponents μ and ν are positive, or if,one of them being positive, the other is contained between the limits 0,−1. On theother hand, we must employ formulas (10) and (11) if the exponents μ and ν areboth negative, or if, one of them being negative, the other is contained between thelimits 0 and 1. Finally, if one of the two exponents being positive, but greater thanunity, and the other is negative, but less than −1, we will make use of formula (6)or formula (7) for the simultaneous reduction of the numerical values of these twoexponents, until one of them is found replaced by a quantity contained between thelimits −1 and +1.

In the particular case where we suppose μ+ν = 0, equations (6) and (7) become

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫tangμ x dx = tangμ−1 x

μ − 1−

∫tangμ−2 x dx,

∫cotν x dx = −cotν−1 x

ν − 1−

∫cotν−2 x dx .

(12)


When the exponents μ and ν have integer numerical values, then, by operatingas has been said above, we finish by reducing each of them to one of the threequantities +1, 0,−1, and the integral (1) is necessarily found replaced by one of thefollowing nine

∫dx = x + C,

∫sin x dx = − cos x + C,

∫cos x dx = sin x + C,

∫sin x cos x dx = 1

2sin2 x + C,

∫sin x dx

cos x= −1

2l(cos2 x

) + C,

∫cos x dx

sin x= 1

2l(sin2 x

) + C,

∫dx

cos x sin x= 1

2l(tang2 x

) + C,

∫dx

sin x=

∫ 12 dx

sin x2 cos

x2

= 1

2l[tang2

( x

2

)]+ C,

∫dx

cos x=

∫d

(x + 1

2π)

sin(x + 1

2π) =1

2l[tang2

( x

2+ π

4

)]+ C.

If we apply these principles to the determination of the integrals∫

sinn x dx,

∫cosn x dx,

∫sinn x

cosn xdx,

∫cosn x

sinn xdx,

∫dx

cosn x,

∫dx

sinn x,

n being an integer number, we will find:1◦ by supposing n even,∫

sinn x dx = −cos x

n

[sinn−1 x + n − 1

n − 2sinn−3 x + · · ·

+ 3 · 5 · · · (n − 3)(n − 1)

2 · 4 · · · (n − 4)(n − 2)sin x

]+ 1 · 3 · · · (n − 3)(n − 1)

2 · 4 · · · (n − 2)nx + C,

∫cosn x dx = sin x

n

[cosn−1 x + n − 1

n − 2cosn−3 x + · · ·

+ 3 · 5 · · · (n − 3)(n − 1)

2 · 4 · · · (n − 4)(n − 2)cos x

]+ 1 · 3 · · · (n − 3)(n − 1)

2 · 4 · · · (n − 2)nx + C,

THIRTY-FIRST LECTURE. 171

∫tangn x dx = tangn−1 x

n − 1− tangn−3 x

n − 3+ tangn−5 x

n − 5− · · · ± tang x ∓ x + C,

∫cotn x dx = −cotn−1 x

n − 1+ cotn−3 x

n − 3− cotn−5 x

n − 5+ · · · ± cot x ∓ x + C,

∫secn x dx = sin x

n − 1

[secn−1 x + n − 2

n − 3secn−3 x + · · ·

+ 2 · 4 · · · (n − 4)(n − 2)

1 · 3 · · · (n − 5)(n − 3)sec x

]+ C,

∫cosecn x dx = − cos x

n − 1

[cosecn−1 x + n − 2

n − 3cosecn−3 x + · · ·

+ 2 · 4 · · · (n − 4)(n − 2)

1 · 3 · · · (n − 5)(n − 3)cosec x

]+ C;

2◦ by supposing n odd,1

∫sinn x dx = −cos x

n

[sinn−1 x + n − 1

n − 2sinn−3 x

+ (n − 1)(n − 3)

(n − 2)(n − 4)sinn−5 x + · · · + 2 · 4 · · · (n − 3)(n − 1)

1 · 3 · · · (n − 4)(n − 2)

]+ C,

∫cosn x dx = sin x

n

[cosn−1 x + n − 1

n − 2cosn−3 x

+ (n − 1)(n − 3)

(n − 2)(n − 4)cosn−5 x + · · · + 2 · 4 · · · (n − 3)(n − 1)

1 · 3 · · · (n − 4)(n − 2)

]+ C,

∫tangn xdx = tangn−1x

n − 1− tangn−3x

n − 3+ tangn−5x

n − 5− · · · ± tang2x

2± 1

2l(cos2 x

)+C,

∫cotn xdx = − cotn−1 x

n − 1+cotn−3 x

n − 3−cotn−5 x

n − 5+· · · ∓ cot2 x

2∓ 1

2l(sin2 x

)+C,

1A typographical error has been corrected here. The original 1823 and the 1899 reprint editionsboth read

∫sinn x dx = − cos x

n

[sinn−1 x + n − 1

n − 2sinn−3 x + (n − 1)(n − 3)

(n − 2)(n − 4)sinn−3 x + · · ·

]+ C.


∫secn x dx = sin x

n − 1

[secn−1 x + n − 2

n − 3secn−3 x + · · ·

+3 · 5 · · · (n − 2)

2 · 4 · · · (n−3)sec2 x

]+1 · 3 · · · (n−2)

2·4 · · · (n−1)

1

2l[tang2

( x

2+π

4

)]+C,

∫cosecn x dx = − cos x

n − 1

[cosecn−1x + n − 2

n − 3cosecn−3x + · · ·

+3 · 5 · · · (n − 2)

2 · 4 · · · (n − 3)cosec2x

]+1 · 3 · · · (n − 2)

2 · 4 · · · (n − 1)

1

2l[tang2

( x

2

)]+C.

We indicate, in finishing, several methods which can be used, like the preceding,for the reduction or for the determination of the integral

∫sin±m x cos±n x dx,

m, n being two integer numbers. First, it is clear that we will reduce the integral∫

sin−m x cos−n x dx

to other simpler integrals by multiplying the function under the∫sign once or sev-

eral times by sin2 x +cos2 x = 1.Moreover, we can render the differential expression

sin±m x cos±n x dx

rational: 1◦ in the case where n is an odd number, by setting sin x = z; 2◦ in thecase where m is an odd number, by setting cos x = z. Finally, note that we will veryeasily obtain the values of the integrals

∫sinm x dx,

∫cosn x dx,

∫sinm x cosn x dx,

fromwhich we will have expanded sinm x, cosn x, and sinm x cosn x into linear func-tions of sin x, sin 2x, sin 3x, . . ., cos x, cos 2x, cos 3x, . . .with the help of the formu-las established in Chapter VII of Algebraic Analysis.2

2Cauchy directly references this earlier book a total of seventeen times throughout his Calculinfinitésimal. There are many more indirect references as well. However, these direct referencesin particular suggest Cauchy did indeed expect his students to have a copy of Cours d’analyse oftheir own, or to visit the library at the École Polytechnique after class.

THIRTY-SECOND LECTURE.

ON THE TRANSITION OF INDEFINITEINTEGRALS TO DEFINITE INTEGRALS.

To integrate the equation

dy = f (x) dx,(1)

or the differential expression f (x) dx, starting from x = x0, is to find a continuousfunction of x which has the double property of giving for a differential, f (x) dx , andvanishing for x = x0. This function, before being included in the general formula

∫f (x) dx =

∫ x

x0

f (x) dx + C,

will necessarily be reduced to the integral∫ x

x0f (x) dx , if the function f (x) is itself

continuous with respect to x between the two limits of this integral. Conceive nowthat, the two functions ϕ(x) and χ(x) being continuous between these limits, thegeneral value of y derived from equation (1) is presented under the form

ϕ(x) +∫

χ(x) dx .

The function sought will obviously be equal to

ϕ(x) − ϕ(x0) +∫ x

x0

χ(x) dx .

By starting from this remark, we will see without difficulty what happens to theformulas established in the preceding lectures when we subject the two members ofeach of them to vanish for a given value of x . Thus, for example, we will easilyrecognize that equations (9) and (12) of the twenty-seventh lecture, namely,

∫f (x) dx =

∫ϒ (z) dz,


173


https://doi.org/10.1007/978-3-030-11036-9_32


and ∫u dv = uv −

∫v du or

∫uv′ dx = uv −

∫vu′ dx,

lead to the following∫ x

x0

f (x) dx =∫ z

z0

ϒ (z) dz(2)

and ∫ x

x0

uv′ dx = uv − u0v0 −∫ x

x0

vu′ dx,(3)

z0, u0, and v0 denoting the values of z, u, and v corresponding to x = x0. If, informulas (2) and (3), we set x = X , we will find, by calling Z , U, V the valuescorresponding to z, u, v,

∫ X

x0

f (x) dx =∫ Z

z0

ϒ (z) dz(4)

and∫ X

x0

uv′ dx = U V − u0v0 −∫ X

x0

vu′ dx .(5)

Equations (4) and (5) are those that we should substitute to formulas (9) and (12)of the twenty-seventh lecture when applying integration by substitution or by parts,to the evaluation or to the reduction of definite integrals, while the integrals of thetype deduced from immediate integration or by decomposition are given by formula(18) of the twenty-sixth lecture or by formula (2) of the twenty-third. These princi-ples being accepted, the methods explained in the preceding lectures can be used todetermine a large number of definite integrals, among which I will cite some of themost remarkable.

If we designate by m an integer number, by a, β, μ, ν positive quantities, byα, A, B, C, . . . any quantities, and finally, by ε an infinitely small number, we willderive from the formulas established in the twenty-seventh and twenty-eighth lec-tures ∫ 1

0xa−1 dx = 1

a,

∫ 1

0x−a−1 dx = ∞,

∫ ∞

0e−x dx = 1,

∫ ∞

0eax dx = ∞,

∫ ∞

0e−ax dx = 1

a,

∫ 1

0(A + Bx + Cx2 + · · · ) dx = A + B

2+ C

3+ · · · ,

∫ 1

0

xm − 1

x − 1dx = 1 + 1

2+ 1

3+ · · · + 1

m,

THIRTY-SECOND LECTURE. 175

∫ ∞

0

dx

1 + x2= π

2,

∫ ∞

−∞dx

x2 + a2= π

a,

∫ 1εν

− 1εμ

x dx

x2 + a2= l

(μ

ν

),

∫ 1ε

− 1ε

x dx

x2 + a2= 0,

∫ a

0

dx√a2 − x2

= π

2,

∫ ∞

−∞dx

(x − α)2 + β2= π

β,

∫ 1εν

− 1εμ

(x − α) dx

(x − α)2 + β2= l

(μ

ν

),

∫ 1ε

− 1ε

(x − α) dx

(x − α)2 + β2= 0,

∫ 1εν

− 1εμ

(A − B

√−1

x − α − β√−1

+ A + B√−1

x − α + β√−1

)dx = 2A l

(μ

ν

)+ 2π B,

∫ 1ε

− 1ε

(A − B

√−1

x − α − β√−1

+ A + B√−1

x − α + β√−1

)dx = 2π B.

Moreover, if we generally represent by ϒ (x)

F(x)a rational fraction whose denominator

does not vanish for any real value of x, by x1, x2, . . . the imaginary roots of theequation F(x) = 0 in which the coefficient of

√−1 is positive, and by A1− B1√−1,

A2−B2√−1, . . . the values of the fraction ϒ (x)

F(x)corresponding to these roots, wewill

obtain the formula∫ 1

εν

− 1εμ

ϒ (x)

F(x)dx = 2(A1 + A2 + · · · )l

(μ

ν

)+ 2π(B1 + B2 + · · · ).(6)

The second member1 of this formula will cease to contain the arbitrary factor l(

μ

ν

),

and we will have, by consequence,∫ ∞

−∞ϒ (x)

F(x)dx = 2π(B1 + B2 + · · · )(7)

whenever the sum A1 + A2 +· · · vanishes. Now, this condition will be fulfilled if thedegree of F(x) exceeds by at least two units the degree of ϒ (x). We reach the sameresult by starting from the remark which terminates the twenty-fifth lecture.

If the degree of the function F(x)were to exceed that ofϒ (x) by only one unit, theintegral

∫ ∞−∞

ϒ (x)

F(x)dx would become indeterminate, and its general value, given by

equation (6), would contain the arbitrary constant μ

ν. But, by reducing this arbitrary

constant to unity, we would end up with equation (7), which in this case would onlyprovide the principal value of the integral in question. Add that this principal valuewould remain the same, if, besides the imaginary roots x1, x2, . . . ,2 the equationF(x) = 0would admit real roots. The reason is that all integrals of the form

∫ ∞−∞

A dxx±a

have null principal values.

1Recall Cauchy uses this phrase when referring to the right-hand side of an equation.2Both the 1823 and the 1899 editions have x here instead of x1.


Examples. – Let m and n be two integer numbers, m being < n. If we make2m+12n = a, we will find3

∫ ∞

−∞x2m dx

1 + x2n= 2π

2n

[sin aπ + sin 3aπ + · · · + sin (2n − 1)aπ

]

= π

n sin aπ= π

n sin (2m+1)π2n

.

We conclude, by setting z = x2n,

∫ ∞

0

za−1 dz

1 + z= 2n

∫ ∞

0

x2m dx

1 + x2n= n

∫ ∞

−∞x2m dx

1 + x2n= π

sin aπ.(8)

Similarly, by reducing each indeterminate integral to its principal value, we willfind

∫ ∞

−∞x2m dx

1 − x2n= 2π

2n

[sin 2aπ + sin 4aπ + · · · + sin (2n − 2)aπ

]

= π

n tang aπ= π

n tang (2m+1)π2n

,

∫ ∞

0

za−1 dz

1 − z= 2n

∫ ∞

0

x2m dx

1 − x2n= n

∫ ∞

−∞x2m dx

1 − x2n= π

tang aπ.(9)

We will also deduce from formulas established in the twenty-ninth and thirtiethlectures

∫ ∞

0

xm−1 dx

(1 + x)n= m − 1

n − m

∫ ∞

0

xm−2 dx

(1 + x)n

= (m − 1) · · · 3 · 2 · 1(n − m) · · · (n − 3)(n − 2)

∫ ∞

0

dx

(1 + x)n

= 1 · 2 · 3 · · · (m − 1) × 1 · 2 · 3 · · · (n − m − 1)

1 · 2 · 3 · · · (n − 1),

∫ ∞

0

dy

(1 + y2)n= 2n − 3

2n − 2

∫ ∞

0

dy

(1 + y2)n−1

= 1 · 3 · 5 · · · (2n − 3)

2 · 4 · 6 · · · (2n − 2)

∫ ∞

0

dy

1 + y2

= 1 · 3 · 5 · · · (2n − 3)

2 · 4 · 6 · · · (2n − 2)

π

2,

3The 1899 edition misprints this equation. It reads,∫ ∞−∞

x2m dx1−x2n = · · · .

THIRTY-SECOND LECTURE. 177

∫ ∞

0zne−z dz = 1 · 2 · 3 · · · n,

∫ ∞

0zne−az dz = 1 · 2 · 3 · · · n

an+1,

∫ ∞

0zne−az


)dz = 1 · 2 · 3 · · · n(

a + b√−1

)n+1 , 4

∫ ∞

0zne−az cos bz dz = 1 · 2 · 3 · · · n(

a2 + b2) n+1

2

cos

[(n + 1) arctang

b

a

],

∫ ∞

0e−az cos bz dz = a

a2 + b2,

∫ ∞

0zne−az sin bz dz = 1 · 2 · 3 · · · n(

a2 + b2) n+1

2

sin

[(n + 1) arctang

b

a

],

∫ ∞

0e−az sin bz dz = b

a2 + b2.

Finally, we will derive from formulas established in the thirty-first lecture: 1◦ bysupposing n even,

∫ π2

0sinn x dx = 1 · 3 · 5 · · · (n − 1)

2 · 4 · 6 · · · n

π

2=

∫ π2

0cosn x dx,

∫ π4

0tangn x dx = 1

n − 1− 1

n − 3+ · · · ∓ 1

3± 1 ∓ π

4;

2◦ by supposing n odd,

∫ π2

0sinn x dx = 2 · 4 · 6 · · · (n − 1)

1 · 3 · 5 · · · (n − 2)n=

∫ π2

0cosn x dx,

∫ π4

0tangn x dx = 1

n − 1− 1

n − 3+ · · · ∓ 1

4± 1

2± 1

2l(1

2

).

The methods of integration that we have indicated often provide the means totransform a given definite integral into a simpler one. Thus, for example, regardless

4The original 1823 text has∫ ∞0 zne−az

(cos bz + sin

√−1 bz)

dz = 1·2·3···n(a+b

√−1)n+1 . This error is

corrected here and in the 1899 edition.


of the function ϒ (x), we will derive from the formulas established in the twenty-seventh lecture

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫ ∞

−∞ϒ (x ± a) dx =

∫ ∞

−∞ϒ (z) dz =

∫ ∞

−∞ϒ (x) dx,

∫ ∞

0ϒ (ax) dx = 1

a

∫ ∞

0ϒ (x) dx, . . . . . . ,

∫ ∞

0xμ−1e−ax dx = 1

aμ

∫ ∞

0xμ−1e−x dx,

∫ ∞

0

sin ax

xdx =

∫ ∞

0

sin x

xdx, . . . . . . .

(10)

When, in an integral relative to the variable x , the function under the∫

signcontains another quantity μ whose value is arbitrary, we can consider this quantityμ as a new variable and the integral itself as a function of μ. Among the functionsof this type, we should note one that Mr. Legendre5 has designated by the letter Γ,

and which is found defined for positive values of μ by the equation

Γ (μ) =∫ 1

0

(l(1

x

))μ−1

dx =∫ ∞

0zμ−1e−z dz.(11)

This function, which Euler and Mr. Legendre were much occupied,6 would satisfy,by virtue of the foregoing, the equations

⎧⎪⎨⎪⎩

Γ (1) = 1, Γ (2) = 1, Γ (3) = 1 · 2, . . . , Γ (n) = 1 · 2 · 3 · · · (n − 1),∫ ∞

0zn−1e−az dz = Γ (n)

an,

(12)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ ∞

0zn−1e−az cos bz dz = Γ (n) cos

(n arctang b

a

)(a2 + b2)

n2

,

∫ ∞

0zn−1e−az sin bz dz = Γ (n) sin

(n arctang b

a

)(a2 + b2)

n2

,

(13)

∫ ∞

0zμ−1e−az dz = Γ (μ)

aμ,

∫ ∞

0

xm−1 dx

(1 + x)n= Γ (m)Γ (n − m)

Γ (n),(14)

in which n denotes an integer number, m another integer number less than n,7 andμ any number.

5Adrien-Marie Legendre (1752–1833) was a well-knownmathematician in Paris who aided Cauchyearly in his career and had important contributions to the field, several of which are named for him.6Legendre is credited with the notation used today and by Cauchy. The function has come to beknown as the Gamma function.7The original 1823 edition has m here instead of n, an error which is corrected in the 1899 version.

THIRTY-THIRD LECTURE.

DIFFERENTIATION AND INTEGRATION UNDERTHE

∫SIGN. INTEGRATION OF DIFFERENTIAL

FORMULAS WHICH CONTAIN SEVERALINDEPENDENT VARIABLES.

Let x, y be two independent variables, f (x, y) a function of these two variables, andx0, X two particular values of x .Wewill find, by settingΔy = α dy, and employingthe notations adopted in the thirteenth lecture,

Δy

∫ X

x0

f (x, y) dx =∫ X

x0

f (x, y + Δy) dx −∫ X

x0

f (x, y) dx

=∫ X

x0

Δy f (x, y) dx;

then, in dividing by α dy, and letting α converge toward the limit zero,

d

dy

∫ X

x0

f (x, y) dx =∫ X

x0

d f (x, y)

dydx .(1)

We will also have1

d

dy

∫ x

x0

f (x, y) dx =∫ x

x0

d f (x, y)

dydx .(2)

It follows from these formulas that, to differentiate the integrals∫ X

x0f (x, y) dx,∫ x

x0f (x, y) dx with respect to y, it is sufficient to differentiate under the

∫sign the

function f (x, y). It also results that the equations⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫ X

x0

f (x, y) dx = F(y),

∫ x

x0

f (x, y) dx = F(x, y),

∫f (x, y) dx = F(x, y) + C,

(3)

1This equation will be referenced in Lecture Thirty-Five assuming f (x, y) is a well-behaved func-tion.© Springer Nature Switzerland AG 2019D. M. Cates, Cauchy’s Calcul Infinitésimal,https://doi.org/10.1007/978-3-030-11036-9_33

179


https://doi.org/10.1007/978-3-030-11036-9_33


always lead to the following2

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫ X

x0

d f (x, y)

dydx = d F(y)

dy,

∫ x

x0

d f (x, y)

dydx = d F(x, y)

dy,

∫d f (x, y)

dydx = d F(x, y)

dy+ C,

(4)

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫ X

x0

dn f (x, y)

dyndx = dn F(y)

dyn,

∫ x

x0

dn f (x, y)

dyndx = dn F(x, y)

dyn,

∫dn f (x, y)

dyndx = dn F(x, y)

dyn+ C.

