amer.math.monthly.120.07.660 (1)

Problems and SolutionsSource: The American Mathematical Monthly, Vol. 120, No. 7 (August–September 2013), pp.660-669Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/10.4169/amer.math.monthly.120.07.660 .

Accessed: 22/10/2013 00:15

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.

http://www.jstor.org

This content downloaded from 149.43.110.17 on Tue, 22 Oct 2013 00:15:43 AMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=maa

http://www.jstor.org/stable/10.4169/amer.math.monthly.120.07.660?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


PROBLEMS AND SOLUTIONS

Edited by Gerald A. Edgar, Doug Hensley, Douglas B. Westwith the collaboration of Mike Bennett, Itshak Borosh, Paul Bracken, Ezra A. Brown,Randall Dougherty, Tamas Erdelyi, Zachary Franco, Christian Friesen, Ira M. Ges-sel, Laszlo Liptak, Frederick W. Luttmann, Vania Mascioni, Frank B. Miles, RichardPfiefer, Dave Renfro, Cecil C. Rousseau, Leonard Smiley, Kenneth Stolarsky, RichardStong, Walter Stromquist, Daniel Ullman, Charles Vanden Eynden, Sam Vandervelde,and Fuzhen Zhang.

Proposed problems and solutions should be sent in duplicate to the MONTHLY

problems address on the back of the title page. Proposed problems should neverbe under submission concurrently to more than one journal. Submitted solutionsshould arrive before November 30, 2013. Additional information, such as gen-eralizations and references, is welcome. The problem number and the solver’sname and address should appear on each solution. An asterisk (*) after the num-ber of a problem or a part of a problem indicates that no solution is currentlyavailable.

PROBLEMS

11719. Proposed by Nicolae Anghel, University of North Texas, Denton, TX. Let f bea twice-differentiable function from [0,∞) into (0,∞) such that

limx→∞

f ′′(x)

f (x)(1+ f ′(x)2)2= ∞.

Show that

limx→∞

∫ x

t=0

√1+ f ′(t)2

f (t)dt∫∞

t=x

√1+ f ′(t)2 f (t) dt = 0.

11720. Proposed by Ira Gessel, Brandeis University, Waltham, MA. Let En(t) be theEulerian polynomial defined by

∞∑k=0

(k + 1)nt k=

En(t)

(1− t)n+1,

and let Bn be the nth Bernoulli number. Show that (En+1(t)− (1− t)n)Bn is a poly-nomial with integer coefficients.

11721. Proposed by Roberto Tauraso, Universita di Roma “Tor Vergata”, Rome, Italy.Let p be a prime greater than 3, and let q be a complex number other than 1 such thatq p= 1. Evaluate

p−1∑k=1

(1− qk)5

(1− q2k)3(1− q3k)2.

http://dx.doi.org/10.4169/amer.math.monthly.120.07.660

660 c© THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 120



11722. Proposed by Nguyen Thanh Binh, Hanoi, Vietnam. Let ABC be an acute trianglein the plane, and let M be a point inside ABC. Let O1, O2, and O3 be the circumcentersof BCM, CAM, and ABM, respectively. Let c be the circumcircle of ABC. Let D, E ,and F be the points opposite A, B, and C , respectively, at which AM, BM, and CMmeet c. Prove that O1 D, O2 E , and O3 F are concurrent at a point P that lies on c.

11723. Proposed by L. R. King, Davidson, NC. Let A, B, and C be three points in theplane, and let D, E , and F be points lying on BC, CA, and AB, respectively. Show thatthere exists a conic tangent to BC, CA, and AB at D, E , and F , respectively, if andonly if AD, BE, and CF are concurrent.

11724. Proposed by Andrew Cusumano, Great Neck, NY. Let f (n) =∑n

k=1 kk and letg(n) =

∑nk=1 f (k). Find

limn→∞

g(n + 2)

g(n + 1)−

g(n + 1)

g(n).

11725. Proposed by Mher Safaryan, Yerevan State University, Yerevan, Armenia. Letm be a positive integer. Show that, as n→∞,∣∣∣∣∣log 2−

n∑k=1

(−1)k−1

k

∣∣∣∣∣ = C1

n+

C2

n2+ · · · +

Cm

nm+ o

(1

nm

),

where

Ck = (−1)kk∑

i=1

1

2i

i∑j=1

(−1) j

(i − 1

j − 1

)j k−1

for 1 ≤ k ≤ m.

