Tabular Integration by PartsDavid Horowitz, Golden West College, Huntington Beach, CA 92647

The College Mathematics Journal,September 1990, Volume 21,Number 4, pp. 307311.

Only a few contemporary calculus textbooks provide even a cursory presentation oftabular integration by parts [see for example, G. B. Thomas and R. L. Finney,Calculusand Analytic Geometry,Addison-Wesley, Reading, MA 1988]. This is unfortunatebecause tabular integration by parts is not only a valuable tool for finding integrals butcan also be applied to more advanced topics including the derivations of some importanttheorems in analysis.

The technique of tabular integration allows one to perform successive integrations byparts on integrals of the form

(1)

without becoming bogged down in tedious algebraic details [V. N. Murty, Integration byparts,Two-Year College Mathematics Journal11 (1980) 90-94]. There are several waysto illustrate this method, one of which is diagrammed in Table 1. (We assumethroughout that F and G are smooth enough to allow repeated differentation andintegration, respectively.)

Table 1

____________________________________________________

Column #1 Column #2____________________________________________________

G

: :. .

- - - -

____________________________________________________

In column #1 list and its successive derivatives. To each of the entries in thiscolumn, alternately adjoin plus and minus signs. In column #2 list and itssuccessive antiderivatives. (The notation denotes the nth antiderivative of G. Donot include an additive constant of integration when finding each antiderivative.) Formsuccessive terms by multiplying each entry in column #1 by the entry in column #2 thatlies just belowit. The integral (1) is equal to the sum of these terms. If is aFsxd

Gs2ndGstd

Fstd

Gs2n21ds21dn11Fsn11dGs2nds21dnFsnd

Gs23d2Fs3dGs22d1 Fs2dGs21d2Fs1d

1 F

EFstdGstd dt

polynomial,then there will be only a finite number of terms to sum. Otherwise theprocess may be truncated at any level by forming a remainder term defined as theintegral of the product of the entry in column #1 and the entry in column #2 that liesdirectly acrossfrom it. Symbolically,

(2)

A proof follows from continued application of the formula for integration by parts [K. W. Folley, Integration by parts,American Mathematical Monthly 54 (1947) 542-543].

The technique of tabular integration by parts makes an appearance in the hit motionpicture Stand and Deliver in which mathematics instructor Jaime Escalante of James A.Garfield High School in Los Angeles (portrayed by actor Edward James Olmos) worksthe following example.

Example.

column #1 column #2____________________________________________________

sin x

cosx

sin x

cosx____________________________________________________

Answer:

The following are some areas where this elegant technique of integration can be applied.

Miscellaneous Indefinite Integrals. Most calculus textbooks would treat integrals ofthe form

where is a polynomial,by time-consuming algebraic substitutions or tedious partialfraction decompositions. However, tabular integration by parts works particularly wellon such integrals (especially if r is not a positive integer).

Pstd

E Pstdsat 1 bdr dt,

2x2 cos x 1 2x sin x 1 2 cos x 1 C

0

21 2

222x

1 x2

ex2 sin x dx

5 on

k50s21dkFskdGs2k21d 1 s21dn11EFsn11dstdGs2n21dstd dt.

1 s21dnFsndGs2n21d 1 s21dn11EFsn11dstdGs2n21dstd dt EFstdGstd dt 5 FGs21d 2 Fs1dGs22d 1 Fs2dGs23d 2 . . .

2

Example.

column #1 column #2____________________________________________________

____________________________________________________

Answer:

Laplace Transforms.Computations involving the Laplace transform

(3)

often require several integrations by parts because of the form of the integrand in (3).Tabular integration by parts streamlines these integrations and also makes proofs ofoperational properties more elegant and accessible. (Many introductory differentialequations textbooks omit formal proofs of these properties because of the lengthy detailinvolved in their derivations.) The following example uses this technique to establish thefundamental formula for the Laplace transform of the nth derivative of a function.