(5)

Examples. – By differentiating each of the integrals∫

dx

x2 + a,

∫ ∞

0

dx

x2 + a,

∫e±ax dx,

∫ ∞

0e−ax dx,

∫ ∞

0xμ−1e−ax dx,

n times in sequence with respect to the quantity a, we will find

∫1 · 2 · · · n dx

(x2 + a)n+1= ±

dn(

1√aarctang

(x√a

))

dan+ C,

∫ ∞

0

1 · 2 · · · n dx

(x2 + a)n+1= ±π

2

dn(

1√a

)

dan= 1 · 3 · 5 · · · (2n − 1)π

2nan√

a,

∫ ∞

0

dx

(1 + x2)n+1= 1 · 3 · 5 · · · (2n − 1)

2 · 4 · 6 · · · (2n)

π

2,

∫xne±ax dx = ±dn(a−1e±ax )

dan+ C,

∫ ∞

0xne−ax dx = ±dn(a−1)

dan= 1 · 2 · 3 · · · n

an+1,

∫ ∞

0xμ+n−1e−ax dx = μ(μ + 1) · · · (μ + n − 1)

aμ+nΓ (μ),

Γ (μ + n) = μ(μ + 1) · · · (μ + n − 1)Γ (μ).

2As differentiation and integration are each limit processes, Cauchy is swapping the order in whichthese limits are taken, a process which is not always allowed.

THIRTY-THIRD LECTURE. 181

Let us now conceive that the function f (x, y) is continuous with respect to thetwo variables x and y whenever x remains contained between the limits x0, X, and ybetween the limits y0, Y . It is easy to see that, for similar values of x and of y, thesecond of the equations in (3) will lead to the following3

∫ x

x0

∫ y

y0

f (x, y) dy dx =∫ y

y0

F(x, y) dy =∫ y

y0

∫ x

x0

f (x, y) dx dy.(6)

In fact, we will derive from formula (2)4

d

dy

∫ x

x0

∫ y

y0

f (x, y) dy dx =∫ x

x0

f (x, y) dx;

then, in multiplying both members by dy, and integrating with respect to y, startingfrom y = 0, we will regain formula (6). We will have, as a result,5

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

x0

∫ y

y0

f (x, y) dy dx =∫ y

y0

∫ X

x0

f (x, y) dx dy,

∫ X

x0

∫ Y

y0

f (x, y) dy dx =∫ Y

y0

∫ X

x0

f (x, y) dx dy.

(7)

It follows from formulas (6) and (7) that, to integrate with respect to y, starting fromy = y0, the expressions∫ x

x0

f (x, y) dx,

∫ X

x0

f (x, y) dx,

multiplied by the differential dy, it is sufficient to integrate under the∫

sign, startingfrom y = y0, the function f (x, y) multiplied by this same differential.

Often, integration under the∫sign makes known the values of certain definite

integrals, even though we do not have any means to evaluate the correspondingindefinite integrals. Thus, although we do not know how to determine, in terms of x,

the indefinite integral ∫xμ − xν

lxdx

x

(μ, ν being two positive quantities), nevertheless, since we have generally, for posi-tive values of μ, ∫ 1

0xμ−1 dx = 1

μ,(8)

3The expression included here is that of the original 1823 edition. The 1899 reprint reverses theorder of the two final differentials.4The 1899 reprint contains another small change by once again reversing the order of the two end-ing differentials within the first integral. The expression included here is that of Cauchy’s original1823 edition.5The equations included here are those appearing in the original 1823 edition. The 1899 reprintreverses the order of dx and dy in each of the four integrals.


we conclude, in multiplying the two members by dμ, then integrating with respectto μ, starting from μ = ν,

∫ 1

0

xμ − xν

lxdx

x= l

(μ

ν

).(9)

Among the formulas of this type, we should again note those that we will establish.If we designate by a, b, c positive quantities, an integration under the

∫sign

relative to the quantity a, performed starting from a = c, and applied to the definiteintegrals

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫ ∞

0e−ax dx = 1

a,

∫ ∞

0e−ax cos bx dx = a

a2 + b2,

∫ ∞

0e−ax sin bx dx = b

a2 + b2,

(10)

will produce the formulas

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫ ∞

0

e−cx − e−ax

xdx = l

(a

c

),

∫ ∞

0

e−cx − e−ax

xcos bx dx = 1

2l(

a2 + b2

c2 + b2

),

∫ ∞

0

e−cx − e−ax

xsin bx dx = arctang

(a

b

)− arctang

( c

b

),

(11)

from which we will derive, by setting c = 0, and a = ∞,

∫ ∞

0

dx

x= ∞,

∫ ∞

0cos bx

dx

x= ∞,

∫ ∞

0sin bx

dx

x= π

2.(12)

Moreover, since we have, for positive values of b (see the thirty-second lecture),∫ ∞

0zb−1e−z(1+x) dz = Γ (b)

(1 + x)b,

and as a result,xa−1

(1 + x)b= 1

Γ (b)

∫ ∞

0xa−1e−zx zb−1e−z dz,

we will deduce, by supposing a and b positive, as well as b − a,

⎧⎪⎪⎨⎪⎪⎩

∫ ∞

0

xa−1 dx

(1 + x)b= Γ (a)

Γ (b)

∫ ∞

0zb−a−1e−z dz

= Γ (a) Γ (b − a)

Γ (b);

(13)

THIRTY-THIRD LECTURE. 183

then, by letting b = 1, taking for a a number of the form 2m+12n , and having regard

to the equation Γ (1) = 1, we will find(see formula (8), thirty-second lecture

)6

⎧⎪⎪⎨⎪⎪⎩

Γ (a) Γ (1 − a) = π

sin aπ,

[Γ

(12

) ]2 = π, Γ(12

) = π12 =

∫ ∞

0z− 1

2 e−z dz =∫ ∞

−∞e−x2

dx .

(14)

Now, let ϕ(x, y), χ(x, y) be two functions that work to satisfy the equation

dϕ(x, y)

dy= dχ(x, y)

dx.(15)

If we successively substitute the twomembers of this equation in the place of f (x, y)

in formula (6), we will obtain the following∫ x

x0

[ϕ(x, y) − ϕ(x, y0)

]dx =

∫ y

y0

[χ(x, y) − χ(x0, y)

]dy.(16)

This expression above remains valid whenever the functions ϕ(x, y), χ(x, y), oneand the other, remain finite and continuous with respect to the variables x and ybetween the limits of integration.

Let us now conceive that we look for a function of u which works to satisfy theequation

du = ϕ(x, y) dx + χ(x, y) dy,(17)

or to what amounts to the same thing, the following two

du

dx= ϕ(x, y),(18)

du

dy= χ(x, y).(19)

Obviously, we can only achieve this in the case where formula (15), every memberof which will be equivalent to d2u

dx dy , is found satisfied. I add that,7 by supposing thiscondition fulfilled, we will easily resolve the proposed question. In fact, let x0 andy0 be particular values of x, y, and let C be an arbitrary constant. To satisfy equation(18), it will suffice to take

u =∫ x

x0

ϕ(x, y) dx + v,(20)

6Recall Γ (μ) = ∫ ∞0 zμ−1e−zdz.

7It is somewhat unusual to find this pronoun employed within a mathematics text. This is not theonly time Cauchy has used the first person within his Calcul infinitésimal text, perhaps illustratinghow personal this effort was for Cauchy at the time.


v designating an arbitrary function of the variable y; and, since we derive fromformula (20)

du

dy=

∫ x

x0

dϕ(x, y)

dydx + dv

dy=

∫ x

x0

dχ(x, y)

dxdx + dv

dy

= χ(x, y) − χ(x0, y) + dv

dy,

it is clear that we will also satisfy equation (19), if we set

dv

dy− χ(x0, y) = 0, v =

∫χ(x0, y) dy =

∫ y

y0

χ(x0, y) dy + C.(21)

By consequence, the general value of u will be

u =∫ x

x0

ϕ(x, y) dx +∫

χ(x0, y) dy =∫ x

x0

ϕ(x, y) dx +∫ y

y0

χ(x0, y) dy + C.

(22)

When, in the preceding equations, we exchange between them the variables x, y,weobtain a second value of u which is obviously in accordance with the first, by virtueof formula (16).

We would integrate with the same ease the differential of a function of three, four,…independent variables, and we would prove, for example, that if the conditions8

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

dχ(x, y, z)

dz= dψ(x, y, z)

dy,

dψ(x, y, z)

dx= dϕ(x, y, z)

dz,

dϕ(x, y, z)

dy= dχ(x, y, z)

dx

(23)

are found fulfilled, the general value of u that works to satisfy the equation

du = ϕ(x, y, z) dx + χ(x, y, z) dy + ψ(x, y, z) dz(24)

will be

u =∫ x

x0

ϕ(x, y, z) dx +∫ y

y0

χ(x0, y, z) dy +∫ z

z0

ψ(x0, y0, z) dz + C,(25)

x0, y0, z0 denoting particular values of the variables x, y, z.

8The third expression included here is that of the 1899 reprint. The original 1823 edition containserrors which have been corrected. The original reads dϕ(x, y,z)

dx = dχ(x, y, z)dy .

THIRTY-FOURTH LECTURE.

COMPARISON OF TWO TYPES OF SIMPLEINTEGRALS WHICH RESULT IN CERTAIN CASESFROM A DOUBLE INTEGRATION.

Consider that equation (15) of the preceding lecture is satisfied. If we integrate thisequation twice, namely once with respect to x between the limits x0, X, and oncewith respect to y between the limits y0, Y, we will find

∫ X

x0

[ϕ(x, Y ) − ϕ(x, y0)

]dx =

∫ Y

y0

[χ(X, y) − χ(x0, y)

]dy.(1)

This last formula establishes a relationship worthy of remark between the integralsthat it contains. But, it ceases to be correct when the functions ϕ(x, y), χ(x, y) be-come infinite for one or several systems of values of x and of y contained betweenthe limits x = x0, x = X, y = y0, y = Y. First, imagine that these systems are re-duced to a single one, namely x = a, y = b. In this particular case, the expressionsdeduced by a double integration of the two members of formula (15) (thirty-thirdlecture) could differ, one from the other. But, they will always again become equal,if in the calculation we have taken care to replace each integral relative to x by itsprincipal value. This observation is sufficient to show by what manner equation (1)should be modified. In fact, if we denote by ε an infinitely small number, we willfind, in the admitted hypothesis,

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ a−ε

x0

[ϕ(x, Y ) − ϕ(x, y0)

]dx +

∫ X

a+ε

[ϕ(x, Y ) − ϕ(x, y0)

]dx

=∫ Y

y0

[χ(X, y) − χ(a + ε, y) + χ(a − ε, y) − χ(x0, y)

]dy;

(2)

then, we will deduce, by letting ε converge toward the limit zero,1

∫ X

x0

[ϕ(x, Y ) − ϕ(x, y0)

]dx =

∫ Y

y0

[χ(X, y) − χ(x0, y)

]dy − Δ,(3)

1An error in equation (3) occurs in both the 1823 and 1899 editions. Each of these texts reads∫ Xx0

[ϕ(x, Y ) − ϕ(x, y0)

]dx= ∫ X

x0

[χ(X, y) − χ(x0, y)

]dy−Δ.This error has been corrected here.


185


https://doi.org/10.1007/978-3-030-11036-9_34


the value of Δ being determined by the formula

Δ = lim∫ Y

y0

[χ(a + ε, y) − χ(a − ε, y)

]dy.(4)

In the general case, Δ will be the sum of several terms similar to the second memberof equation (4).

Example. – If we set

ϕ(x, y) = −y

x2 + y2, χ(x, y) = x

x2 + y2,

x0 = −1, X = 1,

y0 = −1, Y = 1,

equations (3) and (4) will become

∫ 1

−1

−2 dx

1 + x2=

∫ 1

−1

2 dy

1 + y2− Δ, Δ = lim

∫ 1

−1

2 ε dy

ε2 + y2= 2π.

It is easy to see that the functions ϕ(x, y), χ(x, y) will satisfy equation (15) ofthe thirty-third lecture, if we have

ϕ(x, y) dx + χ(x, y) dy = f (u) du,

and as a result,

ϕ(x, y) = f (u)du

dx, χ(x, y) = f (u)

du

dy,(5)

u denoting any function of the variables x, y.

It is again easy to ensure that formulas (1) and (3) remain valid under the statedconditions, even in the case where the functions ϕ(x, y), χ(x, y) become imaginary.For example, conceive that, the function f (x) being algebraic, we set

u = x + y√−1.

We will derive from the equations in (5)

ϕ(x, y) = f(x + y

√−1), χ(x, y) = √−1 f

(x + y

√−1),

and from formula (3)⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

x0

[f(x + Y

√−1) − f

(x + y0

√−1)]

dx

= √−1∫ Y

y0

[f(X + y

√−1) − f

(x0 + y

√−1)]

dy − Δ.

(6)

THIRTY-FOURTH LECTURE. 187

In this last equation, Δ will vanish if the function f(x + y

√−1)remains finite and

continuous for all values of x and of y contained between the limits x = x0, x =X, y = y0, y = Y . But, if, between these same limits, the function f

(x + y

√−1)

becomes infinite for the system of values x = a, y = b, then the value of Δ will begiven by equation (4); and, if we make, for brevity,

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

(x − a − b

√−1)

f (x) = F(x),

y = b + εx,

z0 = −b − y0ε

,

Z = Y − b

ε,

(7)

we will find⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

Δ = √−1 lim∫ Y

y0

[f(a + ε + y

√−1) − f

(a − ε + y

√−1)]

dy

= √−1 lim∫ Z

z0

{F[a + ε + (b + εz)

√−1]

1 + z√−1

− F[a − ε + (b + εz)

√−1]

−1 + z√−1

}dz.

(8)

Now, let

F[a + ε + (b + εz)

√−1]

1 + z√−1

− F[a − ε + (b + εz)

√−1]

−1 + z√−1

= ϖ(ε) + √−1ψ(ε),(9)

⎧⎪⎨⎪⎩

ϖ(ε) − ϖ(0)

ε= α,

ψ(ε) − ψ(0)

ε= β,

(10)

ϖ(ε), ψ(ε), and as a result, α, β, being real quantities. Suppose, moreover, that Yexceeds y0, and that the functions F

(x + y

√−1),F′(x + y

√−1)remain finite and

continuous with respect to the variables x and y between the limits x0, X, y0, Y .Since we will have, by virtue of formula (9),

ϖ ′(ε) + √−1ψ ′(ε) = F′[a + ε + (b + εz)√−1

] − F′[a − ε + (b + εz)√−1

]= F′(a + ε + y

√−1) − F′(a − ε + y

√−1),

it is clear that the numerical values of the quantities ϖ ′(ε), ψ ′(ε) will always re-main very small, as well as those of the two quantities α, β, each of which can bepresented under the form ϖ ′(θε) or ψ ′(θε), θ denoting a number less than unity.


This granted, we will find

lim∫ Z

z0

ε(α + β

√−1)

dz = lim∫ Y

y0

(α + β

√−1)

dy = 0,

lim∫ Z

z0

[ϖ(ε) + √−1ψ(ε)

]dz =

∫ Z

z0

[ϖ(0) + √−1ψ(0)

]dz;

then, by letting ϒ = F(a + b

√−1) = lim ε f

(a + b

√−1 + ε),

Δ = √−1∫ ∞

−∞[ϖ(0) + √−1ψ(0)

]dz = 2ϒ

√−1∫ ∞

−∞dz

1 + z2= 2πϒ

√−1.(11)

If we have y0 = b or Y = b, the integral relative to z in formula (11) should nolonger be taken between the limits z = 0, z = ∞, but rather, between the limitsz = −∞, z = 0, and as a result, the value of Δ would reduce to πϒ

√−1. In thesame hypothesis, the first member of equation (6) would be the principal value ofan indeterminate integral.

It is again essential to observe that a + b√−1 represents a root of the equation

f (x) = ±∞.(12)

If this equation were to admit several roots in which the real parts2 were containedbetween the limits x0, X, and the coefficients of

√−1 between the limits y0, Y, then,in designating by x1, x2, . . . , xm these same roots, and by ϒ1, ϒ2, . . . , ϒm the actualvalues that receive the products

(x − x1) f (x), (x − x2) f (x), . . . , (x − xm) f (x)

while their first factors vanish, we would find

Δ = 2π(ϒ1 + ϒ2 + · · · + ϒm)√−1.(13)

Add that each of the terms ϒ1, ϒ2, . . . , ϒm must be reduced to half whenever, in thecorresponding root, the coefficient of

√−1 coincides with one of the limits y0, Y .When the function f

(x + y

√−1)vanishes: 1◦ for x = ±∞, regardless of y; 2◦

for y = ∞, regardless of x, then by taking x0 = −∞, X = +∞, y0 = 0, Y = ∞,

we derive from formula (6)∫ ∞

−∞f (x) dx = Δ.(14)

When the function f (x) is presented under the form ϒ (x)

F(x), and those of the terms

ϒ1, ϒ2, . . . , ϒm, which do not vanish, correspond to all the roots of the equation

F(x) = 0,(15)

2The 1899 reprint omits the word “real.”

THIRTY-FOURTH LECTURE. 189

the expression Δ can obviously be written as follows

Δ = 2π

[ϒ (x1)

F ′(x1)+ ϒ (x2)

F ′(x2)+ · · · + ϒ (xm)

F ′(xm)

] √−1,(16)

and equation (14) becomes∫ ∞

−∞ϒ (x)

F(x)dx = 2π

[ϒ (x1)

F ′(x1)+ ϒ (x2)

F ′(x2)+ · · · + ϒ (xm)

F ′(xm)

] √−1.(17)

In the second member of this expression, we should only admit the real roots ofequation (15) along with the imaginary roots in which the coefficient of

√−1 ispositive, by taking care to reduce to half all the terms which correspond to realroots. This granted, we will find, for F(x) = 1 + x2, x1 = √−1,

∫ ∞

−∞ϒ (x)

1 + x2dx = πϒ

(√−1);(18)

and for F(x) = 1 − x2, x1 = −1, x2 = +1,∫ ∞

−∞ϒ (x)

1 − x2dx = π

2

[ϒ (−1) − ϒ (1)

]√−1.(19)

This last formula simply gives the principal value of the integral that it contains.

Examples. – Let μ be a number contained between 0 and 2. If we set

ϒ (x) = ( − x√−1

)μ−1,

the imaginary expression

ϒ(x + y

√−1) = (

y − x√−1

)μ−1

will maintain a unique and specified value as long as y remains positive (see Alge-braic Analysis, Chapter VII); and, we will derive from formulas (18) and (19)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ ∞

−∞

( − x√−1

)μ−1

1 + x2dx =

[( − √−1)μ−1 + (√−1

)μ−1],

∫ ∞

0

xμ−1 dx

1 + x2= π,

∫ ∞

0

xμ−1 dx

1 + x2= π

2 sin(12μπ

) ,

(20)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ ∞

−∞

( − x√−1

)μ−1

1 − x2dx = π

2

[(√−1)μ + ( − √−1

)μ],

∫ ∞

0

xμ−1 dx

1 − x2= π cos

(12μπ

)2 sin

(12μπ

) = π

2 tang(12μπ

) .

(21)


If, in the last of the equations in (20)3 and the last of the equations in (21), we replacex2 by z, and μ by 2a, we will reproduce formulas (8) and (9) of the thirty-secondlecture, which are found thus demonstrated, along with the first of the equations in(14) of the thirty-third, for all values of a contained between the limits 0 and 1.

3The 1823 edition has a missing comma between the first two integrals in equation (20) which iscorrected in the 1899 edition.

THIRTY-FIFTH LECTURE.

DIFFERENTIAL OF A DEFINITE INTEGRALWITH RESPECT TO A VARIABLE INCLUDED INTHE FUNCTION UNDER THE

∫SIGN AND IN THE

LIMITS OF INTEGRATION. INTEGRALS OFVARIOUS ORDERS FOR FUNCTIONS OF ASINGLE VARIABLE.