SOLUTIONS

A Sum and Product Inequality

11584 [2011, 558]. Proposed by Raymond Mortini and Jerome Noel, Universite PaulVerlaine, Metz, France. Let 〈a j 〉 be a sequence of nonzero complex numbers inside theunit circle, such that

∏∞

k=1 |ak | converges. Prove that∣∣∣∣∣∣∞∑j=1

1− |a j |2

a j

∣∣∣∣∣∣ ≤ 1−∏∞

j=1 |a j |2∏

∞

j=1 |a j |.

Solution I by Omran Kouba, Higher Institute for Applied Sciences and Technology,Damascus, Syria. Assume that

∏∞

j=1 |a j | converges to a positive number P , and letf (x) = 1

x − x . For a, b ∈ (0, 1), we have

f (ab)− f (a)− f (b) =1

ab(1− a)(1− b)(1− ab) > 0,

so that f (a) + f (b) < f (ab). By induction,∑n

j=1 f (|a j |) < f (|a1a2 · · · an|). Tak-ing the limit as n →∞, we obtain

∑∞

j=1 f (|a j |) ≤ f (P). Thus,∑∞

j=11

a j(1− |a j |

2)

converges absolutely, and the desired result follows.

August–September 2013] PROBLEMS AND SOLUTIONS 661



Solution II by Douglas B. Tyler, Raytheon, Torrence, California. Assume as in thefirst solution that

∏∞

j=1 |a j | converges to P , and let |a j | = e−x j . Thus, x j > 0 and∑∞

j=1 x j = − log P . For x, y > 0, sinh(x + y) = sinh x cosh y + sinh y cosh x ≥sinh x + sinh y. We have∣∣∣∣∣∣

∞∑j=1

1− |a j |2

|a j |

∣∣∣∣∣∣ ≤∞∑j=1

1

|a j |− |a j | = 2

∞∑j=1

sinh x j ≤ 2 sinh

∞∑j=1

x j

= 1

P− P.

Editorial comment. The proposers’ solution used the Schwartz–Pick inequality:

| f ′(z)| ≤ 1−| f (z)|2

1−|z|2for any analytic automorphism of the unit disk. Taking f (z) to

be∏∞

j=1a j

|a j |

a j−z

1−a j z leads to the required conclusion.We of course follow the usual convention: To say that an infinite product with

nonzero factors “converges”, means that the sequence of partial products convergesto a nonzero value.

Also solved by R. Chapman (U. K.), D. Constales (Belgium), W. J. Cowieson, P. P. Dalyay (Hungary), O.Geupel (Germany), M. Goldenberg & M. Kaplan, E. A. Herman, K.-W. Lau (China), J. H. Lindsey II, O. P.Lossers (Netherlands), J. Simons (U. K.), N. C. Singer, A. Stenger, R. Stong, M. Tetiva (Romania), E. I.Verriest, J. Vinuesa (Spain), H. Widmer (Switzerland), Ellington Management Problem Solving Group, andGCHQ Problem Solving Group (U. K.).

An Equivalent of CH

11588 [2011, 653]. Proposed by Taras Banakh, Ivan Franko National University ofLviv, Lviv, Ukraine, and Igor Protasov, Taras Shevchenko National University of Kyiv,Kyiv, Ukraine. Show that R− {0} can be partitioned into countably many subsets, eachof which is linearly independent over Q, if and only if the continuum hypothesis holds.

Editorial comment. Most solvers noted that the result is a theorem of Paul Erdos andShizuo Kakutani, which can be found in “On Non-denumerable Graphs,” Bull. Amer.Math. Soc. 49 (1943) 457–461. O. Guepel and R. Mabry observed that the proposerspublished the result with proof in their article “Partitions of groups and matroids intoindependent subsets,” Algebra Discrete Math. 10 (2010) 1–7, also available at http://arxiv.org/abs/1010.1359.

Solved by N. Caro (Brazil), C. Degenkolb, O. Geupel (Germany), R. Mabry, K. Muthuvel, V. Pambuccian, S.Scheinberg, R. Stong, and the proposers.

Arggh! Eye Factorial . . . Arg(i!)

11592 [2011, 654]. Proposed by Mircea Ivan, Technical University of Cluj-Napoca,Cluj-Napoca, Romania. Find limn→∞

(− log(n)+

∑nk=1 arctan 1

k

).

Solution I by Nora Thornber, Raritan Valley Community College, Somerville, New Jer-sey. Let L be the desired limit. Since limn→∞(− log n +

∑nk=1

1k ) = γ , Euler’s con-

stant, we have

L = γ − limn→∞

n∑k=1

(1

k− arctan

1

k

). (1)

Note that Im log(1+ i/k) = Arg(1+ i/k) = arctan(1/k). Since

0(z) =e−γ z

z

∞∏k=1

(1+

z

k

)−1ez/k, (2)




we have

Im log0(i) = −γ −π

2+

∞∑k=1

1

k− arctan

(1

k

),

and thus L = −π/2−Arg0(i) = −Arg0(1+ i). Since both sides are between 0 and2π , the branch of the logarithm was appropriate.