Theorem. Let n be a positive integer and suppose that is a function such thatis piecewise continuous on the interval Furthermore suppose that there

exist constants A, b, and M such that

for all Then if

(4)LHf snd stdJ 5 2f sn21ds0d 2 sf sn22ds0d 2 ? ? ? 2 sn21f s0d 1 snLHf stdJ.

s > b,k 5 0, 1, 2, . . . , n 2 1.

|f skdstd| Aebt if t M

t 0.f sndstdf std

LHf stdJ 5 E`0

e2stf std dt

s5t2 1 15ds3t 1 2d4y5 2 50t27

s3t 1 2d9y5 1 125567

s3t 1 2d14y5 1 C

12513608

s3t 1 2d14y50

25324

s3t 1 2d9y5241

512

s3t 1 2d4y524t2

s3t 1 2d21y512t2 1 361

E12t2 1 365!3t 1 2 dt

3

Proof.

column #1 column #2____________________________________________________

: :. .

____________________________________________________

Taylor s Formula. Tabular integration by parts provides a straightforward proof ofTaylors formula with integral remainder term.

Theorem. Suppose has continuous derivatives throughout an intervalcontaining a. If x is any number in the interval then

(5)1 E xa

f sn11dstd

n!sx 2 tdn dt.

f sxd 5 f sad 1 f s1d(adsx 2 ad 1 fs2dsad2!

sx 2 ad2 1 ? ? ? 1f sndsad

n!sx 2 adn

n 1 1fstd

5 2f sn21ds0d 2 sf sn22ds0d 2 ? ? ? 2 sn21f s0d 1 snLHf stdJ.

1 E`0

sne2stf std dt

LHf snd stdJ 5 3e2stf sn21dstd 1 se2stf sn22dstd 1 ? ? ? 1 sn21e2stf std4t5`

t50

f stds21dns21dnsne2stf s1dstds21dn21s21dn21sn21e2st

f sn22dstds2e2st1f sn21dstd2se2st2

f sndstde2st1

4

Proof.

column #1 column #2____________________________________________________

: :. .

____________________________________________________

Equation (5) follows immediately.

Residue Theorem for Meromorphic Functions. Tabular integration by parts alsoapplied to complex line integrals and can us used to prove the following form of theresidue theorem.

Theorem. Suppose is analytic in $ and has a pole oforder m at Then if

(6)f szd dz 5 2pism 2 1d! limzz 0 3dm21

dzm21sz 2 z0dmfszd4.r|z2z0|5r

0 < r < Rz0.5 Hz: 0 < |z 2 z0| < RJf szd

1 E xa

f sn11dstd

n!sx 2 tdn dt

5 32f s1dstdsx 2 td 2 fs2dstd2!

sx 2 td2 2 ? ? ? 2f sndstd

n!sx 2 tdn4

t5x

t5a

E xa

f2f s1dstdgf21g dt

s21dn11 sx 2 tdn

n!s21dn122f sn11dstd

s21dn sx 2 tdn21

sn 2 1d!s21dn112f sndstd

2sx 2 td2

2!2f s3dstd1

sx 2 td2f s2dstd2212f s1dstd1

5

1^

Proof.

column #1 column #2____________________________________________________

: :. .

____________________________________________________

Thus,

(7)

where the summation term in (7) must be evaluated for the closed circle However, each of the functions in column #1 has a removable singularity at and istherefore single-valued in $. Furthermore, each of the functions in column #2 is alsosingle-valued in $. Thus,the summation term in (7) must vanish. The integral term onthe right side of (7) is evaluated by the Cauchy integral formula and the result (6)follows directly. (The right-hand side of (6) can be recognized as the formula for theresidue of at the pole ) See [Ruel V. Churchill and James W. Brown, ComplexVariables and Applications,McGraw-Hill, New York, 1984].

z0.fszd

z0|z 2 z0| 5 r.

dm21

dzm21sz 2 z0dmf szd

z 2 z0 dz,r

|z2z0|5r1

1sm 2 1d!

|z2z0|5r5 32 om22k50

sm 2 k 2 2d! dk

dzksz 2 z0dmf szd

sm 2 1d!sz 2 z0dm2k21 4sz 2 z0dmf szdsz 2 z0d2m dzr|z2z0|5r

s21dm21sz 2 z0d21

m 2 1!s21dm21 d

m21

dzm21sz 2 z0dmf szd

sz 2 z0d2m12sm 2 1dsm 2 2d

d 2

dz2sz 2 z0dmf szd1

2sz 2 z0d2m11

m 2 1ddz

sz 2 z0dmf szd2

sz 2 z0d2msz 2 z0dmf szd1

6