Let

A =∫ Z

z0

f (x, z) dz(1)

be a definite integral relative to z. If, in this integral, we vary separately and inde-pendently, one and the other, the three quantities Z , z0, x, we will find, by virtue ofthe formulas in (5) (twenty-sixth lecture)1 and of formula (2) (thirty-third lecture),2

d A

d Z= f (x, Z),

d A

dz0= − f (x, z0),

d A

dx=

∫ Z

z0

d f (x, z)

dxdz.(2)

As a result, if the two quantities z0, Z become functions of the variable x, we willhave, by considering A as a function of this single variable,

d A

dx=

∫ Z

z0

d f (x, z)

dxdz + f (x, Z)

d Z

dx− f (x, z0)

dz0dx

.(3)

In the particular case where z0 is reduced to a constant and f (x, Z) to zero, we havesimply

d

dx

∫ Z

z0

f (x, z) dz =∫ Z

z0

d f (x, z)

dxdz.(4)

1The first version of the Fundamental Theorem of Calculus,

d

dx

∫ x

x0f (x) dx = f (x).

2This second formula is,d

dy

∫ x

x0f (x, y) dx =

∫ x

x0

d f (x, y)

dydx .

Without realizing it, Cauchy is again assuming his function is well behaved. As discussed earlier,the exchange of limits that occurs by reversing the order of integration and differentiation he istaking for granted here is not always allowed.


191


https://doi.org/10.1007/978-3-030-11036-9_35


Example. – Let z0 = x0 (x0 denoting a particular and constant value of x), Z = x,and f (x, z) = (x − z)m f (z); 3 we will obtain the formula

d

dx

∫ x

x0

(x − z)m f (z) dz = m∫ x

x0

(x − z)m−1 f (z) dz,(5)

from which we will deduce

∫ x

x0

∫ x

x0

(x − z)m−1 f (z) dz dx = 1

m

∫ x

x0

(x − z)m f (z) dz(6)

and∫∫ x

x0

(x − z)m−1 f (z) dz dx = 1

m

∫ x

x0

(x − z)m f (z) dz + C,(7)

C being an arbitrary constant. If m is reduced to unity, formula (6) will produce∫ x

x0

∫ x

x0

f (z) dz dx =∫ x

x0

(x − z) f (z) dz.(8)

It is now easy to resolve the following question.

PROBLEM. – Find the general value of y that works to satisfy the equation

dn y

dxn= f (x).(9)

Solution. – As we can put equation (9) in the form

d

(dn−1y

dxn−1

)= f (x) dx,

we will deduce, by integrating both sides with respect to x,

dn−1y

dxn−1=

∫f (x) dx =

∫ x

x0

f (x) dx + C,


dn−1y

dxn−1=

∫ x

x0

f (z) dz + C.(10)

3The reader may recognize this function as one similar to the important special case Cauchyinvestigates in his Lecture Nineteen and where his conditions leading to equation (4) have beensatisfied.

THIRTY-FIFTH LECTURE. 193

By integrating again, and several times to follow, with respect to the variable x,between the limits x0, x, having regard to formulas (6) and (8), then adding to theresult of each integration a new arbitrary constant, we will find successively,4

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

dn−2y

dxn−2=

∫ x

x0

(x − z) f (z) dz + C(x − x0) + C1,

dn−3y

dxn−3=

∫ x

x0

(x − z)2

1 · 2 f (z) dz + C(x − x0)2

1 · 2 + C1(x − x0) + C2,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,

dy

dx=

∫ x

x0

(x − z)n−2

1 · 2 · 3 · · · (n − 2)f (z) dz + C

(x − x0)n−2

1 · 2 · · · (n − 2)

+ C1(x − x0)n−3

1 · 2 · · · (n − 3)+ C2

(x − x0)n−4

1 · 2 · · · (n − 4)+ · · · + Cn−2,

(11)

and finally,5

⎧⎪⎪⎪⎨⎪⎪⎪⎩

y=∫ x

x0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)f (z)dz+C

(x − x0)n−1

1 · 2 · · · (n − 1)

+ C1(x−x0)n−2

1 · 2 · · · (n − 2)+C2

(x − x0)n−3

1 · 2 · · · (n − 3)+ · · ·+Cn−2(x − x0)+Cn−1,

(12)

C, C1, C2, . . . , Cn−1 being various arbitrary constants. It is important to observe thatthe definite integral included in the second member of equation (12) can be easilytransformed with the help of formula (17) (twenty-second lecture). In fact, if in thisformula we replace x by z and X by x, we will derive

∫ x

x0

f (z) dz =∫ x−x0

0f (x0 + z) dz =

∫ x−x0

0f (x − z) dz,(13)

and as a result,⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ x

x0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)f (z)dz =

∫ x−x0

0

(x − x0 − z)n−1

1 · 2 · 3 · · · (n − 1)f (x0 + z)dz

=∫ x−x0

0

zn−1

1 · 2 · 3 · · · (n − 1)f (x − z)dz.

(14)

4There is an error in both the original 1823 and 1899 reprint editions which has been correctedhere. Cauchy assigns the last arbitrary constant in the final equation as Cn−1.5Interestingly, the original 1823 edition of equation (12) is also in error, but it is corrected in

the 1899 reprint. The 1823 version reads, y = · · · +C2(x−x0)n−3

1·2···(n−3) + · · · +Cn−1(x − x0)+Cn . Theoriginal edition includes the additional arbitrary constant, Cn, in the subsequent list, but this is alsocorrected in the reprint.


If we take, for simplicity, x0 = 0, the value of y given by equation (12) wouldreduce to

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

y =∫ x

0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)f (z) dz + C

xn−1

1 · 2 · · · (n − 1)

+ C1xn−2

1 · 2 · · · (n − 2)+ C2

xn−3

1 · 2 · · · (n − 3)

+ · · · + Cn−2x + Cn−1,

(15)

and formula (14) would become∫ x

0

(x − z)n−1

1 · 2 · 3 · (n − 1)f (z) dz =

∫ x

0

zn−1

1 · 2 · 3 · · · (n − 1)f (x − z) dz.(16)

When we use indefinite integrals and we are content to indicate the successiveintegrations, the values of the functions

dn−1y

dxn−1,

dn−2y

dxn−2,

dn−3y

dxn−3, . . . , y

derived from equation (9), are presented under the form∫

f (x) dx,∫

·∫

f (x) dx · dx,∫

·∫

·∫

f (x) dx · dx · dx,

. . . . . . . . . . . . . . . . . . . . . . . . ,∫·∫

·∫

· · ·∫

f (x) dx · · · dx · dx · dx .

These last expressions are what we call the integrals of first, of second, of third, …order, and finally of order n, relative to the variable x . For brevity, we will denotethem from now on by the notations

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫f (x) dx,

∫∫f (x) dx2,

∫∫∫f (x) dx3,

. . . . . . . . . . . . . . . . . . . . . . . . ,∫∫· · · f (x) dxn,

(17)

to which we substitute the following⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

∫ x

x0

f (x) dx,∫ x

x0

∫ x

x0

f (x) dx2,

∫ x

x0

∫ x

x0

∫ x

x0

f (x) dx3,

. . . . . . . . . . . . . . . . . . . . . . . . ,∫ x

x0

∫ x

x0

· · · f (x) dxn

(18)

THIRTY-FIFTH LECTURE. 195

when we assume each integration relative to x is performed between the limits x0, x .This granted, we will obviously6 have7

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫ x

x0

∫ x

x0

· · · f (x)dxn =∫ x

x0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)f (z)dz

= 1

1 · 2 · · · (n − 1)

[xn−1

∫ x

x0

f (z)dz − n − 1

1xn−2

∫ x

x0

z f (z)dz

+ (n − 1)(n − 2)

1 · 2 xn−3∫ x

x0

z2 f (z)dz − · · · ±∫ x

x0

zn−1 f (z)dz

],

(19)

or to what amounts to the same thing,8

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∫ x

x0

∫ x

x0

· · · f (x)dxn

= 1

1 · 2 · · · (n − 1)

[xn−1

∫ x

x0

f (x)dx − n − 1

1xn−2

∫ x

x0

x f (x)dx

+ · · · ±∫ x

x0

xn−1 f (x)dx

].

(20)

We can directly verify formula (20) with the help of several integration by parts.9

Now let F(x) be a particular value of y that works to satisfy equation (9) in sucha way that we have

F (n)(x) = f (x).(21)

If the function F(x) and its successive derivatives, up to that of order n, remaincontinuous between the limits x0, x, then, by setting x = x0 in formulas (10), (11),and (12), we will find

{C = F (n−1)(x0), C1 = F (n−2)(x0), C2 = F (n−3)(x0),

. . . . . . , Cn−2 = F ′(x0), Cn−1 = F(x0),(22)

6These next results are anything but obvious. The phrasing Cauchy uses here is similar to themodern usage of, “It can easily be shown.” One can picture Cauchy most assuredly smiling at thethought of what his students and any future readers will need to go through to verify the claimspresented in (19) and (20).7Cauchy’s original 1823 text and the 1899 reprint have a clear misprint that has been corrected

here. The texts both read, · · · = 11·2···(n−1)

[· · · + (n−1)(n−2)

1·2∫ x

x0z2 f (z) dz − · · ·

].

8Another two errors in Cauchy’s texts occur here. The equation has been corrected, but the texts

both have · · · = 11·2···(n−1)

[xn−1

∫ xx0

f (x) dx − n−11 xn−2

∫ xx0

x2 f (x) dx − · · ·±∫ xx0

xn−1 f (x) dx].

9Certainly not a trivial alternative derivation method.


and formula (12) will give10

⎧⎪⎪⎪⎨⎪⎪⎪⎩

F(x) = F(x0) + (x − x0)

1F ′(x0) + · · ·

+ (x − x0)n−1

1 · 2 · · · (n − 1)F (n−1)(x0) +

∫ x

x0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)f (z) dz.

(23)

From this latter equation, combined with equation (19), we deduce the following

⎧⎪⎪⎨⎪⎪⎩

∫ x

x0

∫ x

x0

· · · f (x) dxn = F(x) − F(x0) − (x − x0)

1F ′(x0)

− (x − x0)2

1 · 2 F ′′(x0) − · · · − (x − x0)n−1

1 · 2 · 3 · · · (n − 1)F (n−1)(x0),

(24)

which includes as a special case, formula (17) of the twenty-sixth lecture.11 Whenwe suppose x0 = 0, equation (24) is reduced to

⎧⎪⎪⎨⎪⎪⎩

∫ x

0

∫ x

0· · · f (x) dxn = F(x) − F(0) − x

1F ′(0)

− x2

1 · 2 F ′′(0) − · · · − xn−1

1 · 2 · 3 · · · (n − 1)F (n−1)(0).

(25)

Example. – Let F(x) = ex ; we will have

f (x) = F (n)(x) = ex ,

and by consequence,

⎧⎪⎪⎨⎪⎪⎩

∫ x

0

∫ x

0· · · ex dxn = ex − 1 − x

1− x2

1 · 2 − · · · − xn−1

1 · 2 · 3 · · · (n − 1)

=∫ x

0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)ez dz.

(26)

10Cauchy’s original text, as well as the reprint, has the final integral as∫ x

x0(x−z)n

1·2·3···n f (z) dz.11Cauchy is noting a particular case we have seen from the referenced earlier lecture, namely onewhich reduces to the second version of the Fundamental Theorem of Calculus, when n = 1.

THIRTY-SIXTH LECTURE.

TRANSFORMATION OF ANY FUNCTIONS OF xOR OF x + h INTO ENTIRE FUNCTIONS OF xOR OF h TOWHICH WE APPEND DEFINITEINTEGRALS. EQUIVALENT EXPRESSIONSTO THESE SAME INTEGRALS.

If, in equation (23) of the preceding lecture, we replace f (z) by the value F (n)(z)derived from formula (21), we will find, under the same conditions,1

(1)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

F(x) = F(x0) + x − x01

F ′(x0) + (x − x0)2

1 · 2 F ′′(x0) + · · ·

+ (x − x0)n−1

1 · 2 · · · (n − 1)F (n−1)(x0) +

∫ x

x0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)F (n)(z) dz;

then, in setting x0 = 0,

(2)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

F(x) = F(0) + x

1F ′(0) + x2

1 · 2 F ′′(0) + · · ·

+ xn−1

1 · 2 · 3 · · · (n − 1)F (n−1)(0) +

∫ x

0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)F (n)(z) dz.

If we let, in this expression above, F(x) = f (x + h), and then we exchange betweenthem the two letters x and h, we will obtain the equation

(3)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

f (x + h) = f (x) + h

1f ′(x) + h2

1 · 2 f ′′(x) + · · ·

+ hn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x)

+∫ h

0

(h − z)n−1

1 · 2 · 3 · · · (n − 1)f (n)(x + z) dz,

1Here Cauchy has expressed Taylor’s Theorem with an integral remainder term. The basic ideaof Taylor’s Theorem had been stated as early as 1671 by James Gregory (1638–1675), but it isnamed after Brook Taylor (1685–1731), an English mathematician who originally wrote somethingclose in the early 18th century. However, the theorem did not become widely known until LeonhardEuler used the result in his work several decades later. The theorem was not more fully developedto include a remainder term until much later by Lagrange and Cauchy.


197


https://doi.org/10.1007/978-3-030-11036-9_36


in which the last term of the second member2 can be presented under several dif-ferent forms, since we have, by virtue of formulas (14) and (19) of the thirty-fifthlecture,

(4)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫ h

0

(h − z)n−1

1 · 2 · 3 · · · (n − 1)f (n)(x + z) dz

=∫ h

0

zn−1

1 · 2 · · · (n − 1)f (n)(x + h − z) dz

=∫ x+h

x

(x + h − z)n−1

1 · 2 · 3 · · · (n − 1)f (n)(z) dz

=∫ h

0

∫ h

0· · · f (n)(x + z) dzn.

Equation(3)assumesthat thefunctions f (x + z), f ′(x + z), . . . , f (n)(x + z) remaincontinuous between the limits z = 0, z = h. We could immediately deduce it fromformula (1) by taking x = x0 + h, then replacing x0 by x, and F by f. Only thelast term of the second member would then be the third of the integrals included informula (4).

Moreover, we can demonstrate equation (3) directly with the help of several in-tegration by parts, by operating more or less as Mr. de Prony has done in a paperpublished in 1805.3 In fact, if, in formula (13) of the preceding lecture, we firstreplace x by x0 + h, and then x0 by x, we will derive

(5)∫ h

0f (x + z) dz =

∫ h

0f (x + h − z) dz.

We will have, therefore, by consequence4

(6) f (x + h) − f (x) =∫ h

0f ′(x + z) dz =

∫ h

0f ′(x + h − z) dz.

2The definite integral term in equation (3).3Gaspard de Prony (1755–1839) and Cauchy shared a mutual distaste for one another, datingback to when Cauchy was a student in de Prony’s mechanics class at the École Polytechnique.Unfortunately for Cauchy, much of the period in which Cauchy himself was teaching at the ÉcolePolytechnique, de Prony was responsible for overseeing the quality of Cauchy’s teaching. DuringCauchy’s tenure, it is clear de Prony was generally dissatisfied with what he observed, feelingmuch of Cauchy’s teachings were too abstract. Mr. de Prony would continue producing unfavor-able reports of Cauchy up until the time Cauchy leaves the school in 1830. To make matters evenworse for Cauchy, de Prony was also a highly influential member of the Conseil d’Instruction (thecommittee responsible for overseeing the curriculum) while Cauchy was battling over the contentof his analysis course at the École Polytechnique, which created additional friction. It must havebeen a nightmare for Cauchy.4The Fundamental Theorem of Calculus.

THIRTY-SIXTH LECTURE. 199

Moreover, in integrating by parts several times in sequence, we find

(7)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫

f ′(x + h − z) dz

= z

1f ′(x + h − z) +

∫z

1f ′′(x + h − z) dz

= z

1f ′(x + h − z) + z2

1 · 2 f ′′(x + h − z) +∫

z2

1 · 2 f ′′′(x + h − z) dz

= · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·= z

1f ′(x + h − z) + z2

1 · 2 f ′′(x + h − z) + · · ·

+ zn−1

1 · 2 · · · (n − 1)f (n−1)(x + h − z)

+∫

zn−1

1 · 2 · · · (n − 1)f (n)(x + h − z) dz;

then, by assuming that each integration is performed between the limits z = 0, z = h,

and that the functions f (x + z), f ′(x + z), . . . , f (n)(x + z) remain continuousbetween these same limits,

(8)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫ h

0f ′(x + h − z) dz = h

1f ′(x) + h2

1 · 2 f ′′(x) + · · ·

+ hn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x)

+∫ h

0

zn−1

1 · 2 · 3 · · · (n − 1)f (n)(x + h − z) dz.

This granted, we will obviously deduce from formula (6) an equation which will bein accordance, by virtue of formula (4), with equation (3). The same method couldagain be used to directly establish equation (2).

Not only can the integrals contained in the second members of formulas (2) and(3) be replaced by several others similar to those that comprise formula (4), but weshould deduce from equation (13) (twenty-third lecture)5 that they are also equiva-lent to two products of the form

(9) F (n)(θx)

∫ x

0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)dz = xn

1 · 2 · 3 · · · nF (n)(θx),

(10) f (n)(x + θh)

∫ h

0

(h − z)n−1

1 · 2 · 3 · · · (n − 1)dz = hn

1 · 2 · 3 · · · nf (n)(x + θh),

5The general form of the Mean Value Theorem for Definite Integrals with χ(z) = (x−z)n−1

1·2·3···(n−1) and

ϕ(z) = F (n)(z).


θ denoting an unknown number that may vary from one product to another by al-ways remaining less than unity. We will have as a result6

(11)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

F(x) = F(0) + x

1F ′(0) + x2

1 · 2 F ′′(0) + · · ·

+ xn−1

1 · 2 · 3 · · · (n − 1)F (n−1)(0) + xn

1 · 2 · 3 · · · nF (n)(θx),

(12)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f (x + h) = f (x) + h

1f ′(x) + h2

1 · 2 f ′′(x) + · · ·

+ hn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x) + hn

1 · 2 · 3 · · · nf (n)(x + θh).

It is essential to observe that the function, F(x), along with its successive deriva-tives, shall remain continuous in formula (11) between the limits 0, x, and the func-tion f (x + z), along with its successive derivatives in formula (12), between thelimits z = 0, z = h.7

Now, let u = f (x, y, z, . . . ) be a function of several independent variablesx, y, z, . . . , and let

(13) F(α) = f (x + α dx, y + α dy, z + α dz, . . . ).

We will derive from formula (11), in replacing x by α, then having regard to theprinciples established in the fourteenth lecture,

(14)

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

f (x + α dx, y + α dy, z + α dz, . . . )

= u + α

1du + α2

1 · 2 d2u + · · ·

+ αn−1

1 · 2 · · · (n − 1)dn−1u + αn

1 · 2 · 3 · · · nF (n)(θα).

If the quantity α becomes infinitely small, it will be the same for the difference

F (n)(θα) − F (n)(0), or F (n)(θα) − dnu;

6Equation (12) is one of the more common forms to express Taylor’s Theorem with remainder.The remainder in this case is called the Lagrange form. Cauchy has shown as long as the first nderivatives of f (x) are continuous in the neighborhood of x , there exists a θ between zero and onein which this expansion is valid—essentially one way of stating the Taylor’s Theorem of today. Hehas now completed an integral development of Taylor’s Theorem. Beautiful.7Cauchy adds a final derivation of Taylor’s formula in his ADDITION at the end of his originaltext. In this later derivation, he develops the formula without the integration techniques employedin this and previous lectures, using only differential calculus methods and established theorems.

THIRTY-SIXTH LECTURE. 201

and, in denoting this difference by β, we will find

(15)

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

f (x + α dx, y + α dy, z + α dz, . . . )

= u + α

1du + α2

1 · 2 d2u + · · ·

+ αn−1

1 · 2 · · · (n − 1)dn−1u + αn

1 · 2 · 3 · · · n(dnu + β).

When the independent variables are reduced to a single variable x, then, by settingy = f (x), we obtain the formula

(16)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f (x + α dx) = y + α

1dy + α2

1 · 2 d2y + · · ·

+ αn−1

1 · 2 · · · (n − 1)dn−1y + αn

1 · 2 · 3 · · · n(dn y + β).

Consider now that, for a particular value x0 attributed to the variable x, the func-tion f (x) and its successive derivatives up to that of order n − 1 vanish. In this case,we will derive from formula (12)

(17) f (x0 + h) = hn

1 · 2 · 3 · · · nf (n)(x0 + θh);

then, by substituting for the finite quantity h, an infinitely small quantity denoted byi,

(18) f (x0 + i) = i n

1 · 2 · 3 · · · nf (n)(x0 + θ i).

When, among the functions f (x), f ′(x), . . . , f (n−1)(x), the first is the only one thatdoes not vanish for x = x0, equation (18) should obviously be replaced by the fol-lowing

(19) f (x0 + i) − f (x0) = i n

1 · 2 · 3 · · · nf (n)(x0 + θ i).