Solution II by Omran Kouba, Higher Institute for Applied Sciences and Technol-ogy, Damascus, Syria. As before, the limit is given by (1). Since ei arctan(1/k)

=

(k + i)/√

1+ k2, we have

ei L= eiγ

∞∏k=1

ei arctan(1/k)−i/k=

(∞∏

k=1

(1+

1

k2

))−1/2

· eiγ∞∏

k=1

(1+

i

k

)e−i/k .

Thus, i L = −(1/2) log C − log0(1+ i), where C =∏∞

k=1(1+ 1/k2). Note that C isreal. Taking imaginary parts and using (2), we have L = −Arg0(1+ i).

Solution III by Denis Constales, Ghent University, Ghent, Belgium. Consider thedigamma function, defined by

ψ(x) =0′(x)

0(x).

Since ψ(x + 1) = ψ(x)+ 1/x and ψ(n) = log(n)+ O(1/n) as n→∞, we have

n∑k=1

k

x2 + k2=

1

2

n∑k=1

(1

k + i x+

1

k − i x

)

=1

2(ψ(n + 1+ i x)− ψ(1+ i x)+ ψ(n + 1− i x)− ψ(1− i x))

= −1

2(ψ(1+ i x)+ ψ(1− i x))+ log(n)+ O(1/n).

Hence,∫ 1

0

n∑k=1

k

x2 + k2dx =

n∑k=1

arctan1

k

=i

2

[log0(1+ i x)− log0(1− i x)

]1

x=0+ log(n)+ O(1/n).

Thus, the desired limit is

limn→∞

(− log(n)+

n∑k=1

arctan1

k

)=

i

2log

0(1+ i)

0(1+ i)= −Arg0(1+ i).

Editorial comment. Many partial solutions were submitted, for example γ − ζ(3)/3+ζ(5)/5− ζ(7)/7+ · · · . D. Beckwith and others noted formula 6.1.27 in Abramowitzand Stegun’s Handbook of Mathematical Functions, which states Arg0(x + iy) =yψ(x) +

∑∞

k=0(y

x+k − arctan yx+k ), where ψ(z) = 0′(z)

0(z) . The result follows by takingx = 1 and y = −1, sinceψ(1) = −γ . D. Bailey, D. Borwein, and J. Borwein observedthat limn→∞(− log n +

∑nk=2 arctanh(1/k)) is an easier problem, in which the limit is

− log√

2.




Also solved by T. Amdeberhan & V. H. Moll, D. H. Bailey & D. Borwein & J. M. Borwein (Canada &Canada & Australia), D. Beckwith, R. Chapman (U. K.), P. J. Fitzsimmons, O. Furdui (Romania), O. Geupel(Germany), M. Goldenberg & M. Kaplan, J. Grivaux (France), E. A. Herman, K.-W. Lau (China), O. P. Lossers(Netherlands), N. C. Singer, A. Stenger, I. Sterling, R. Stong, M. Vowe (Switzerland), S. Wagon & M. Trott,T. Wiandt, M. Wildon (U. K.), GCHQ Problem Solving Group (U. K.), and the proposer.

A Limit of an Integral

11611 [2011, 937]. Proposed by Ovidiu Furdui, Technical University of Cluj-NapocaCluj, Romania. Let f be a continuous function from [0, 1] into [0,∞). Find

limn→∞

n∫ 1

x=0

(∞∑

k=n

x k

k

)2

f (x) dx .

Solution by Byron Schmuland, University of Alberta, Edmonton, AB, Canada. Be-gin by noting that f may be replaced by a constant function. Indeed, let gn(x) =(∑∞

k=n x k/k)2. Suppose that n∫ 1

0 gn(x) dx converges to a finite value c. For any con-tinuous f and positive ε, let ρ be less than 1 but near enough to 1 that supρ≤x≤1 | f (x)−f (1)| < ε. For 0 ≤ x ≤ ρ, we have the crude bound

gn(x) ≤

(∞∑

k=n

ρk

)2

=ρ2n

(1− ρ)2.

Thus∣∣∣∣n ∫ 1

0gn f (x) dx − n

∫ 1

0gn f (1) dx

∣∣∣∣ ≤ 2‖ f ‖∞ · n∫ ρ

0gn(x) dx + εn

∫ 1

ρ

gn(x) dx

≤ 2‖ f ‖∞nρ2n

(1− ρ)2+ εn

∫ 1

0gn(x) dx .

Take lim sup in n and then let ε→ 0 to conclude

limn→∞

n∫ 1

0gn(x) f (x) dx = lim

n→∞n∫ 1

0gn(x) f (1) dx = c f (1),

as claimed.Now we must compute c. For m ≥ 1, let Am = [m,∞)× [m,∞) and write∫ 1

0gn(x) dx =

∫ 1

0

∞∑j=n

∞∑k=0

x j+k

jkdx =

∞∑j=n

∞∑k=n

1

jk( j + k + 1).

Now let n > 1 and take j, k ≥ n. On the square (x, y) ∈ [ j, j + 1] × [k, k + 1],(j + k

j + k + 1

)1

xy(x + y)≤

1

jk( j + k + 1)≤

1

(x − 1)(y − 1)(x + y − 2).

Summing over all these squares yields(2n

2n + 1

)∫An

dx dy

xy(x + y)≤

∞∑j=n

∞∑k=n

1

jk( j + k + 1)≤

∫An−1

dx dy

xy(x + y).




However, ∫∞

n

dx

xy(x + y)=

[1

y2log

x

x + y

]∞n

=1

y2log

n + y

y,

so ∫An

dx dy

xy(x + y)=

[1

nlog

y

n + y+

1

ylog

n

n + y

]∞n

=2 log 2

n,

and we conclude that c = 2 log 2.

Editorial comment. Several solvers noted that f need not have nonnegative values; infact, the result holds for all bounded integrable f that are continuous at x = 1.

Also solved by N. Caro (Brazil), R. Chapman (U. K.), D. Constales (Belgium), P. J. Fitzsimmons, J. Grivaux(France), E. J. Ionascu, B. Karaivanov, J. C. Kieffer, O. Kouba (Syria), J. H. Lindsey II, O. P. Lossers (Nether-lands), U. Milutinovic (Slovenia), M. Omarjee (France), P. Perfetti (Italy), M. A. Prasad (India), N. C. Singer,A. Stenger, R. Stong, D. B. Tyler, J. Vinuesa (Spain), T. Viteam (Uruguay), BSI Problems Group (Germany),GCHQ Problem Solving Group (U. K.), and the proposer.

Concurrent Lines Defined by a Triangle

11615 [2012, 68]. Proposed by Constantin Mateescu, Zinca Golescu High School,Pitesti, Romania. Let A, B, and C be the vertices of a triangle, and let K be a point inthe plane distinct from these vertices and the lines connecting them. Let M , N , and Pbe the midpoints of BC, CA, and AB, respectively. Let D, E , and F be the intersectionsof the lines through MK and NP, NK and PM, and PK and MN, respectively. Prove thatthe parallels from D, E , and F to AK, BK, and CK, respectively, are concurrent.

Solution by C. R. Pranesechar, Indian Institute of Science, Bangalore, India. We usevectors. Let K be the origin, and use the same letters A, B,C,M, N , P, D, E, F forthe position vectors of the corresponding points. We describe lines by a point, a direc-tion, and a parameter ‘t’, so that the line through, say, Q and R is given by the set of allpoints of the form Q + t (R − Q), or, for short, by the equation r = Q + t (R − Q).

Since K is not on any of the lines BC, CA, AB, we see that any two of the vectorsA, B,C are linearly independent. Write 0 = l A + m B + nC where l, m, and n are allnonzero real numbers. Thus A = −(m/ l)B − (n/ l)C . The equation of line N P is

r = P + t1(P − N ) =1

2(A + B)+

t1

2(B − C),

and the equation of line MK is

r = t2 D =t2

2(B + C).

Next we find the point D, the intersection of these two lines. Proceeding from thesetwo equations, we write A in terms of B and C , equate the coefficients of B and C ,solve for t1 and t2 in terms of l,m, n, and substitute back for t1 to get

D =l − m − n

4l(B + C).

Similarly,

E =m − n − l

4m(C + A).




The line L1 through D and parallel to AK has equation

r = D + t3 A =l − m − n

4l(B + C)+ t3 A.

The line L2 through E and parallel to BK is

r = E + t4 B =m − n − l

4m(C + A)+ t4 B.

Let T be the point of intersection of L1 and L2. As before, substitute for A in termsof B and C , equate coefficients of B and C in the two equations, solve for t2 and t4 interms of l,m, n, and substitute back for t3. This yields

T =(m − l)(m − n + l)

4nlB +

(n − l)(n − m + l)

4lmC.