If, in the same hypothesis, we write x in place of x0, and if we set

f (x) = y, Δx = i = αh,

equation (19) will take the form

(20) Δy = αn

1 · 2 · 3 · · · n(dn y + β),

β denoting, as well as α, an infinitely small quantity. We could again deduce for-mula (20) from equation (16) by observing that the value attributed to x makes the


differentials dy, d2y, . . . , dn−1y vanish at the same time as the derived functions,f ′(x), f ′′(x), . . . , f (n−1)(x).

Equation (20) provides the means to resolve the fourth problem of the sixth lec-ture8 in several cases where the method that we have proposed is insufficient. Infact, assuming y and z designate two functions of the variable x, the particular valuex0 attributed to this variable reduces to the form 0

0 , not only the fraction s = zy , but

also the following z′y′ ,

z′′y′′ , . . . ,

z(m−1)

y(m−1) . Then, by letting Δx = α dx, and denoting byβ, γ two infinitely small quantities, we will have, for x = x0,

(21)

⎧⎪⎪⎨

⎪⎪⎩

Δy = αm

1 · 2 · 3 · · · m(dm y + β),

Δz = αm

1 · 2 · 3 · · · m(dm z + γ ),

(22) s = limz + Δy

y + Δy= lim

Δz

Δy= lim

dm z + γ

dm y + β= dm z

dm y= z(m)

y(m).

Example. –We will have, for x = 0,

sin2 x

1 − cos x= d2(sin2 x)

d2(1 − cos x)= 2(cos2 x − sin2 x)

cos x= 2.

8Cauchy is referring to his first serious inquiry into how to resolve the true value of the indeter-minate fraction 0

0 . The solution offered in Lecture Six determines the ratio of the first derivatives.However, this Lecture Six solution fails if this new ratio is also found to be indeterminate. Here,Cauchy looks to the ratio of higher and higher order derivatives until a determinate ratio is located.

Although it would have certainly cleaned up many of the expressions in the latter lectures ofCauchy’s text, the factorial notation we use today, m!, to indicate the operation 1 · 2 · 3 · · · m wasnot in common use at the time Cauchy wrote Calcul infinitésimal. Christian Kramp (1760–1826),a French mathematician, is credited with the introduction of the modern notation as early as 1808;however, it was not widely adopted until later in the century.

THIRTY-SEVENTH LECTURE.

THEOREMS OF TAYLOR AND OF MACLAURIN.EXTENSION OF THESE THEOREMS TOFUNCTIONS OF SEVERAL VARIABLES.

We call a series an indefinite sequence of terms

(1) u0, u1, u2, . . . , un, . . . ,

that are derived, one after the other, according to a known rule. Let

sn = u0 + u1 + u2 + · · · + un−1

be the sum of the first n terms, n denoting any integer number. If, for the values ofn always increasing, the sum sn indefinitely approaches a certain limit s, the serieswill be said to be convergent, and the limit s represented by the notation

u0 + u1 + u2 + u3 + · · ·will be called the sum of the series.1 If, on the other hand, while n increases indefi-nitely, the sum sn does not approach any fixed limit, the series will be divergent, andwill no longer have a sum.2 In one case and the other, the term corresponding to theindex n, namely, un, is called the general term. In addition, if in the first hypothesis,we let s = sn + rn,

3 rn will be what we name the remainder of the series, startingfrom the nth term.

1Cauchy is careful with his wording here. He is defining s, the sum of a convergent infinite series,as the limit of the partial sums, sn, as n gets large. Specifically, he is not defining s to be any of thesn’s. Cauchy defines the value of s as a limit, when the limit exists.2The concept of a divergent series having no sum, in other words having no meaning whatsoever,was not as obvious an idea at the time of Cauchy as we see it today. Many mathematicians upto his day had avoided any such conclusion and had considered divergent series just as valid asconvergent series. There was a belief, even into the 19th century, that these divergent series wouldturn out to be justified in some manner, and so, there was a great deal of reluctance to excludethem from a general theoretical analysis of infinite series. Cauchy stands strong in this regard andmakes the statement that unless an infinite series converges to a sum, none of his analysis to followapplies. This is a bold leap for Cauchy to make in 1823, one he outlines at the beginning of thistext in his FOREWORD.3The 1899 reprint has s − sn = rn .


203


https://doi.org/10.1007/978-3-030-11036-9_37


These definitions being admitted, it obviously follows from formulas (2) and (3)of the thirty-sixth lecture that the series

(2) F(0),x

1F ′(0),

x2

1 · 2 F ′′(0),x3

1 · 2 · 3 F ′′′(0), . . . ,

(3) f (x),h

1f ′(x),

h2

1 · 2 f ′′(x),h3

1 · 2 · 3 f ′′′(x), . . . ,

will be convergent, and will have for sums the two functions, F(x), f (x + h), re-spectively, whenever the two integrals

(4)∫ x

0

(x − z)n−1

1 · 2 · 3 · · · (n − 1)F (n)(z) dz = xn

1 · 2 · 3 · · · nF (n)(θx),

(5)∫ h

0

(h − z)n−1

1 · 2 · 3 · · · (n − 1)f (n)(x + z) dz = hn

1 · 2 · 3 · · · nf (n)(x + θh)

are convergent for increasing values of n toward the limit zero.4 We will find, byconsequence,5

(6) F(x) = F(0) + x

1F ′(0) + x2

1 · 2 F ′′(0) + x3

1 · 2 · 3 F ′′′(0) + · · · ,

if expression (4) vanishes for infinite values of n, and

(7) f (x + h) = f (x) + h

1f ′(x) + h2

1 · 2 f ′′(x) + h3

1 · 2 · 3 f ′′′(x) + · · · ,

if expression (5) satisfies the same condition. Formulas (6) and (7) contain the The-orems of Maclaurin and of Taylor. They are used, when the integrals (4) and (5)fulfill the prescribed conditions, to expand the two functions F(x) and f (x + h) intoa series ordered according to ascending and integer powers of the quantities x andh. The remainder of these series are precisely the two integrals which we have justspoken about.

Now, suppose that we denote by u = f (x, y, x, . . . ), a function of several inde-pendent variables, and that to equations (2) and (3) of the preceding lecture we

4A sufficient condition for convergence (as will be shown in a subsequent example). Cauchy isquite aware that in general it is not enough for the ending terms of a sequence to be approachingzero to guarantee the corresponding series converges. The classic counterexample illustrating thispossibility is the harmonic series, which diverges despite its terms becoming smaller and smaller.What is different in Cauchy’s case here is that by knowing the remainder of the series is becomingsmaller, it is known the sum of all the ending terms combined is also approaching zero. This lattercondition is enough to guarantee convergence as Cauchy correctly argues.5 The 1899 text has a clear misprint in equation (6). It reads, F(x) = F(0) + 1

x F ′(0) + x21·2 F ′′(0) +

x31·2·3 F ′′′(0) + · · · .

THIRTY-SEVENTH LECTURE. 205

substitute equation (14). We will deduce from this last equation

(8)

⎧⎪⎨⎪⎩

f (x + α dx, y + α dy, z + α dz, . . . )

= u + α

1du + α2

1 · 2 d2u + α3

1 · 2 · 3 d3u + · · ·

whenever the termαn

1 · 2 · 3 · · · nF (n)(θα),

or rather, the integral that this term represents, and that we can write under the form

(9)∫ α

0

(α − v)n−1

1 · 2 · 3 · · · (n − 1)F (n)(v) dv,

will vanish for infinite values of n. We will find as a result, by setting α = 1,

(10) f (x + dx, y + dy, z + dz, . . . ) = u + du

1+ d2u

1 · 2 + d3u

1 · 2 · 3 + · · · ,

provided that the integral

(11)∫ 1

0

(1 − v)n−1

1 · 2 · 3 · (n − 1)F (n)(v) dv

satisfies the stated condition. When the independent variables x, y, z, . . . are re-duced to a single variable x, equation (10) becomes

(12) f (x + dx) = u + du

1+ d2u

1 · 2 + d3u

1 · 2 · 3 + · · · .

This expression above coincides with equation (7), that is to say, with the formulaof Taylor. By replacing x by zero, and dx by x, we would regain the Theorem ofMaclaurin. Let us add that equation (10), and that which we deduce when we replacex, y, z, . . . by zero, thendx, dy, dz, . . . by x, y, z, . . . ,provides themeans to extendthe Theorems of Taylor and of Maclaurin to functions of several variables. Finally,we remark that equations (6), (8), (10), (12) coincide with equations (4), (6), (7), (8)of the nineteenth lecture in the case where F(x) and f (x) represent entire functionsof degree n.

Since, by virtue of formula (19) (twenty-second lecture),6 the integral (4) isequivalent to a product of the form7

(13) x(x − θx)n−1

1 · 2 · 3 · · · (n − 1)F (n)(θx),

6The Mean Value Theorem for Definite Integrals.7Known today as the remainder in Cauchy form.


θ denoting a number less than unity, it is clear that the infinite values of n will causethis integral to vanish, if they reduce the function

(14)(x − z)n−1

1 · 2 · 3 · · · (n − 1)F (n)(z)

to zero, for all values of z contained between the limits 0 and x . This latter condi-tion will obviously be fulfilled if the numerical value of the expression F (n)(θx),

assumed real, or the magnitude of the same expression, assumed imaginary, is notincreasing indefinitely as n increases.8 In fact, since the quantity

m(n − m) =(n

2

)2 −(n

2− m

)2

grows along with the number m between the limits m = 1, m = n2 , and that we have

as a result1(n − 1) < 2(n − 2) < 3(n − 3) < · · · ,

1 · 2 · 3 · · · (n − 1) > (n − 1)n−12 ,

we can affirm that the numerical value or the magnitude of the expression (14), willalways remain less than the numerical value or the magnitude of the product

(15)

(x − z√n − 1

)n−1

F (n)(z).

Now, this product will become null in the admitted hypothesis, for n = ∞.

Examples. – If we take for successive values of the function F(x),

ex , sin x, cos x,

we will find for the corresponding values of F (n)(θx),

eθx , sin(nπ

2+ θx

), cos

(nπ

2+ θx

).

Since these last quantities remain finite, regardless of x, while n increases, we mustconclude that the Theorem of Maclaurin9 is always applicable to these three pro-posed functions. We will have, by consequence, for any values of x, and for positive

8In today’s nomenclature, as long as the nth derivative is bounded. Using this condition for F (n)(z),to verify the expression in (14) tends to zero Cauchy only needs to show the preceding term,

(x−z)n−1

1·2·3···(n−1) , vanishes as n gets large. It would be interesting to be sitting in a desk at the back of

Cauchy’s École Polytechnique classroom in 1823, given the descriptive set of subsequent steps heprovides as clues here, to have him explain his demonstration of these valid claims.9This theorem, as well as the Maclaurin series, are named after Colin Maclaurin (1698–1746) ofScotland, one of Isaac Newton’s most successful followers.

THIRTY-SEVENTH LECTURE. 207

values of A,

(16)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

ex = 1 + x

1+ x2

1 · 2 + x3

1 · 2 · 3 + · · · ,

Ax = ex lA = 1 + x lA1

+ x2(lA)2

1 · 2 + x3(lA)3

1 · 2 · 3 + · · · ,

(17)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

sin x = sin (0) + x

1sin

(π2

) + x2

1 · 2 sin(2π2

) + x3

1 · 2 · 3 sin(3π2

) + · · ·

= x

1− x3

1 · 2 · 3 + x5

1 · 2 · 3 · 4 · 5 − · · · ,

(18)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

cos x = cos (0) + x

1cos

(π2

) + x2

1 · 2 cos(2π2

) + x3

1 · 2 · 3 cos(3π2

) + · · ·

= 1 − x2

1 · 2 + x4

1 · 2 · 3 · 4 − · · · .

When the function F (n)(θx) becomes infinite for infinite values of n, expression(14) can still converge toward the limit zero. This will occur, for example, if we takeF(x) = l(1 + x), and if at the same time, we attribute to x a numerical value smallerthan unity. In fact, we will find in this case, by assuming z = θx, θ < 1, x2 < 1,

(19)(x − z)n−1

1 · 2 · 3 · · · (n − 1)F (n)(z) = ± (x − z)n−1

(1 + z)n= ± xn−1

1 − θ

(1 − θ

1 + θx

)n

;

and, since the fraction 1−θ1+θx will obviously be less than unity, it is clear that expres-

sion (19) will vanish for n = ∞. We will find, by consequence, for all values of xcontained between the limits of −1 and +1,

(20) l(1 + x) = x

1− x2

2+ x3

3− x4

4+ · · · .

THIRTY-EIGHTH LECTURE.

RULES ON THE CONVERGENCE OF SERIES.APPLICATION OF THESE RULES TO THE SERIESOF MACLAURIN.

As equations (6) and (7) (thirty-seventh lecture) are only valid in the case where theseries (2) and (3) are convergent, it is important to establish the conditions for theconvergence of the series.1 This is the objective we will now occupy ourselves.

One of the simplest series is the geometric progression

(1) a, ax, ax2, ax3, . . . ,

which has for a general term, axn . Now, the sum of its first n terms, namely,

a(1 + x + x2 + · · · + xn−1

) = a1 − xn

1 − x= a

1 − x− axn

1 − x

will obviously converge,2 for increasing values of n, toward the fixed limit a1−x , if

the numerical value of the variable x, assumed real, or the magnitude of the samevariable, assumed imaginary, is a number less than unity, while in the other casethis sum will cease to converge toward a similar limit. The series (1) will be, there-fore, always convergent in the first case and always divergent in the second. Thisconclusion remains valid even when the factor a becomes imaginary.

Now consider the series

(2) u0, u1, u2, u3, . . . , un, . . .

composed of any real or imaginary terms. To decide whether it is convergent ordivergent we will not need to examine its first terms; we can even omit these terms

1Most of Cauchy’s contemporaries continued to treat infinite series as if they were very largefinite series that carried with them the same properties. This tradition dated back through the daysof Euler to at least the time of Isaac Newton. As an example, all through the 18th century andinto Cauchy’s days an infinite power series was generally treated as though it was simply a largepolynomial. However, Cauchy now demands the convergence of an infinite series be demonstratedbefore any properties can be assigned.2Properties of the common geometric series are covered thoroughly in Chapter VI of Cauchy’sCours d’analyse. Even though this earlier text was never officially used in practice, one mustassume his students would have been expected to be quite familiar with this fundamental series.


209


https://doi.org/10.1007/978-3-030-11036-9_38


in such a way as to replace this series by the following

(3) um, um+1, um+2, . . . ,

m denoting a number as large as we will wish. Moreover, let ρn be the numericalvalue or the magnitude of the general term un. It is clear that the series (3) will beconvergent if the magnitudes of its different terms, namely,

(4) ρm, ρm+1, ρm+2, . . . ,

in their turn form a convergent series,3 and it will become divergent if ρn does notdecrease indefinitely for increasing values of n. This granted, we will easily estab-lish the two theorems that follow.

THEOREM I. – Find the limit or the limits toward which the expression (ρn)1n

converges, while n increases indefinitely, and let λ be the largest of these limits. Theseries (2) will be convergent if we have λ < 1, divergent if we have λ > 1.4

Proof. – First, suppose λ < 1, and arbitrarily choose between the two numbers 1and λ a third number μ, so that we have λ < μ < 1. n coming to grow beyond anyassignable limit, the largest values of (ρn)

1n cannot indefinitely approach the limit

λ without finally being constantly less than μ. As a result, it will be possible toattribute an integer number m a value considerable enough such that, for n becominggreater than or equal to m,we constantly have (ρn)

1n < μ, ρn < μn. Then, the terms

of the series (4) will be numbers less than those of the terms corresponding to thegeometric progression

(5) μm, μm+1, μm+2, . . . ;and, since this last series will be convergent (due toμ < 1), we must accord as muchto the series (4), and by consequence, of the series (2).

In the second place, suppose λ > 1, and again place between the two numbers 1and λ a third number μ, such that we have λ > μ > 1. If n comes to grow beyondall limits, the largest values of (ρn)

1n , by indefinitely approaching λ, will eventually

surpassμ. We can, therefore, satisfy the condition (ρn)1n > μ or ρn > μn > 1,with

values of n as large as we will wish; and as a result, we will find in the series (4)an indefinite number of terms larger than unity, which will be sufficient to note thedivergence of the series (2), (3), and (4).

3Cauchy seems to be taking for granted that absolute convergence of a series implies its con-vergence. Absolute convergence is not a concept Cauchy ever defines specifically within Calculinfinitésimal, but this current assumption is one he essentially proves in his Cours d’Analyse Chap-ter VI, §III.4Cauchy is stating what we now know as the Root Convergence Test. He has also demonstratedsimilar versions of this theorem in his Cour d’analyse Chapter VI, §II, Theorem I and Chapter VI,§III, Theorem I.

THIRTY-EIGHTH LECTURE. 211

THEOREM II. – If, for the increasing values of n, the ratio ρn+1

ρnconverges to-

ward a fixed limit λ, the series (2) will be convergent whenever we have λ < 1, anddivergent whenever we have λ > 1.5

Proof. – Choose an arbitrary number ε less than the difference that exists between1 and λ. It will be possible to attribute a value to m large enough such that, for n be-coming greater than or equal to m, the ratio ρn+1

ρnalways remains contained between

the two limits λ − ε, λ + ε. Then, the different terms of the series (4) will be foundcontained between the terms corresponding to the two geometric progressions

ρm, ρm(λ − ε), ρm(λ − ε)2, ρm(λ − ε)3, . . . ,

ρm, ρm(λ + ε), ρm(λ + ε)2, ρm(λ + ε)3, . . . ,

whichwill both be convergent if we haveλ < 1, and both divergent if we haveλ > 1.Therefore, etc.

Scholium. – It would be easy to prove that the limit of the ratio ρn+1

ρn, in the case

where this limit exists, is at the same time that of the expression (ρn)1n . (See Alge-

braic Analysis, Chapter VI.)6

By applying Theorems I and II to the series of Maclaurin, namely,

(6) F(0),x

1F ′(0),

x2

1 · 2 F ′′(0),x3

1 · 2 · 3 F ′′′(0), . . . ,

we obtain the following proposition.

THEOREM III. – Let ρn be the numerical value or the magnitude of the ex-pression F (n)(0)

1·2·3···n , and λ the limit toward which the largest values of (ρn)1n , or else

the unique limit (if this limit exists) of the ratio ρn+1

ρn, converges while n grows in-

definitely. The series (6) will be convergent whenever the numerical value or themagnitude of the variable x will be less than 1

λ, and divergent whenever the numer-

ical value or the magnitude will exceed 1λ.7

Examples. – If we take for successive values of F(x)

ex , sin x, cos x, l(1 + x), (1 + x)μ,

5This theorem is today known as the Ratio Convergence Test. Cauchy has stated similar versionsof this result in his Cours d’analyse Chapter VI, §II, Theorem II and Chapter VI, §III, Theorem II.6Cauchy is pointing out his Theorem I result (the Root Convergence Test) is stronger and moregeneral than his Theorem II result (the Ratio Convergence Test).7This theorem has been corrected. Both the 1823 and 1899 editions incorrectly define ρn to beF (n)(0). The error must clearly be a typographical omission dating back to 1823, as Cauchy knewbetter. He even demonstrates a related result in his Cours d’analyse Chapter VI, §IV, Theorem II,Corollary IV from 1821.


μ being a constant quantity, the corresponding values of 1λwill be

∞, ∞, ∞, 1, 1.

As a result, the series included in equations (16), (17), (18) of the thirty-seventhlecture will remain convergent between the limits x = −∞, x = +∞, that is to say,for any values of x . On the other hand, the series

(7)

⎧⎪⎪⎨

⎪⎪⎩

1,μ

1x,

μ(μ − 1)

1 · 2 x2,μ(μ − 1)(μ − 2)

1 · 2 · 3 x3, . . . ,

μ(μ − 1) · · · (μ − n + 1)

1 · 2 · 3 · · · nxn, . . . ,

and that contained in formula (20) (thirty-seventh lecture) will only be convergentif the variable x is real, and is between the limits8 x = −1, x = +1.

We have already remarked that the series (6) is real and that it has F(x) for a sumwhenever the variable x being real, the variable z being contained between the limitsof 0, x, and the expression (14) (thirty-seventh lecture) itself vanishes for infinitevalues of n. Now, this latter condition will obviously be satisfied if the expressionin question is the general term of a convergent series, which will occur, by virtue ofTheorem III, if, for values of i ncreasing n, the magnitude or the numerical value ofthe product

(8)x − z

n

F (n+1)(z)

F (n)(z)

converges toward a limit less than unity.

Example. – LetF(x) = (1 + x)μ,

μ denoting a constant quantity. If, in expression (8) we replace z by θx, this expres-sion will become

x1 − θ

1 + θx

μ − n

n= −x

1 − θ

1 + θx

(1 − μ

n

),

and will converge for increasing values of n toward a limit of the form −x 1−θ1+θx , a

limit whose numerical value will be less than unity, if we suppose x2 < 1. We willhave, therefore, under this condition,

(9) (1 + x)μ = 1 + μ

1x + μ(μ − 1)

1 · 2 x2 + μ(μ − 1)(μ − 2)

1 · 2 · 3 x3 + · · · .

8Cauchy is presenting here what today we would refer to as the radius of convergence of theMaclaurin series for each of these functions.

THIRTY-EIGHTH LECTURE. 213

We would also prove that the equation

(10) (1 + ax)μ = 1 + μ

1ax + μ(μ − 1)

1 · 2 a2x2 + μ(μ − 1)(μ − 2)

1 · 2 · 3 a3x3 + · · ·

remains valid, for real or imaginary values of the constant a, while the numericalvalue of x is less than the magnitude of 1

a .

We might believe9 that the series (6) always has F(x) for a sum when it is con-vergent, and that, in the case where its different terms vanish, one after the other, thefunction F(x) would itself vanish. But, to ensure the contrary, it suffices to observethat the second condition will be fulfilled, if we assume10

F(x) = e−( 1x )

2

,

and the first, if we assume11

F(x) = e−x2 + e−( 1x )

2

.

However, the function e−( 1x )

2

is not ever identically null, and the series deduced

from the latter assumption has for a sum, not the binomial e−x2 + e−( 1x )

2

, but onlyits first term e−x2

.

9Cauchy is likely referring to the work of Lagrange.10The example to follow demonstrates the Taylor series for the two functions, H(x) = 0 and

F(x) = e−

(1x

)2

are identical, even though the functions themselves are certainly not. This example

will also illustrate the Taylor series for F(x) = e−

(1x

)2

does not accurately represent the originalfunction, as the series will suggest F(x) is identically zero for all values of x, when the functionitself is never zero, as Cauchy points out. With this single example, Cauchy clearly shows thatnot every function can be written as a convergent Taylor series and that a Taylor series does notuniquely identify a function.

11This second counterexample of F(x) = e−x2 + e−

(1x

)2

produces a Taylor series identical to thatof the distinct function G(x) = e−x2 , demonstrating once again a Taylor series representation doesnot uniquely identify a function.

As discussed earlier, one ofCauchy’s early contemporaries andmentors, Joseph-LouisLagrange,had previously proposed (years before) that the derivative of a function be defined as the coef-ficient of the linear term of its Taylor series expansion. This theory had even been a part of theÉcole Polytechnique curriculum in its early years. Cauchy uses these last two examples to clearlyillustrate this theory is not valid. Among other problems, such as the fact some Taylor series donot converge, he shows different functions can produce the same series. These last two examplesplay a pivotal role in the demolishment of Lagrange’s foundational assumptions of functions andthe Lagrangean theory of the calculus. Ironically, even though Cauchy helps dismantle Lagrange’stheory, one of Cauchy’s main references while developing his own theory was Lagrange’s Théoriedes fonctions analytiques published in installments beginning in 1797. In this latter text, Lagrangepresents his work without any diagrams; a practice Cauchy continued.

THIRTY-NINTH LECTURE.

IMAGINARY EXPONENTIALS ANDLOGARITHMS. USE OF THESE EXPONENTIALSAND OF THESE LOGARITHMS IN THEDETERMINATION OF DEFINITE OR INDEFINITEINTEGRALS.

We have proven in the thirty-seventh lecture that the exponential Ax (A denoting apositive constant, and x a real variable) is always equivalent to the sum of the series

1,x lA

1,

x2(lA)2

1 · 2 ,x3(lA)3

1 · 2 · 3 , . . . ,(1)

so that we have, for all real values of x,

Ax = 1 + x lA

1+ x2(lA)2

1 · 2 + x3(lA)3

1 · 2 · 3 + · · · .(2)

On the other hand, since, by virtue of Theorem III of the thirty-eighth lecture, theseries (1) remains convergent for any imaginary values of the variable x , we agree toextend equation (2) to all possible cases, and to use it in the case where the variable xbecomes imaginary, to settle the meaning of the notation Ax . This convention beingadmitted, we easily deduce from equation (2) several remarkable formulas that wewill now make known.

First, if we take A = e, equation (2) will become

ex = 1 + x

1+ x2

1 · 2 + x3

1 · 2 · 3 + · · · .(3)

If we set, in this last formula, x = z√−1 (z denoting a real variable), we will find

ez√−1 = 1 + z

√−1

1− z2

1 · 2 − z3√−1

1 · 2 · 3 + · · ·

= 1 − z2

1 · 2 + z4

1 · 2 · 3 · 4 − · · · +(

z

1− z3

1 · 2 · 3 + · · ·) √−1,

and as a result,

ez√−1 = cos z + √−1 sin z.(4)


215


https://doi.org/10.1007/978-3-030-11036-9_39


We will also find1

e−z√−1 = cos z − √−1 sin z;(5)

then, we will deduce from equations (4) and (5) combined together

cos z = ez√−1 + e−z

√−1

2, sin z = ez

√−1 − e−z√−1

2√−1

.(6)

Let, in the second place, x = (a + b

√−1)z, a, b denoting two real constants.

Then, the series included in the second member of formula (3) will be precisely thatwhich we deduce from the Theorem of Maclaurin, applied to the imaginary functioneaz


). We will therefore have

e(a+b√−1 )z = eaz


) = eazebz√−1.(7)

This last formula is analogous to the identical equation

e(a+b)z = eazebz,

from which we would deduce, but only by induction, in substituting for the realconstant b an imaginary constant b

√−1. We add that by relying on formula (7), weextend without difficulty the equation

ex+y = ex ey(8)

to any imaginary values of the variables x, y; and, by comparing formula (2) toformula (3) we derive, for any value of x,

Ax = ex lA.(9)

Let us now conceive that, u and v designating two real quantities, we look for thevarious values of x that work to resolve the two equations

Ax = u + v√−1,(10)

ex = u + v√−1.(11)

These various values will be the different logarithms of u +v√−1 : 1◦ in the system

whose base is A; 2◦ in the Napierian system whose base is e. Moreover, since, byvirtue of formula (9), the logarithms of the expression u + v

√−1 in the first systemwill be equal to the Napierian logarithms of this same expression divided by lA,

it will suffice to resolve equation (11). This granted, we let x = α + β√−1, α, β

1The result in equation (4) is commonly known as Euler’s formula. The relationships in (4) and (5)are the source of the incredible equation eiπ +1 = 0, by allowing z = ±π. This single relationshiplinks five of the most important numbers in mathematics.

THIRTY-NINTH LECTURE. 217

denoting two real quantities. Formula (11) will become

eα+β√−1 = u + v

√−1;then, we will derive eα cosβ = u, eα sin β = v, and by consequence,

eα = (u2 + v2

) 12 ,(12)

cosβ = u√u2 + v2

, sin β = v√u2 + v2

.(13)

Now, we would satisfy equation (12) with only one real value of α, namely α =12 l

(u2 + v2

). Moreover, in denoting by n an arbitrary integer number, we will satisfy

the equations in (13) with every value of β included in the formula

β = 2nπ + arctangv

u,(14)

if u is positive, or in the following

β = (2n + 1)π + arctangv

u,(15)

if u becomes negative. Therefore, there exists an infinite number of imaginary log-arithms of the expression u + v

√−1. The simplest of all these logarithms, in thecase where the quantity u remains positive, is that which we obtain by setting n = 0,namely 1

2 l(u2 + v2

) + √−1 arctang vu . This same logarithm, which for a null value

of v is reduced to a real logarithm of u, will be that which we denote by the nota-tion l

(u + v

√−1)(see Algebraic Analysis, Chapter IX),2 so that we will have, for

positive values of u,

l(u + v

√−1) = 1

2l(u2 + v2

) + √−1 arctangv

u.(16)

As a result, if r represents a positive quantity, and t a real arc contained between thelimits −π

2 , +π2 , the equation

x = r(cos t + √−1 sin t

) = ret√−1(17)

will lead to the following

lx = lr + t√−1.(18)

The formulas which are used to differentiate real exponentials and real loga-rithms remain valid when these exponentials and these logarithms become imagi-nary. Thus, for example, we will recognize without difficulty that we have: 1◦ for

2This chapter of Cauchy’s earlier text deals primarily with convergent imaginary series.


imaginary values of the variable x,

d ex = ex dx,(19)

d l(±x) = dx

x;(20)

2◦ for real values of the variables x, y, z, and the constants α, β, a, b,

d ex+y√−1 = ex+y

√−1(dx + dy√−1

),(21)

d l[ ± (

x + y√−1

)] = dx + dy√−1

x + y√−1

,(22)

(23)

⎧⎪⎪⎨⎪⎪⎩

d l[ ± (

x − α − β√−1

)] = dx

x − α − β√−1

,

d l[ ± (

x − α + β√−1

)] = dx

x − α + β√−1

,

d e(a+b√−1 )z = e(a+b

√−1 )z(a + b

√−1)

dz.(24)

In these various formulas, we must adopt, after the letter l, the sign + or the sign −,

according to whether the imaginary expression whose Napierian logarithm is taken,has a positive or negative real part. From these same formulas, we will immediatelydeduce the following

(25)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ (A − B

√−1)

dx

x − α − β√−1

= (A − B

√−1)l[ ± (

x − α − β√−1

)] + C,

∫ (A + B

√−1)

dx

x − α + β√−1

= (A + B

√−1)l[ ± (

x − α + β√−1

)] + C,

(26)⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫e(a+b

√−1 )z dz = e(a+b√−1 )z

a + b√−1

+ C,

∫zne(a+b

√−1 )z dz

= zne(a+b√−1 )z

a + b√−1

[1 − n(

a + b√−1

)z

+ n(n − 1)(a + b

√−1)2

z2− · · ·

],

THIRTY-NINTH LECTURE. 219

which are in accordance with the formulas established in the twenty-eighth andthirtieth lectures.

Imaginary exponentials and imaginary logarithms can again be employed withadvantage in the determination of definite integrals. In this way, for example, itresults from the second of the equations in (26) that the second formula of page1283 remains valid when we replace the constant a, assumed real, by the imaginaryconstant a + b

√−1. We then obtain the equation∫ ∞

0zne(a+b

√−1 )z dz = 1 · 2 · 3 · · · n(a + b

√−1)n+1 ,(27)

which coincides with the third of the cited page.4 Moreover, it is clear that formula(18) of the thirty-fourth lecture will still remain valid, if instead of taking an alge-braic function for Υ (x), we successively set

Υ (x) = eax√−1, Υ (x) = (− x

√−1)μ−1

eax√−1, Υ (x) =

( − x√−1

)eax

√−1

l(1 − r x

√−1) ,

μ, a, r designating three positive constants, the first of which remains containedbetween the limits 0 and 2. We will find, by consequence,

∫ ∞

−∞eax

√−1

1 + x2dx =

∫ ∞

−∞cos ax dx

1 + x2= πe−a,(28)

∫ ∞−∞

( − x√−1

)μ−1

1 + x2eax

√−1 dx = 2∫ ∞0

xμ−1 sin(μπ

2− ax

) dx

1 + x2= πe−a,(29)

(30)

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫ ∞

−∞

( − x√−1

)eax

√−1

l(1 − r x

√−1) dx

1 + x2

=∫ ∞

0

sin ax l(1 + r2x2) + 2 cos ax arctang r x[12 l

(1 + r2x2

)]2 + (arctang r x)2

x dx

1 + x2= πe−a

l(1 + r).

3The 1899 edition reads, “…the formula given in line 13 of page 192.” Cauchy is referencing theformula ∫ ∞

0zne−az dz = 1 · 2 · 3 · · · n

an+1 ,

from his thirty-second lecture.4The 1899 edition reads, “…the formula from line 14 of the cited page.” Here, Cauchy is refer-encing the formula∫ ∞

0zne−az( cos bz + √−1 sin bz

)dz = 1 · 2 · 3 · · · n(

a + b√−1

)n+1 ,

from the same thirty-second lecture.

FORTIETH LECTURE.

INTEGRATION BY SERIES.

Consider a series

(1) u0, u1, u2, u3, . . . , un, . . .

whose different terms are functions of the variable x, that remain continuous be-tween the limits x = x0, x = X. If, after having multiplied these same terms by dx,we integrate between the limits in question, we will obtain a new series1 composedof the definite integrals

(2)∫ X

x0

u0 dx,∫ X

x0

u1 dx,∫ X

x0

u2 dx,∫ X

x0

u3 dx, . . . ,

∫ X

x0

un dx, . . . .

By comparing this new series to the first, we will establish2 without difficulty thetheorem that we now state.

THEOREM I.3 – Assume that, the two limits, x0, X being finite quantities, theseries (1) is convergent, not only for x = x0 and for x = X, but also for all the valuesof x contained between x0 and X. The series (2) will itself be convergent; and, if wecall s the sum of the series (1), the series (2) will have the integral

∫ X

x0

s dx

1The lists in (1) and (2) are, of course, both sequences, but as we have seen before, Cauchy isthinking of these terms eventually being added together to form a series.2The 1899 edition replaces the term établira with obtiendra, which would change the translationfrom will establish to will obtain. Cauchy’s original wording is used here.3It has been suggested Cauchy may be making an error here. The imprecise wording he uses,similar to his imprecise idea, description, and definition of continuity from well back in LectureTwo, has generated debate for years on whether Cauchy’s view of convergence is what we refer totoday as uniform convergence. This theorem deals with an exchange of limits operation and is onlytrue when applied to a sequence of uniformly convergent functions.


221


https://doi.org/10.1007/978-3-030-11036-9_40


for a sum. In other words, the equation

(3) s = u0 + u1 + u2 + u3 + · · ·will lead to the following

(4)∫ X

x0

s dx =∫ X

x0

u0 dx +∫ X

x0

u1 dx +∫ X

x0

u2 dx +∫ X

x0

u3 dx + · · · .

Proof. – Let

(5) sn = u0 + u1 + u2 + · · · + un−1

be the sum of the first n terms of the series (1), and rn the remainder starting fromthe nth term. We will have

(6)

{s = sn + rn

= u0 + u1 + u2 + · · · + un−1 + rn,

and we will infer

(7)

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∫ X

x0

s dx =∫ X

x0

u0 dx +∫ X

x0

u1 dx +∫ X

x0

u2 dx + · · ·

+∫ X

x0

un−1 dx +∫ X

x0

rn dx .

Now, since, by virtue of formula (14) (twenty-third lecture),4 the integral

∫ X

x0

rn dx

will be a particular value of the product rn(X − x0) corresponding to a value of xcontained between the limits x0, X, and that, in the admitted hypothesis, this prod-uct will become null for infinite values of n, it is clear that we will obtain equation(4) by setting, in formula (7), n = ∞.

Corollary I. – If, in formula (4), we replace X by x, we will obtain the following

(8)∫ x

x0

s dx =∫ x

x0

u0 dx +∫ x

x0

u1 dx +∫ x

x0

u2 dx + · · · ,

which will remain valid, like equation (3), between the limits x = x0, x = X.

Corollary II. – Assume that the series (1), being convergent for x = x0, and forall values of x contained between the limits x0, X, ceases to be for x = X. In this

4The Mean Value Theorem for Definite Integrals.

FORTIETH LECTURE. 223

hypothesis, equations (3) and (8) continue to still be correct between the limits inquestion. Let me add that equation (4) will itself remain valid, if the integrals con-tained in the second member form a convergent series. In fact, we will recognizewithout trouble that, if this condition is fulfilled, the two members of equation (8)will be continuous functions of the variable x in the neighborhood of the particu-lar value x = X (see Algebraic Analysis, page 131), and that it will suffice to letx converge toward this same value to obtain the two members of equation (4). Onthe other hand, equation (4) will not be valid if the integrals that are included in itssecond member form a divergent series.

Corollary III. – Assume that the series (1), being convergent between the limitsx = x0, x = X, becomes divergent for the first of these two limits, or for both. Then,in denoting by ξ0, ξ two quantities contained between x0 and X, we will obtain theequation

(9)∫ ξ

ξ0

s dx =∫ ξ

ξ0

u0 dx +∫ ξ

ξ0

u1 dx +∫ ξ

ξ0

u2 dx + · · · ;

by then letting ξ0 converge toward the limit x0, and ξ toward the limit X, we willregain equation (4) any time that the integrals included in the second member forma convergent series.

This remark extends itself to similar cases where the quantities x0, X becomeinfinite separately or simultaneously, for example, to the case where we would havex0 = −∞, X = ∞.

Corollary IV. – If we take un = anxn, an being a real or imaginary coefficient,and if, in addition, we denote by ρn ρn the numerical value or the magnitude of an,and by λ the largest value that receives the expression (ρn)

1n when the number n

becomes infinite, the series (1) will be convergent (see Theorem III of the thirty-eighth lecture) between the limits x = − 1

λ, x = + 1

λ. Therefore, by leaving the

variable x contained between these limits, and setting

(10) s = a0 + a1x + a2x2 + · · · ,

we will find

(11)∫ x

0s dx = a0x + a1

x2

2+ a2

x3

3+ · · · .

This last equation still remains valid (see Corollary II) for the particular values x =− 1

λ, x = + 1

λ, if these particular values do not cease to render the series a0x, 1

2a1x2,

13a2x

3, . . . convergent.

Using the principles that we have just established, we can expand a large numberof integrals by convergent series that produce values of these integrals approximated


as close as we will wish. It is this that consists of integration by series.5 We can alsoemploy this method of integration with advantage to expand by series all kinds ofquantities; and often, in order to achieve this, it is best to express the given quantitiesby definite integrals to which we can then apply the method in question.

Examples. – To expand by series the functions

l(1 + x), arctang x, arcsin x,

we will turn to the formulas

l(1 + x) =∫ x

0

dx

1 + x,

arctang x =∫ x

0

dx

1 + x2,

arcsin x =∫ x

0

dx√1 − x2

=∫ x

0(1 − x2)−

12 ;

and, since we will find, between the limits x = −1, x = +1,

1

1 + x= 1 − x + x2 − · · · ,

1

1 + x2= 1 − x2 + x4 − · · · ,

(1 − x2)−12 = 1 + 1

2x2 + 1 · 3

2 · 4 x4 + 1 · 3 · 5

2 · 4 · 6 x6 + · · · ,

integration by series will produce, between these same limits,

(12)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

l(1 + x) = x − x2

2+ x3

3− · · · ,

arctang x = x − x3

3+ x5

5− · · · ,

arcsin x = x + 1

2

x3

3+ 1 · 3

2 · 4x5

5+ 1 · 3 · 5

2 · 4 · 6x7

7+ · · · .

5Without being nearly as concerned about the convergence or divergence of the resulting series,Isaac Newton used this very method to integrate a multitude of various functions. For more details,be sure to investigate his De analysi per æquationes numero terminorum infinitas (On Analysisby Equations with an Infinite Number of Terms) of 1669, or simply De analysi for short, to seehim employ this integration method and for a wonderful demonstration of Newton’s use of theFundamental Theorem of Calculus.

FORTIETH LECTURE. 225

If, in the equations in (12), we set x = 1, the series contained in the second membersremain convergent, and we will have (by virtue of Corollary II),6

(13)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

l 2 = 1 − 1

2+ 1

3− · · · ,

π

4= 1 − 1

3+ 1

5− · · · ,

π

2= 1 + 1

2· 13

+ 1 · 32 · 4 · 1

5+ 1 · 3 · 5

2 · 4 · 6 · 17

+ · · · .

We can easily demonstrate (see Algebraic Analysis, page 163) that two conver-gent series ordered according to ascending and integer powers of x can only producethe same sum, for very small numerical values of x, so far as the coefficients of likepowers of x are equal in both series. From this remark and from Theorem III (thirty-eighth lecture), it results that, if the two series remain convergent and generate thesame sum for the real values of x contained between the limits −r, +r (r denotinga positive quantity), they will fulfill the same conditions for the imaginary values ofx whose magnitudes will be less than r. This granted, we will deduce the followingtheorem without difficulty from the principles established above.

THEOREM II. – If, for the real values of z contained between the limits z0, Z ,

and for the real values of x contained between the limits −r, +r, the functions

f (x, z)

and

(14) F(x) =∫ Z

z0

f (x, z) dz

can be expanded by the Theorem of Maclaurin into convergent series ordered ac-cording to ascending and integer powers of x; if, moreover, the sums of theseseries, when x becomes imaginary, continue to be represented by the notationsf (x, z), F(x), equation (14) will remain valid for the imaginary values of x whosemagnitudes will be less than r.

Example. – Since we have, for any values of x,

π12 =

∫ ∞

−∞e−z2 dz =

∫ ∞

−∞e−(z+x)2 dz = e−x2

∫ ∞

0e−z2(e−2zx + e2zx ) dz,

6The second series in equation (13) is the famous Leibniz series (or Leibniz formula)—certainlyone of the most beautiful and simple relationships in all of mathematics. In an interesting way, theseries connects the most important irrational number in geometry with the integers.


and as a result,

(15)∫ ∞

0e−z2 e

2zx + e−2zx

2dz = 1

2π12 ex

2,

we will deduce, in replacing x by x√−1,

(16)∫ ∞

0e−z2 cos 2zx dz = 1

2π12 e−x2 .

This last formula, that we owe to Mr. Laplace,7 is very useful in the solution to sev-eral problems.

THE END.8

7Pierre-Simon Laplace (1749–1827), a family friend, was an important figure in the early devel-opment of Cauchy’s education and career. Laplace is one of the most well-known and importantmathematicians and physicists during this time period (although Laplace himself would have ar-gued these two disciplines were inseparable). Laplace was highly influential to Cauchy throughmuch of his life in many ways—as a teacher, mentor, and resource. One of Cauchy’s most impor-tant mathematics references was Laplace’s own Traité de Mécanique céleste (Treatise on CelestialMechanics) published in 1797. Cauchy held a deep regard and indebtedness to Laplace for his helpand teaching during his early career and wrote a special thanks to him in the Introduction of his1821 Cours d’analyse book.8It turns out Cauchy was not actually finished with his text before the publication date. He addsone more section before the printing of the original 1823 edition commenced, what he will title hisADDITION –an incredible appendix which includes new proofs for some of the most importantresults in calculus.

ADDITION.

Since the printing of this work, I recognized that with the help of a very simple for-mula we could bring back to the differential calculus the solution of several prob-lems that I returned to the integral calculus. I will, in the first place, give this formula;then, I will indicate its main applications.

From what has been said in the seventh lecture, if we denote by x0, X two valuesof x between which the functions f (x) and f ′(x)1 remain continuous, and by θ anumber less than unity, we will have

f (X) − f (x0)

X − x0= f ′[x0 + θ(X − x0)

].

Now, it is easy to see that reasoning entirely similar to that which we used to demon-strate the preceding equation will suffice to establish the formula

f (X) − f (x0)

F(X) − F(x0)= f ′[x0 + θ(X − x0)

]

F ′[x0 + θ(X − x0)] ,(1)

This final portion of Calcul infinitésimal was attached to the text by our Mr. Cauchy afterhis initial submission, and while it awaited printing. This important addendum includes proofs ofthe Generalized Mean Value Theorem for Derivatives, Cauchy’s version of l’Hôpital’s Rule, andhis conditional differential calculus derivation of Taylor’s formula. One may wonder why suchan important portion of the text is left for the end of the book and included as an apparentafterthought. Unfortunately, no record exists to fully explain the added notes and proofs, but a pos-sible explanation could have to do with the situation Cauchy found himself. He and Ampère (thetwo professors responsible for the analysis course at the École Polytechnique) were under constantpressure from the Conseil d’Instruction to construct a course on analysis in which they (the Conseil)would approve – one which developed the calculus quicker and included more applications. Giventhe importance of Taylor series expansions in the curriculum at the time, Cauchy may have felt itnecessary to include these new derivations so that Taylor’s formula could be presented in a courseimmediately following the development of differential calculus. However, the true story is notknown and is likely lost to history.

1The 1899 text has “ f (x) and f (x),” an obvious typographical error that does not occur in theoriginal 1823 edition.

© Springer Nature Switzerland AG 2019D. M. Cates, Cauchy’s Calcul Infinitésimal,https://doi.org/10.1007/978-3-030-11036-9

227

https://doi.org/10.1007/978-3-030-11036-9


θ still denoting a number less than unity, and F(x) a new function that, always grow-ing or decreasing from the limit x = x0 up to the limit x = X, remains continuous,along with its derivative F ′(x), between these same limits.2

We can also demonstrate formula (1) directly with the help of principles estab-lished in the sixth lecture (page 21).3 In fact, it results from these principles that, inthe admitted hypothesis, the function F ′(x)will constantly retain the same sign fromx = x0 up to x = X. As a result, if A and B represent the smallest and the largest ofthe values that the ratio f ′(x)

F ′(x) receives in this interval, the two products

F ′(x)[f ′(x)F ′(x)

− A

]= f ′(x) − A F ′(x),

F ′(x)[B − f ′(x)

F ′(x)

]= B F ′(x) − f ′(x)

will remain, one and the other, constantly positive or constantly negative betweenthe limits x0, X of the variable x . Therefore, the two functions

f (x) − A F(x), B F(x) − f (x),

that have these same products for derivatives, will grow or will decline simultane-ously from the first limit up to the second. So, the difference between the extremevalues of the first function, namely

f (X) − f (x0) − A[F(X) − F(x0)

],

and the difference between the extreme values of the last, namely

B[F(X) − F(x0)

] − [f (X) − f (x0)

],

will be two quantities of the same sign;4 hence, we can conclude that the difference

f (X) − f (x0)

will be contained between the two products

A[F(X) − F(x0)

], B

[F(X) − F(x0)

],

and the fractionf (X) − f (x0)

F(X) − F(x0),

2Cauchy realizes this simple first proof of the General Mean Value Theorem is not adequate, as heimmediately begins a second, more convincing proof. The first proof does not guarantee the twoθ’s are identical. It is not clear why Cauchy even includes this flawed proof in his text.3Cauchy seems to be referring to Problem I of his sixth lecture where he demonstrates, amongother things, that a constantly increasing or decreasing function has a nonzero derivative that doesnot change sign on the interval of interest.4Because the two functions are known to either both be increasing or both be decreasing.

ADDITION. 229

between the limits A and B. Moreover, the two functions f ′(x), F ′(x), being con-tinuous by hypothesis between the limits x = x0, x = X, any quantity containedbetween A and B will be equivalent to an expression of the form5

f ′[x0 + θ(X − x0)]

F ′[x0 + θ(X − x0)] ,

θ denoting a number less than unity.6 There will exist, therefore, a number of thiskind that works to satisfy equation (1); it is this that is necessary to demonstrate.7

If we let X = x0 + h, equation (1) will become

f (x0 + h) − f (x0)

F(x0 + h) − F(x0)= f ′(x0 + θh)

F ′(x0 + θh).(2)

This latter equation, which includes as a particular case equation (6) of the seventhlecture, is open to several important applications, as we will prove in a few words.

First, consider that the functions f (x) and F(x) vanish, one and the other, forx = x0, and let, for brevity, θh = h1. In this case, we will derive from formula (2)

f (x0 + h)

F(x0 + h)= f ′(x0 + h1)

F ′(x0 + h1),(3)

h1 being a quantity of the same sign as h, but a value numerically less. If the func-tions

f (x), f ′(x), f ′′(x), . . . , f (n−1)(x),

F(x), F ′(x), F ′′(x), . . . , F (n−1)(x)

all vanish for x = x0, and remain continuous, aswell as those of f (n)(x) and F (n)(x),between the limits x = x0, x = x0 + h, then, by supposing each of the functions

F(x), F ′(x), F ′′(x), . . . , F (n−1)(x)

always increasing or always decreasing from the first limit up to the second, anddenoting by

h1, h2, . . . , hn

quantities of the same sign, but whose numerical values are from most to least,we would obtain along with equation (3), a sequence of similar equations which

5It helps to recall A <f ′(x)F ′(x) < B, for all x values within the interval. The 1899 version contains a

typographical error here. The reprint uses x in place of the correct X. This error does not occur inthe 1823 edition.6These two θ ’s are identical.7Cauchy has just ended his proof of what is known today as a result very close to the GeneralizedMean Value Theorem, sometimes called Cauchy’s Mean Value Theorem, or the Extended MeanValue Theorem. Most modern versions of the theorem restrict F(x) in a similar way by requiringF(x) �= 0, for all x in (x0, X).


together would comprise the formula

f (x0 + h)

F(x0 + h)= f ′(x0 + h1)

F ′(x0 + h1)= f ′′(x0 + h2)

F ′′(x0 + h2)= · · · = f (n)(x0 + hn)

F (n)(x0 + hn).(4)

If, in formula (4), we are content to set the first fraction equal to the last, the equationto which we will arrive can be written as follows

f (x0 + h)

F(x0 + h)= f (n)(x0 + θh)

F (n)(x0 + θh),(5)

θ always being a number less than unity. In conclusion, if in equation (5) we sub-stitute for the finite quantity h, an infinitely small quantity8 denoted by i, we willhave

f (x0 + i)

F(x0 + i)= f (n)(x0 + θ i)

F (n)(x0 + θ i).(6)

When, in formulas (5) and (6), we set9

F(x) = (x − x0)n,

we findF (n)(x) = 1 · 2 · 3 · · · n,

and by consequence,

f (x0 + h)

hn= f (n)(x0 + θh)

1 · 2 · 3 · · · n ,(7)

f (x0 + i)

i n= f (n)(x0 + θ i)

1 · 2 · 3 · · · n .(8)

These last equations are in accordance with formulas (4) and (5) of the fifteenthlecture and coincide with formulas (17), (18) of the thirty-sixth. They can be em-ployed with advantage, not only in the study of maxima and minima, but again inthe determination of the values of fractions which are presented under the form 0

0 .

Moreover, to resolve this last problem, it will most often suffice to resort to formula(6). Suppose, in fact, that the two terms of the fraction

f (x)

F(x),

8Recall from Cauchy’s Lecture One, an “infinitely small quantity” is one whose limit is zero, notnecessarily one that ever becomes zero.9Cauchy stops short of taking limits here, interrupting his derivation with a specific examplewhich he will employ later.

ADDITION. 231

and their successive derivatives, up to those of order n − 1, vanish for x = x0. For-mula (6) will generally remain valid for very small numerical values of i, because ingeneral, each of the functions

F(x), F ′(x), F ′′(x), . . . , F (n−1)(x)

will grow or will decline steadily from the particular value of x represented by x0,up to a value very close; and, we will derive from this formula, by letting i convergetoward the limit zero,10

limf (x0 + i)

F(x0 + i)= lim

f (n)(x0 + θ i)

F (n)(x0 + θ i)= f (n)(x0)

F (n)(x0).(9)

If we replace in formula (7), x0 by zero, and the letter f by F, we will infer

F(h) = hn

1 · 2 · 3 · · · nF(n)(θh).(10)

This last formula assumes that the functions

F(h), F′(h), F′′(h), . . . , F(n)(h),

being continuous starting from the limit h = 0, all vanish, with the exception ofF(n)(h), at the same time as the quantity h.11

Now, let f (x) be an arbitrary function of the variable x, but such that

f (x + h), f ′(x + h), f ′′(x + h), . . . , f (n)(x + h)

remain continuous with respect to h, starting from h = 0. With the help of formula(10), we can easily extract from f (x + h), or to what amounts to the same thing,from the difference f (x + h) − f (x), a sequence of terms proportional to the integerpowers of h; andfirst, since the difference f (x + h) − f (x), considered as a functionof h, vanishes with h and has f ′(x + h) for its first order derivative, it is clear that bysubstituting this function for F(h), and setting n = 1, we will derive from formula(10)

f (x + h) − f (x) = h

1f ′(x + θh).(11)

When, in the second member of the preceding equation, we replace θ by zero, weobtain the term h

1 f′(x), and by subtracting this term from the first member, we find

10After now taking limits, Cauchy has developed his form of l’Hôpital’s Rule as an intermediateresult on his way to Taylor’s formula. This is the final of several inquiries within this text into howto evaluate an indeterminate fraction that presents itself in the form 0

0 .11This condition will need to be satisfied throughout this derivation.


for the remainder a new function of h, namely

f (x + h) − f (x) − h

1f ′(x).

As this new function of h vanishes with h, along with its first order derivative, andit has f ′′(x + h) for its second order derivative, by substituting this new function ofh for F(h), and setting n = 2, we will derive from formula (10)12

f (x + h) − f (x) − h

1f ′(x) = h2

1 · 2 f ′′(x + θh).(12)

If, in the second member of equation (11), we replace θ by zero, we will obtain theterm h2

1·2 f′′(x), and by subtracting this term from the first member, we will find for

the remainder a third function of h, namely

f (x + h) − f (x) − h

1f ′(x) − h2

1 · 2 f ′′(x).

As this third function of h vanishes with h, along with its first and second orderderivatives, and it has f ′′′(x + h) for its third order derivative, by substituting thisthird function of h for F(h), and setting n = 3, we will derive from formula (10)

f (x + h) − f (x) − h

1f ′(x) − h2

1 · 2 f ′′(x) = h3

1 · 2 · 3 f ′′′(x + θh),(13)

etc. By continuing in this same manner, we will generally establish the formula⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f (x + h) − f (x) − h

1f ′(x) − h2

1 · 2 f ′′(x) − · · ·

− hn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x) = hn

1 · 2 · 3 · · · n f (n)(x + θh),

(14)

which coincides with equation (12) of the thirty-sixth lecture. If, in this formula, wereplace x by zero, h by x, and f by F, F(x) denoting an arbitrary function of x, wewill find

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

F(x) − F(0) − x

1F ′(0) − x2

1 · 2 F′′(0) − · · ·

− xn−1

1 · 2 · 3 · · · (n − 1)F (n−1)(0) = xn

1 · 2 · 3 · · · n F(n)(θx).

(15)

This last equation coincides with formula (11) of the thirty-sixth lecture, and we canagain directly achieve this in the following manner.

12The1899 reprint has a typographical error. It reads f (x + h) − f (x) − h1 f ′(x) = h2

1·2 f ′(x + θh).

This error is not present in the 1823 edition.

ADDITION. 233

Let F(x) be any function of x, and let �(x) be an entire polynomial of degreen − 1, subject to satisfy the condition equations

�(0) = F(0), � ′(0) = F ′(0),

� ′′(0) =F ′′(0), . . . , � (n−1)(0) = F (n−1)(0).

� (n)(x) then being identically null, if in formula (10) we replace h by x, and F(x)by F(x) − �(x), we will find

F(x) − �(x) = xn

1 · 2 · 3 · · · n F(n)(θx);(16)

and, moreover, as we will have (see the nineteenth lecture)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

�(x) = �(0) + x

1� ′(0) + x2

1 · 2� ′′(0) + · · · + xn−1

1 · 2 · 3 · · · (n − 1)�(n−1)(0),

= F(0) + x

1F ′(0) + x2

1 · 2 F′′(0) + · · · + xn−1

1 · 2 · 3 · · · (n − 1)F (n−1)(0),

(17)

it is clear that formula (16) will lead to equation (15).It is important to observe that, in all the cases where the second members of equa-

tions (14) and (15) converge toward zero for increasing values of n, we immediatelydeduce from these formulas the Theorems of Taylor and of Maclaurin.13

If, in formula (8), we have x0 = 0, it would simply become

f (i)

i n= f (n)(θ i)

1 · 2 · 3 · · · n .(18)

This equation above assumes that the functions

f (i), f ′(i), f ′′(i), . . . , f (n−1)(i), f (n)(i),

being continuous for very small numerical values of i, all vanish, with the exceptionof the last, for i = 0. In this hypothesis, the ratios

f (i)

i,

f (i)

i2, . . . ,

f (i)

i n−1,

being themselves equivalent to expressions of the form

f ′(θ i)1

,f ′′(θ i)1 · 2 , . . . ,

f (n−1)(θ i)

1 · 2 · 3 · · · (n − 1),

13Cauchy’s earlier work on Taylor’s Theorem from Lectures Thirty-Five through Thirty-Seveninvolves integral calculus. Here, he shows the relationships can likewise be derived through differ-ential calculus only with the observed limitations.


will all vanish with i. By consequence, i and f (i) representing two infinitely smallquantities,

f (i)

i n

will be the first term of the geometric progression

f (i),f (i)

i,

f (i)

i2,

f (i)

i3, . . .(19)

which ceases to be an infinitely small quantity, if f (n)(0) is the first of the quantities

f (0), f ′(0), f ′′(0), f ′′′(0), . . .(20)

which ceases to be null. Add that, in the admitted hypothesis,

f (n)(0)

1 · 2 · 3 · · · nwill be, by virtue of formula (18), the actual value of the ratio f (i)

i n corresponding toi = 0.

The preceding considerations naturally lead us to partition the infinitely smallquantities into different classes. Conceive, in fact, that all the quantities of this typethat enter in a calculation are functions of one from among them designated by i, andwe name f (i) one of these functions. Several consecutive terms of the progression(19), counted starting from the first term, may be infinitesimals; and, depending onwhether the number of these terms will be 1, 2, 3, . . . , we will say that the quantityf (i) is an infinitesimal of first, of second, of third class, etc. This granted, f (i) willbe an infinitesimal of the nth class, if f (i)

i n is the first term of the progression (19)which ceases to vanish with i. In the same hypothesis, f (i) will become what wecall an infinitesimal of order n, if for decreasing numerical values of i, the ratio f (i)

i n

converges toward a finite limit different from zero.These definitions being accepted, we will immediately deduce the following

propositions from the principles established above.

THEOREM I. – When f (i) is an infinitesimal of nth class, f (n)(0) is the firstterm of the series (20) which ceases to be null. In the same case, f (i) will be aninfinitesimal of order n, if f (n)(0) obtains a finite value different from zero.

THEOREM II. – When, f (i) being an infinitesimal of nth class, the functionf (x) and its successive derivatives, up to order n, remain continuous between thelimits x = 0, x = h, we have, in designating by m an integer number less than orequal to n,

f (h) = hm

1 · 2 · 3 · · ·m f (m)(θh).(21)

ADDITION. 235

If, in this last formula, we replace h by i, and m by n, we will end up with equa-tion (18), with the help of which we can establish the theorem that we will now state.

THEOREM III. – Let f (i) be an infinitely small quantity of order n. This quan-tity will change its sign with i if n is an odd number, and will be constantly affectedby the same sign as f (n)(0) if n is an even number.

Theorem III assumes, like formula (18), that the function f (i) and its successivederivatives, up to that of order n, remain continuous with respect to i in the neigh-borhood of the particular value i = 0. If this condition has not been satisfied, thequantity designated by f (n)(0) could admit several values, and if these values arenot all of the same sign, the theorem in question would cease to exist. It is this thatwould happen, for example, if we were to take for f (i) the infinitely small quantity√i2. In this hypothesis, the derived function14

f ′(i) = i√i2

would admit one solution of continuity15 corresponding to i = 0, and it would some-times reduce to +1, sometimes to −1, depending on whether the value of i waspositive or negative. Moreover, it is obvious that the quantity

√i2, although we are

naturally inclined to consider it as an infinitesimal of first order, constantly remainspositive and does not change sign with i. The same remark applies to the infinitelysmall quantity

√i6 that we are naturally led to regard as an infinitesimal of third

order, etc.Theorem I provides a very simple means to recognize the class or order of an

infinitely small quantity. Thus, for example, we will conclude from this theoremthat the quantities

1

li,

√i, i

23 , sin i

are four infinitesimals of first class, the last being only of first order. We will ensureby the same manner that the four quantities

i

li, i

32 , sin2 i, 1 − cos i

are infinitesimals of second class, the last two being of second order; that the threequantities

i2

li, i3, i − sin i

are infinitesimals of third class, the last two being of third order; and so on.When we multiply an infinitesimal of nth class or of nth order by a constant quan-

tity or by a function of i which has for a limit a finite quantity different from zero,

14Recall this is Cauchy’s term, adopted from Lagrange, to denote our modern derivative.15Another use of Cauchy’s older term for a point of discontinuity.


we obviously obtain for a product another infinitesimal of the same class or of thesame order as the first.

It is again easy to prove that, among infinitely small quantities, those which per-tain to upper classes end up constantly obtaining the smallest numerical values. Infact, let ϕ(i), χ(i) be two infinitely small quantities, the first of nth class, the secondof mth, m being < n. The first of the two fractions ϕ(i)

im ,χ(i)im will be the only one

which converges with i toward the limit zero; and as a result, the ratio that we obtainin dividing one by the other, or the fraction ϕ(i)

χ(i) , will equivalently converge towardzero, which it cannot do without its numerical value dropping below unity, or inother words, without the numerical value of the numerator becoming less than thatof the denominator.

Finally, we will easily establish the following proposition.

THEOREM IV. –Designate by i and by f (i) two infinitely small quantities. Zerowill be the single value, or one of the values, that the ratio

f (i)

f ′(i)(22)

receives when we will cause the quantity i to vanish.

Proof. – It is obviously sufficient to demonstrate Theorem IV in the case wherethe derived function f ′(i) vanishes at the same time as f (i) for i = 0, expecting thatthe limit of the ratio f (i)

f ′(i) , to be null in all other hypotheses, is then presented only

under the indeterminate form 00 . Now, we will achieve this without difficulty with

the help of formula (18), at least when the two functions f (i), f ′(i) are continuouswith respect to i in the vicinity of the particular value i = 0. In fact, if this conditionis fulfilled, we will derive from formula (18), by setting n = 1,

f (i) = i f ′(θ i),(23)

and we will have, by consequence,

f (i)

f ′(i)= i

f ′(θ i)f ′(i)

,(24)

θ always denoting a number less than unity. Let us now conceive that, in formula(24), we allow the numerical value of i to decrease indefinitely. f ′(0) being null byhypothesis, and θ i denoting a quantity contained between zero and i, f ′(θ i) willconverge toward the limit zero more rapidly than f ′(i), where it results that thefraction f ′(θ i)

f ′(i) will obtain a multitude of numerical values less than unity, and the

product i f ′(θ i)f ′(i) a multitude of values essentially null. Therefore, the limit, or one of

the limits, toward which this same product will converge, and the ratio it represents,will be equal to zero.

ADDITION. 237

Scholium I. – Theorem IV can be easily verified with regard to the functions

sin i, 1 − cos i, e−( 1i )

2

, i3 sin1

i, . . . .

It remains valid even in the case where the function f (i) does not remain real andinfinitely small as we attribute to the variable i values affected by a certain sign, asit happens, for example, when we take for f (i) one of the functions

li,√i, e− 1

i , e−( 1i )

3

, . . .

which cease to be real or infinitely small when we give negative values to i. Finally,this theorem can remain valid, even though the function f ′(i) becomes discontinu-ous for i = 0. Thus, by supposing

f (i) = i sin1

i,(25)

we will find that the function

f ′(i) = sin1

i− 1

icos

1

i(26)

becomes indeterminate, and by consequence, discontinuous for i = 0; and, if wethen let i converge toward the limit zero, the value of the ratio (22) derived fromequations (25) and (26), namely

f (i)

f ′(i)= i

1 − 1i cot

1i

,(27)

will admit an infinite number of limits, one of which will be equal to zero.

Scholium II. – Suppose that, the function f (i) and its successive derivatives, up tothat of order n − 1, being continuous with respect to i in the vicinity of the particularvalue i = 0, the n quantities

f (0), f ′(0), f ′′(0), . . . , f (n−1)(0)(28)

vanish; and let us conceive that the numerical value of i comes to decrease indefinitely.Zero will be the limit, or one of the limits, toward which each of the ratios

f (i)

f ′(i),

f ′(i)f ′′(i)

,f ′′(i)f ′′′(i)

, . . . ,f (n−1)(i)

f (n)(i)(29)

will converge, and by consequence, their product, or the ratio

f (i)

f (n)(i).(30)


We can say as much of the expressions

f ′(i)f (n)(i)

,f ′′(i)f (n)(i)

, . . . ,f (n−2)(i)

f (n)(i)(31)

that we obtain by multiplying, one by the others, some of the ratios in question.

This page ends the published 1823 version of Mr. Cauchy’s original Calcul infinitésimal. How-ever, the 1899 reprint includes an added short research paper entitled “On the Formulas of Taylorand Maclaurin,” which can be found in Appendix D.

APPENDIX A.

COURS D’ANALYSE – CHAPTER II, §III.

THEOREM I.1 – If, for increasing values of x, the difference

f (x + 1) − f (x)

converges toward a certain limit k, the fraction

f (x)

x

will converge at the same time toward the same limit.

Proof. – First, we suppose that the quantity k has a finite value, and designateby ε a number as small as we will wish. Since the increasing values of x make thedifference

f (x + 1) − f (x)

converge toward the limit k, we can give to the number h a value considerableenough, such that, x being greater than or equal to h, the difference in question isconstantly contained between the limits

k − ε, k + ε.

Included in this Appendix A are excerpts fromCauchy’sCours d’analyseChapter II, §III, wherehe proves two straightforward convergence tests. These two theorems are referenced by Cauchy inhis Calcul infinitésimal Lecture Seven by page number. However, they are particularly importantand interesting because within Cauchy’s actual proofs of these theorems we clearly see how hetransforms his verbal definition of the limit into an algebra of inequalities squeeze.

1This is the precursor of Cauchy’s Theorem II in his Lecture Thirty-Eight and eventually leads tothe Ratio Convergence Test for infinite series.


239

https://doi.org/10.1007/978-3-030-11036-9

240 COURS D’ANALYSE – CHAPTER II.

This granted, if we designate by n any integer number, each of the quantities

f (h + 1) − f (h),

f (h + 2) − f (h + 1),

. . . . . . . . . . . . . . . . . . . . . ,

f (h + n) − f (h + n − 1),

and as a result, their arithmetic average, namely

f (h + n) − f (h)

n,

will be found contained between the limits k − ε, k + ε. We will have, therefore,

f (h + n) − f (h)

n= k + α,

α being a quantity contained between the limits −ε, +ε. Now, let

h + n = x .

The previous equation will become

f (x) − f (h)

x − h= k + α,(1)

and we will infer

f (x) = f (h) + (x − h)(k + α),

f (x)

x= f (h)

x+

(1 − h

x

)(k + α).(2)

In addition, to make the value of x grow indefinitely, it will suffice to let theinteger number n grow indefinitely without changing the value of h. We suppose,by consequence, that in equation (2) we consider h as a constant quantity, and x asa variable quantity which converges toward the limit ∞. The quantities

f (h)

x,

h

x

contained in the second member will converge toward the limit zero, and the secondmember itself toward a limit of the form

k + α,

APPENDIX A. 241

α being always contained between −ε and +ε. As a result, the ratio

f (x)

x

will have for a limit a quantity contained between k − ε and k + ε. This conclusionabove remains valid, regardless of the smallness of the number ε, it follows that thelimit in question will be precisely the quantity k. In other words, we will have

limf (x)

x= k = lim

[f (x + 1) − f (x)

].(3)

In the second place, we suppose k = ∞. Then, by denoting with H a number aslarge as we will wish, we can always attribute to the number h a value considerableenough, such that, x being greater than or equal to h, the difference

f (x + 1) − f (x),

which converges toward the limit ∞, becomes constantly greater than H ; and, byreasoning like that above, we will establish the formula

f (h + n) − f (h)

n> H.

If we now set h + n = x,wewill find, in place of equation (2), the following formula

f (x)

x>

f (h)

x+ H

(1 − h

x

),

from which we will deduce, by letting x converge toward the limit ∞,

limf (x)

x> H.

The limit of the ratiof (x)

x

will be, therefore, greater than the number H, however large it is. This limit, greaterthan any assignable number, can only be that of positive infinity.

In finishing, we suppose k = −∞. To return this last case to the previous one, itwill suffice to observe that the difference

f (x + 1) − f (x),

having for a limit −∞, the following

[ − f (x + 1)] − [ − f (x)

],


will have +∞ for a limit. We will conclude that the limit of − f (x)x is equal to +∞,

and as a result, that of f (x)x to −∞.

...

THEOREM II.2 – If, the function f (x) being positive for very large values of x,the ratio

f (x + 1)

f (x)

converges, while x grows indefinitely, toward the limit k, the expression

[f (x)

] 1x

will converge at the same time toward the same limit.

Proof. – First, we suppose that the quantity k, necessarily positive, has a finitevalue, and designate by ε a number as small as we will wish. Since the increasingvalues of x make the ratio

f (x + 1)

f (x)

converge toward the limit k, we can give to the number h a value considerableenough, such that, x being greater than or equal to h, the ratio in question is con-stantly contained between the limits

k − ε, k + ε.

This granted, if we designate by n any integer number, each of the quantities

f (h + 1)

f (h),

f (h + 2)

f (h + 1), . . . ,

f (h + n)

f (h + n − 1),

and as a result, their geometric average, namely

[f (h + n)

f (h)

] 1n

,

will be found contained between the limits k − ε, k + ε. We will have, therefore,

[f (h + n)

f (h)

] 1n

= k + α,

2This is the precursor of Cauchy’s Theorem I in his Lecture Thirty-Eight and eventually leads tothe Root Convergence Test for infinite series.

APPENDIX A. 243

α being a quantity contained between the limits −ε, +ε. Now, let

h + n = x .

The previous equation will become

[f (x)

f (h)

] 1x−h

= k + α,(4)

and we will conclude

f (x) = f (h)(k + α

)x−h,

[f (x)

] 1x = [

f (h)] 1

x (k + α)1−hx .(5)

In addition, to make the value of x grow indefinitely, it will suffice to let the integernumber n grow indefinitely without changing the value of h. We suppose, by conse-quence, that in equation (5) we consider h as a constant quantity, and x as a variablequantity which converges toward the limit ∞. The quantities

[f (h)

] 1x , 1 − h

x

contained in the second member will converge toward the limit 1, and the secondmember itself toward a limit of the form

k + α,

α being always contained between −ε and +ε. As a result, the expression

[f (x)

] 1x

will have for a limit a quantity contained between k − ε and k + ε.

This conclusion above remains valid, regardless of the smallness of the numberε, it follows that the limit in question will be precisely the quantity k. In otherwords, we will have

lim[f (x)

] 1x = k = lim

f (x + 1)

f (x).(6)

In the second place, we suppose the quantity k infinite, that is to say, since thisquantity is positive, k = ∞. Then, in denoting by H a number as large as we willwish, we can always attribute to the number h a value considerable enough, suchthat, x being greater than or equal to h, the ratio

f (x + 1)

f (x),


which converges toward the limit ∞, becomes constantly greater than H ; and, byreasoning like that above, we will establish the formula

[f (h + n)

f (h)

] 1n

> H.

If we now set h + n = x,wewill find, in place of equation (5), the following formula

[f (x)

] 1x >

[f (h)

] 1x H 1− h

x ,

from which we will deduce, by letting x converge toward the limit ∞,

lim[f (x)

] 1x > H.

The limit of the relationship[f (x)

] 1x

will be, therefore, greater than the number H, however large it is. This limit, greaterthan any assignable number, can only be that of positive infinity.

...

APPENDIX B.

COURS D’ANALYSE – NOTE II.

We now turn to theorems on averages. As we have already said (Preliminaries,p. 14), we call an average among several given quantities a new quantity containedbetween the smallest and the largest of those that we consider.1 From this definition,the quantity h will be an average between the two quantities g, k, or between severalquantities among which one of the two that we just mentioned would be the largestand the other the smallest, if the two differences g − h and h − k are of the samesign. This granted, if, to denote an average among the quantities a, a′, a′′, . . . weemploy, as in the Preliminaries, the notation

M(a, a′, a′′, . . .

),

we will establish without difficulty the following propositions.

...

THEOREM XII.2 – Let b, b′, b′′, . . . be several quantities of the same sign, nin number, and a, a′, a′′, . . . any quantities equal in number to those of the first. Wewill have

a + a′ + a′′ + · · ·b + b′ + b′′ + · · · = M

(a

b,a′

b′ ,a′′

b′′ , . . .)

.(17)

Note II is a large addendum of Cauchy’s Cours d’analyse comprised of seventeen theorems. Ofparticular interest to us are his definition of an average and his Theorem XII, both of which playpivotal roles in his Calcul infinitésimal.

1This unusual definition of an average is used throughout the Calcul infinitésimal text, where itplays a prominent role. Cauchy’s definition is nothing at all like the modern notion.2Cauchy’s Theorem XII is a result he calls upon in his Calcul infinitésimal Lectures Seven andTwenty-One. It also provides a wonderful detailed example of one of his proof techniques.


245

https://doi.org/10.1007/978-3-030-11036-9

246 COURS D’ANALYSE – NOTE II.

Proof. – Let g be the largest and k the smallest of the quantities

a

b,

a′

b′ ,a′′

b′′ , . . . .

The differences

g − a

band

a

b− k,

g − a′

b′ anda′

b′ − k,

g − a′′

b′′ anda′′

b′′ − k,

. . . . . . . . . . . . . . . ,

will all be positive. In multiplying the first two by b, the following two by b′, etc.,we will obtain the products

gb − a and a − kb,

gb′ − a′ and a′ − kb′,gb′′ − a′′ and a′′ − kb′′,. . . . . . . . . . . . . . . ,

which will all be of the same sign, as well as the quantities b, b′, b′′, . . . .As a result,the sums of these two kinds of products, namely

g(b + b′ + b′′ + · · · ) − (

a + a′ + a′′ + · · · ),a + a′ + a′′ + · · · − k

(b + b′ + b′′ + · · · ),

and the quotients of these sums by b + b′ + b′′ + · · · , namely

g − a + a′ + a′′ + · · ·b + b′ + b′′ + · · · ,

a + a′ + a′′ + · · ·b + b′ + b′′ + · · · − k,

will again be quantities of the same sign, from which we will conclude

a + a′ + a′′ + · · ·b + b′ + b′′ + · · · = M(g, k) = M

(a

b,a′

b′ ,a′′

b′′ , . . .)

(see in the Preliminaries, Theorem I and formula (6)

).

...

APPENDIX B. 247

Corollary III.3 – If we designate by α, α′, α′′, . . . new quantities which are allof the same sign, we will have, by virtue of equation (17),

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

αa + α′a′ + α′′a′′ + · · ·αb + α′b′ + α′′b′′ + · · · = M

(αa

αb,α′a′

α′b′ ,α′′a′′

α′′b′′ , . . .)

= M

(a

b,a′

b′ ,a′′

b′′ , . . .)

.

(20)

This latter formula is sufficient to establish Theorem III of the Preliminaries.

3This obscure result is used by Cauchy within Lecture Twenty-One in his effort to show thedefinite integral exists independent of the derivative.

APPENDIX C.

COURS D’ANALYSE – NOTE III.

ON THE NUMERICAL RESOLUTION OF EQUATIONS.

To numerically resolve one or several equations is to find the numerical valuesof the unknowns which they contain, which obviously requires that the constantsincluded in the equations in question are themselves reduced to numbers. We willonly occupy ourselves here with equations which contain one unknown, and we willstart by establishing, to their regard, the following theorems.

THEOREM I. – Let f (x) be a real function of the variable x, which remainscontinuous with respect to this variable between the limits x = x0, x = X.1 If the twoquantities f (x0), f (X) are of different signs, we will be able to satisfy the equation

f (x) = 0(1)

by one or several real values of x contained between x0 and X.2

Cauchy’s Note III is again located in the ending material of his 1821 Cours d’analyse and ingeneral deals with finding the numerical solutions to equations. It is also a fairly lengthy addendumto his original work which includes multiple theorems and problems. Only his Theorem I from thisNote III is included in our Appendix C, as it is here where we find his wonderful proof of theIntermediate Value Theorem (a result absolutely fundamental to the entire Calcul infinitésimaltext).

1Cauchy intuitively recognizes one of the most important properties of continuous single-vari-able functions, namely their preservation of sequential limits. Given a sequence {xk}∞k=1 ={x1, x2, x3, . . . } such that limk→∞ xk = a, if f (x) is continuous,wewillhave limx→a f (x) = f (a).

In other words, a limit can be brought inside a continuous function.2This result is today known as Bolzano’s Theorem and is a special case of the Intermediate ValueTheorem. It is named for Bernard Bolzano (1781–1848), a talented Bohemian monk who devel-oped work similar to Cauchy’s on this subject during roughly the same period. Bolzano’s work isgreatly underappreciated and not nearly as well known today as it deserves.


249

https://doi.org/10.1007/978-3-030-11036-9

250 COURS D’ANALYSE – NOTE III.

Proof. – Let x0 be the smallest of the two quantities x0, X. Allow

X − x0 = h,

and designate by m any integer number greater than unity. Since, of the two quan-tities f (x0), f (X), one is positive, the other is negative, if we form the sequence3

f (x0), f

(x0 + h

m

), f

(x0 + 2h

m

), . . . , f

(X − h

m

), f (X),

and that, in this sequence, we successively compare the first term with the second,the second with the third, the third with the fourth, etc., we will necessarily end byfinding one or several times, two consecutive terms that will have different signs.Let

f (x1), f (X ′)

be two terms of this type, x1 being the smaller of the two corresponding values of x .We will obviously have

x0 < x1 < X ′ < X

and

X ′ − x1 = h

m= 1

m(X − x0).

Having determined x1 and X ′ like we have just said, we can do the same betweenthese two new values of x, by placing two others, x2 and X ′′, which, substituted inf (x), produce the results of opposite signs, and which work to satisfy the conditions

x1 < x2 < X ′′ < X ′,

X ′′ − x2 = 1

m(X ′ − x1) = 1

m2(X − x0).

By continuing thusly, we will obtain: 1◦ a series4 of increasing values of x, namely

x0, x1, x2, . . . ;(2)

2◦ a series of decreasing values

X, X ′, X ′′, . . . ,(3)

which, surpassing the first set by quantities, respectively, equal to the products

1 × (X − x0),1

m× (X − x0),

1

m2× (X − x0), . . . ,

3Cauchy uses the phrase la suite here, meaning sequence.4Cauchy uses the phrase une série, meaning series, to describe the lists in (2) and (3). He surelyknows these are both sequences and is likely using the word “series” in an informal sense.

APPENDIX C. 251

will eventually differ from these first values by as little as we will want. We must de-duce that the general terms of the series (2) and (3) will converge toward a commonlimit.5 Let a be this limit. Since the function f (x) remains continuous from x = x0up to x = X, the general terms of the following series

f (x0), f (x1), f (x2), . . . ,

f (X), f (X ′), f (X ′′), . . .

will equally converge toward the common limit f (a); and, since by approaching thislimit they will always keep different signs, it is clear that the quantity f (a), neces-sarily finite, cannot differ from zero. By consequence, we will satisfy the equation

f (x) = 0(1)

by attributing to the variable x the particular value a contained between x0 and X.

In other words,

x = a(4)

will be a root of equation (1).6

Scholium I. – If, after having extended the series (2) and (3) up to the terms

xn and X (n)

(n denoting any integer number), we take the half-sum of these two terms for theapproximated value of the root a, the error committed will be smaller than theirhalf difference, namely

1

2

X − x0mn

.

5Cauchy is arguing this must be a common limit, otherwise there would be two distinct limits, a1and a2, where it would always be possible to find an n such that 1

mn (X − x0) < |a1 − a2|.6Cauchy is doing something quite bold here. He is showing the existence of a quantity, a, bydemonstrating one can get closer and closer to it. Cauchy is doing something extraordinary forhis time, a method which is now very common in analysis, but not in his day. He is assigning avalue to the limit of a converging sequence. This technique is also employed by Cauchy in a loosesense within his Lecture One when he defines e, and later when he defines his definite integralin Lecture Twenty-One. In short, with this analysis, Cauchy is attempting to prove the existenceof a limit. However, it is certainly true there is a great deal Cauchy does not say here. Althoughhe makes an attempt at showing a exists, he does not prove it rigorously by today’s standards.The final details took a great deal of work in later years, culminating with what is known as thecompleteness property of the real numbers. Cauchy assumes this completeness property withoutknowing it (which was common for the time) through this entire text. The rigorous treatment ofcompleteness was not dealt with properly until the late 19th century, approximately fifty years afterthe publication of Cauchy’s Calcul infinitésimal text, by many mathematicians including RichardDedekind (1831–1916) and Georg Cantor (1845–1918).

252 COURS D’ANALYSE – NOTE III.

Since this latter expression decreases indefinitely as the measure of n increases, itfollows that, by calculating a sufficient number of terms of the two series, we willfinally obtain approximated values of the root a as close as we will want.

Scholium II. – If there exists between the limits x0, X several real roots of equa-tion (1), the preceding method will make known a part of them and sometimes willfurnish all of them. Then, we will find for x1 and X ′, as well for x2 and X ′′, . . .

several systems of values that enjoy the same properties.

Scholium III. – If the function f (x) is constantly increasing or constantly decreas-ing from x = x0 up to x = X, therewill only exist between these limits a single valueof x that works to satisfy equation (1).

Corollary I. – If equation (1) does not have real roots contained between thelimits x0, X, the two quantities

f (x0), f (X)

will be of the same sign.

Corollary II. – If, in the wording of Theorem I, we replace the function f (x) by

f (x) − b

(b denoting a constant quantity), we will obtain precisely Theorem IV of ChapterII (§II). In the same hypothesis, by following the method indicated above, we willnumerically determine the roots of the equation

f (x) = b(5)

contained between x0 and X.7

...

7One of his most beautiful proofs, Cauchy has just proven the Intermediate Value Theorem.

APPENDIX D.

ON THE FORMULAS OF TAYLOR AND MACLAURIN.

We easily prove that, in the case where the fraction

F(h)

hn−1(1)

vanishes for h = 0, we have

F(h) = hn

1 · 2 · 3 · · · n F(n)(θh),(2)

θ denoting an unknown number, but less than unity. Now, equation (2), with thehelp of which we can directly establish the theory of maxima or minima, and settlethe values of the fractions which present under the form 0

0 , also leads very simply tothe series of Taylor and to the determination of the remainder which must completethis series. In fact, we will successively derive from equation (2):

1◦ By setting F(h) = f (x + h) − f (x), and n = 1,

f (x + h) − f (x) = h

1f ′(x + θh);(3)

then, by setting f ′(x + h) = f ′(x) + H1,

H1 = f (x + h) − f (x) − h1 f

′(x)h

;

2◦ By setting F(h) = f (x + h) − f (x) − h f ′(x), and n = 2,

f (x + h) − f (x) − h f ′(x) = h2

1 · 2 f ′′(x + θh);(4)

This research paper is NOT included in the published edition of Cauchy’s original 1823 Calculinfinitésimal. However, the editors of Œuvres complètes d’Augustin Cauchy attached it at the endof their published reprint in 1899. It is only included here for the sake of thoroughness.


253

https://doi.org/10.1007/978-3-030-11036-9

254 ON THE FORMULAS OF TAYLOR AND MACLAURIN.

then, by setting f ′′(x + θh) = f ′′(x) + H2,

1

1 · 2H2 = f (x + h) − f (x) − h f ′(x) − h2

1·2 f′′(x)

h2;

3◦ By setting F(h) = f (x + h) − f (x) − h f ′(x) − h2

1·2 f′′(x), and n = 3,

f (x + h) − f (x) − h f ′(x) − h2

1 · 2 f ′′(x) = h3

1 · 2 · 3 f ′′′(x + θh);(5)

then, by setting f ′′′(x + θh) = f ′′′(x) + H3,

1

1 · 2 · 3H3 = f (x + h) − f (x) − h f ′(x) − h2

1·2 f′′(x) − h3

1·2·3 f′′′(x)

h3.

By continuing in the same manner, and observing that the quantities

H1,1

1 · 2H2,1

1 · 2 · 3H3, . . .

all vanish along with h, we will generally establish the equation⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f (x + h) − f (x) − h f ′(x) − h2

1 · 2 f ′′(x) − · · ·

− hn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x) = hn

1 · 2 · · · n f (n)(x + θh)

(6)

or⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

f (x + h) = f (x) + h

1f ′(x) + h2

1 · 2 f ′′(x)

+ · · · + hn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x)

+ hn

1 · 2 · · · n f (n)(x + θh).

(7)

If we replace x by 0, and h by x, we will find⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

f (x) = f (0) + x

1f ′(0) + x2

1 · 2 f ′′(0)

+ · · · + xn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(0)

+ xn

1 · 2 · 3 · · · n f (n)(θx).

(8)

APPENDIX D. 255

It follows from formula (7) that the function f (x + h) can be considered as composedof an entire function of h, namely

f (x) + h

1f ′(x) + h2

1 · 2 f ′′(x) + · · · + hn−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x),(9)

and of a remainder, namely

hn

1 · 2 · 3 · · · n f (n)(x + θh).(10)

When this remainder becomes infinitely small for infinitely large values of thenumber n, we can assert that the series

f (x), h f ′(x),h2

1 · 2 f ′′(x), . . .(11)

is convergent, and that it has for a sum, f (x + h). Therefore, we can then write theequation

f (x + h) = f (x) + h

1f ′(x) + h2

1 · 2 f ′′(x) + · · · ,(12)

which is precisely the formula of Taylor. Similarly, if the remainder

xn

1 · 2 · 3 · · · n f (n)(θx)(13)

becomes infinitely small for the infinite values of n, equation (8) will lead to thefollowing

f (x) = f (0) + x

1f ′(0) + x2

1 · 2 f ′′(0) + · · · ,(14)

which is precisely the formula of Maclaurin.It is often useful to substitute for the expressions (10) and (13) other equivalent

expressions. We can achieve this as follows.Denote by ϕ(z) that which becomes the first member of equation (6) when we

replace h by h − z, and x by x + z, or in other words, the remainder that we obtainwhen we expand f (x + h) according to ascending and integer powers of h − z, andthat we stop at the power of degree n − 1; so that we would have

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f (x + h) = f (x + z) + h − z

1f ′(x + z) + · · ·

+ (h − z)n−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x + z) + ϕ(z).

(15)

256 ON THE FORMULAS OF TAYLOR AND MACLAURIN.

ϕ(0) will represent the common value of each of the members of equation (6).Moreover, by differentiating formula (15) with respect to z, we will find

ϕ′(z) = − (h − z)n−1

1 · 2 · 3 · · · (n − 1)f (n)(x + z),(16)

and we will deduce

ϕ(h) − ϕ(0)

h= − (h − θh)n−1

1 · 2 · 3 · · · (n − 1)f (n−1)(x + θh),(17)

or, because ϕ(h) is obviously reduced to zero,

ϕ(0) = (h − θh)n−1

1 · 2 · 3 · · · (n − 1)h f (n)(x + θh).(18)

The preceding value of ϕ(0) is nothing other than the remainder of the series ofTaylor presented in a new form. If, in this remainder, we replace x by 0, and h by x,we will obtain the remainder of the series of Maclaurin under the following form

x(x − θx)n−1

1 · 2 · 3 · · · (n − 1)f (n)(θx).(19)

It is sufficient, in several cases, to substitute this last product into expression (13) toestablish formula (14). Suppose, for example,

f (x) = (1 + x)μ,(20)

μ denoting a real constant. Expressions (13) and (19) will become, respectively,

μ(μ − 1) · · · (μ − n + 1)

1 · 2 · 3 · · · n xn(1 + θx)μ−n(21)

and

μ(μ − 1) · · · (μ − n + 1)

1 · 2 · 3 · · · (n − 1)xn(1 − θ)n−1(1 + θx)μ−n.(22)

This granted, we will easily prove: 1◦ with the help of expression (21) that theequation

(1 + x)μ = 1 + μ

1x + μ(μ − 1)

1 · 2 x2 + · · ·(23)

remains valid when the numerical value of the ratio

x

1 + θx(24)

APPENDIX D. 257

is less than unity; 2◦ with the help of expression (22), that equation (23) remainsvalid when the product

x1 − θ

1 + θx(25)

is contained between the limits −1 and 1. As a result, it will suffice to employexpression (21) to establish formula (23) between the limits x = 0, x = 1. But, willrevert back to expression (22) if we want to extend the same formula to all values ofx contained between the limits x = −1, x = +1.

APPENDIX E.

PAGINATION OF THE 1823 & 1899 EDITIONS.

1823 1899

Foreword v 9Errata viiiTable of Contents ix

DIFFERENTIAL CALCULUS

Lecture One 1 13Lecture Two 5 17Lecture Three 9 22Lecture Four 13 27Lecture Five 17 32Lecture Six 21 37Lecture Seven 25 42Lecture Eight 29 47Lecture Nine 33 53Lecture Ten 37 58Lecture Eleven 41 63Lecture Twelve 45 69Lecture Thirteen 49 76Lecture Fourteen 53 82Lecture Fifteen 57 88Lecture Sixteen 61 93Lecture Seventeen 65 98Lecture Eighteen 69 104Lecture Nineteen 73 110Lecture Twenty 77 116


259

https://doi.org/10.1007/978-3-030-11036-9

260 PAGINATION OF THE 1823 & 1899 EDITIONS.

1823 1899

INTEGRAL CALCULUS

Lecture Twenty-One 81 122Lecture Twenty-Two 85 128Lecture Twenty-Three 89 134Lecture Twenty-Four 93 140Lecture Twenty-Five 97 145Lecture Twenty-Six 101 151Lecture Twenty-Seven 105 157Lecture Twenty-Eight 109 164Lecture Twenty-Nine 113 170Lecture Thirty 117 176Lecture Thirty-One 121 182Lecture Thirty-Two 125 188Lecture Thirty-Three 129 195Lecture Thirty-Four 133 202Lecture Thirty-Five 137 208Lecture Thirty-Six 141 214Lecture Thirty-Seven 145 220Lecture Thirty-Eight 149 225Lecture Thirty-Nine 153 231Lecture Forty 157 237

ENDING MATERIAL

Addition 161 243

On the Formulas of Taylor and Maclaurin 257

One cannot help but take notice of the symmetry of the original published edition in 1823. Eachand every one of Cauchy’s forty lectures is exactly four pages in length.

References

1. Belhoste, B., Augustin-Louis Cauchy: A Biography, Trans. Frank Ragland, Springer-Verlag,New York, 1991.

2. Birkhoff, G. (editor), A Source Book in Classical Analysis, Harvard University Press, Cam-bridge, Massachusetts, 1973.

3. Bottazzini, U., The Higher Calculus: A History of Real and Complex Analysis from Euler toWeierstrass, Springer-Verlag, New York, 1986.

4. Boyer, C. B., The History of The Calculus and its Conceptual Development (The Concepts ofthe Calculus), Dover Publications, Inc., New York, 1949.

5. Bradley, R. E. and Sandifer, C. E., Cauchy’s Cours d’analyse: An Annotated Translation,Springer, New York, 2009.

6. Bressoud, D. M., A Radical Approach to Real Analysis, The Mathematical Association ofAmerica, Washington, DC, 1994.

7. Cauchy,A.L.,Cours d’analyse de l’École royale polytechnique, deBure, Paris, 1821.Reprintedin Œuvres complètes d’Augustin Cauchy, Series II, Volume III, Gauthier-Villars, Paris, 1897.

8. Cauchy, A. L., Résumé des leçons données à l’École royale polytechnique sur le calcul in-finitésimal, de Bure, Paris, 1823. Reprinted in Œuvres complètes d’Augustin Cauchy, SeriesII, Volume IV, Gauthier-Villars, Paris, 1899.

9. Dunham, W., The Calculus Gallery: Masterpieces from Newton to Lebesgue, Princeton Uni-versity Press, Princeton, New Jersey, 2005.

10. Edwards, C. H., The Historical Development of the Calculus, Springer-Verlag, New York,1979.

11. Grabiner, J. V., The Origins of Cauchy’s Rigorous Calculus, The MIT Press, Cambridge,Massachusetts, 1981.

12. Grattan-Guiness, I., The Development of the Foundations of Mathematical Analysis from Eulerto Riemann, M.I.T. Press, Cambridge, Massachusetts, 1970.

13. Hairer, E. and Wanner, G., Analysis by Its History, Springer, New York, 1996.14. Iacobacci, R. F.,Augustin-Louis Cauchy and the Development of Mathematical Analysis, Ph.D.

Dissertation, New York University, 1965.15. Jahnke, H. N. (editor), A History of Analysis, American Mathematical Society, Providence,

Rhode Island, 2003.16. Smith, D. E. (editor), A Source Book in Mathematics, McGraw-Hill Book Company, Inc., New

York, 1929.17. Smithies, F., Cauchy and the Creation of Complex Function Theory, Cambridge University

Press, Cambridge, United Kingdom, 1997.18. Spivak, M., Calculus, Fourth Edition, Publish or Perish, Inc., Houston, Texas, 2008.19. Struik, D. J.,ASource Book inMathematics, 1200–1800, HarvardUniversity Press, Cambridge,

Massachusetts, 1969.


261

https://doi.org/10.1007/978-3-030-11036-9

Index

Aabsolute convergence, 210Algebraic Analysis, ixalgebraic function, 149Ampère, André-Marie, 32, 227Analyse algébrique, ixAnalysis Course at the Royal Polytechnic

School, ixAnalytical Mechanics, xvantiderivative, xvi, 141applications of the derivative, 27arbitrary constant, 143arc, 19, 93, 161, 162, 217arcs of a circle, 161arithmetic progression, 118average, 34, 112, 113, 125, 126, 245average fraction, 34average notation, 245

BBernoulli, Jakob, 114Bernoulli, Johann, 30, 114Bibliotèque nationale de France, xbinomial differential, 155binomial expansion, 98, 212Bolzano’s Theorem, 249Bolzano, Bernard, 249Bonnet, Pierre Ossian, 33Borel, Émile, 9

CCalcul infinitésimal, viiCantor, Georg, 251Cauchy Principal Value, 132

Cauchy Ratio Test, 211Cauchy remainder, 205Cauchy Root Test, 210Cauchy sequence, 114Cauchy’s Mean Value Theorem, 227, 229Cauchy, Augustin-Louis, viiChain Rule, 14, 20change of independent variable, 64circular function, 19, 161Complete Works of Augustin Cauchy, ixcompleteness property, 114, 251complex logarithm, 217composed function, 12, 43composite function, 14condition equation, 54constant, 3construct the curves, 10continuity, 9continuous function, 9convergence test, 239convergent, 203convergent series, 210Cours d’analyse, ix, xCours d’analyse average theorems, 239Cours d’analyse Chapter II, 238Cours d’analyse de l’École royale polytech-

nique, ixCours d’analyse Note II, 34, 112, 126, 245Cours d’analyse Note III, 34, 249

Dd’Alembert, Jean le Rond, viii, 15Darboux’s Theorem, 35Darboux, Jean-Gaston, 35


263

https://doi.org/10.1007/978-3-030-11036-9

264 Index

De analysi per æquationes numero termino-rum infinitas, 224

de Moivre’s formula, 162de Moivre, Abraham, 162de Prony, Gaspard, 198decomposition of definite integral, 124decomposition of rational fraction, 103Dedekind, Richard, 251definite integral, xvi, 111, 114definite integral existence proof, xvi, 114definite integral notation, 115definite integral properties, 143definite integration, 144degree of an entire function, 98degree of homogeneous function, 49Delta (�) notation, 8derivative applications, 27derivative as slope, 29derivative function, 12derivative notation, 12derivative of various orders, 59derivative properties, 21derived function, xv, 12Descartes, René, 7develop, xv, 54difference quotient, 11differential, 17, 38differential coefficient, 18, 60differential coefficient of order n, 60differential equation, 45, 139differential of a function, 38differential of various orders, 60differentiate, 18differentiation, 18differentiation order exchange, 66Dirichlet Function, 148Dirichlet, Lejeune, 148discontinuity, 10discontinuous, 9Divergence Test, 210divergent, 203divergent series, 210

Ee, 5ellipse, 56entire function, 22Euler’s formula, 215, 216Euler, Leonhard, ix, 7, 114, 178, 197exchange of differentiation order, 65

exchange of limits operation, 123, 180, 184,191, 221

existence of definite integral, xvi, 114explicit function, 7exponential function, 161exponential infinite series, 207Extended Mean Value Theorem, 229

Ffactorial notation, 202finite difference, 9, 32, 38First Derivative Increasing/Decreasing The-

orem, 27First Derivative Test, 27Formula of Maclaurin, 253, 255Formula of Taylor, xv, 253, 255Fourier coefficients, 114Fourier integral notation, 115Fourier series, 114Fourier, Joseph, 114, 115function, 7function of a function, 14Fundamental Theorem of Algebra, 96, 103Fundamental Theorem of Calculus, 137,

138, 140, 141, 196

GGamma function, 178Gauss, Carl Friedrich, 24General Mean Value Theorem for Definite

Integrals, 120, 126general term, 203, 209general value, 134Generalized Mean Value Theorem, 227, 229geometric progression, 118, 209geometric series, 209Gregory, James, 197

Hhalf-sum, 121harmonic series, 204Heine, Eduard, 9Heine-Borel Theorem, 9homogeneous function, 49hyperbola, 57hyperbolic cosine, 216hyperbolic logarithm, 13hyperbolic sine, 216

Index 265

Ii used as

√−1, 24imaginary conjugate, 95imaginary exponential, 215imaginary function, 24imaginary logarithm, 215immediate integration, 144implicit differentiation, 14implicit function, 8, 43, 45improper integral, 133indefinite integral, 140indefinite integration, 144independent variable, 7indeterminate factor, 54indeterminate form, 30, 31indeterminate integral, 132indeterminate ratio, 231Indicator Function, 148infinite series, 203infinite series tail, 210infinitely large, 6infinitely small quantity, 6infinitesimal, 6infinitesimal class, 234infinitesimal of order n, 234infinity symbol, 6integral general value, 134integral principal value, 134integral properties, 143integrals are continuous, 137integrals of higher order, 194integrate, 144integration by decomposition, 147integration by parts, 147integration by series, 224integration by substitution, 145integration of algebraic functions, 149integration of binomial functions, 155integration of circular functions, 161integration of exponential functions, 161integration of logarithmic functions, 161integration of multiple variable functions,

179integration of trigonometric functions, 167intermediate value property, 249Intermediate Value Theorem, 34, 249, 252interval of convergence, 211interval partition, 111interval refinement, 112, 114irrational function, 149irreducible quadratic factor, 105

JJohn Napier, 62

KKramp, Christian, 202

LL’Hôpital’s Rule, 30, 105, 202, 227, 231L’Hôpital, Guillaume de, 30L’Huilier, Simon Antoine Jean, 3La géométrie, 7Lagrange Mean Value Theorem, 35Lagrange remainder, 200Lagrange Taylor series counterexamples,

213Lagrange, Joseph-Louis, viii, xv, 3, 12, 99,

197, 213, 235Laplace, Pierre-Simon, 226Legendre, Adrien-Marie, 178Leibniz formula, 225Leibniz integration, 115Leibniz series, 225Leibniz, Gottfried Wilhelm, 11, 115, 138limit, 3limit of imaginary function, 25limit superior, 210linearization, 41, 50linearization of an entire function, 98linearization of multiple variable function,

71logarithm base A, 7logarithm base e, 13logarithm notation, 13logarithmic function, 161logarithmic infinite series, 207

MMaclaurin series, 97, 204Maclaurin series remainder, 256Maclaurin’s Theorem, 197, 204, 233Maclaurin, Colin, 206maximum, 28, 50Mean Value Theorem for Definite Integrals,

120, 126MeanValueTheorem forDerivatives, 31, 33,

35minimum, 28, 50modulus, 93

266 Index

NNapierian logarithm, 13, 62natural logarithm, 13nested intervals, 250Newton, Isaac, 11, 138, 206, 224null derivative function, 138number, 4numerical integration, 121numerical integration error, 121

OOn Analysis by Equations with an Infinite

Number of Terms, 224open vs. closed interval, 118optimization, 28, 79, 81optimization of multiple variable function,

51, 53Oresme, Nicole, 4origin, 111

Ppartial derivative, 37, 43partial derivative of higher order, 67partial differential, 40, 65partial differential notation, 40partial differential of various orders, 65partial differentiation order exchange, 66partial fraction decomposition, 96, 104, 105partial sum, 111, 203partition, 111pathological functions, 148point of discontinuity, 10polygons, 3Polytechnic School, viiPower Rule for differentiation, 22preservation of sequential limits, 249primitive function, xviprincipal value, xvi, 132, 134, 185Product Rule for differentiation, 23

Qquantity, 4Quotient Rule for differentiation, 23

RRésumé des leçons données a l’École royale

polytechnique sur le calcul infinitési-mal, vii

radius of convergence, 211, 212radius vector, 3, 56

Ratio Convergence Test, 211, 239ratio of differences, 11rational fraction, 24, 103rational function, 103, 149refinement, 112, 114remainder of the series, 203Riemann pathological function, 148Riemann, Bernhard, 111, 148Root Convergence Test, 210, 242root of equation, 251Ruler Function, 148

SSandwich Theorem, 4Second Derivative Test, 80, 83semi-axis, 56series, 203simple function, 12simultaneous increment, 8singular definite integral, xvi, 133slope of a curve, 29solution of continuity, 10Squeeze Theorem, 4subinterval, 111sum of the series, 203Summary of Lectures given at the Royal

Polytechnic School on the Infinites-imal Calculus, vii

symbol, 73symbolic equation, 73symbolic expression, 73

Ttail of series, 210tangent to the curve, 29Taylor series, 99, 204, 253Taylor series remainder, 256Taylor’s formula, 227Taylor’s Theorem, xvi, 197, 200, 203, 204,

233Taylor, Brook, 197Théorie des fonctions analytiques, ix, 213Theorem of Homogeneous Functions, 49Theory of Analytic Functions, ix, 213Thomae, Carl Johannes, 148total differential, 38, 41, 65, 85total differential of various orders, 65Traité de Mécanique céleste, 226Treatise on Celestial Mechanics, 226trigonometric arc, 19trigonometric function, 161trigonometric infinite series, 206

Index 267

trigonometric line, 19, 161trigonometric substitution, 14

Uuniform continuity, 9, 33, 34, 113, 114uniform convergence, 221uniformly continuous, 34, 113

Vvariable, 3

WWeierstrass Function, 148Weierstrass, Karl, 3, 148

isidore.co Infinitesimal... · Translator’s Preface...

Documents

Transcript of isidore.co Infinitesimal... · Translator’s Preface...