This may be rearranged as

A + B + C

4+

l3 A + m3 B + n3C

4lmn,

so it is symmetric with respect to the pairs (A, l), (B,m), (C, n). Thus T also lies onthe line L3 through F and parallel to C K . This proves the concurrency of the linesL1, L2, L3, as desired.

Also solved by G. Apostolopoulos (Greece), M. Bataille (France), R. Chapman (U. K.), C. Delorme (France),O. Geupel (Germany), J.-P. Grivaux (France), O. Kouba (Syria), J. H. Lindsey II, R. Stong, D. Stout, T. Viteam(Germany), Z. Zhang, GCHQ Problem Solving Group (U. K.), and the proposer.

A Point of Zero Net Force

11616 [2012, 68]. Proposed by Stefano Siboni, University of Trento, Trento, Italy. Letx1, . . . , xn be distinct points in R3, and let k1, . . . , kn be positive real numbers. A testobject at x is attracted to each of x1, . . . , xn with a force along the line from x to x j ofmagnitude k j‖x − x j‖

2, where ‖u‖ denotes the usual Euclidean norm of u. Show thatwhen n ≥ 2, there is a unique point x∗ at which the net force on the test object is zero.

Solution by Jeff Boersema, Seattle University, Seattle, WA. We will prove a strongerstatement: Let x1, . . . , xn be distinct points in Rm , with n ≥ 2 and m ≥ 2. Letg1, . . . , gn be continuous strictly increasing functions from [0,∞) to [0,∞) withg j (0) = 0. If a test object at x is attracted to each of x1, . . . , xn with a force along theline from x to x j of magnitude g j (‖x − x j‖), then there is a unique point at which thenet force is zero.

Let F(x) be the vector field on Rm defined by

F(x) =n∑

j=1

g j

(‖x j − x‖

) x j − x

‖x j − x‖,

where the j th term is zero if x = x j . We must show that there is a unique point x∗ suchthat F(x∗) = 0. To prove existence, let K be the convex hull of the points x1, . . . , xn .For each x ∈ K , there is a positive number ε such that the segment from x to x +εF(x) is contained in K . Indeed, at a point on the boundary of K , the direction ofF(x) is toward the interior of K . Since F is continuous and K is compact, there is apositive number ε∗ such that x + ε∗F(x) ∈ K for all x ∈ K . Now K is homeomorphicto a simplex (of some dimension less than or equal to m), so by the Brouwer FixedPoint Theorem, the function x + ε∗F(x) has a fixed point x∗. Thus F(x∗) = 0.




For uniqueness, suppose that F has distinct zeros u and v. This leads to a contradic-tion as follows: Write z = v − u. Define G : R→ R by G(t) = F(u + t z) · z, where‘·’ denotes the dot product. Note that G is the sum of the n strictly decreasing functionsof t given by

g j

(‖x j − u − t z‖

) x j − u − t z

‖x j − u − t z‖· z.

However, F has distinct zeros u and v, so G(0) = G(1) = 0, a contradiction.

Also solved by R. Chapman (U. K.), D. Constales (Belgium), P. J. Fitzsimmons, E. J. Ionascu, J. H. LindseyII, O. P. Lossers (Netherlands), A. Plaza & J. Sanchez-Reyes (Spain), C. R. Pranesachar (India), M. B. Y.Ranorovelonalohotsy (South Africa), J. G. Simmonds, N. C. Singer, R. Stong, R. Tauraso (Italy), E. I. Verriest,and the proposer.

A Quadruple Integral

11621 [2012, 161]. Proposed by Z. K. Silagadze, Budker Institute of Nuclear Physicsand Novosibirsk State University, Novosibirsk, Russia. Find∫

∞

s1=−∞

∫ s1

s2=−∞

∫ s2

s3=−∞

∫ s3

s4=−∞

cos(s21 − s2

2) cos(s23 − s2

4) ds4 ds3 ds2 ds1.

Solution by Richard Stong, Center for Communications Research, San Diego, CA . LetI denote this value. We will show that I = π2/16.

Let Fc(x) =∫ x−∞

cos(t2) dt and Fs(x) =∫ x−∞

sin(t2) dt . These are essentially theFresnel integrals; in particular

limx→∞

Fc(x) = limx→∞

Fs(x) =

√π

2.

Integrating by parts, we see that in fact

Fs(x) =

√π

2−

cos(x2)

2x+ O

(1

x3

)as x →+∞,

and

Fc(x) =

√π

2+

sin(x2)

2x+ O

(1

x3

), as x →+∞.

Noting that Fc(x)+ Fc(−x) =√π/2 and Fs(x)+ Fs(−x) =

√π/2, we have

Fc(x) = O

(1

x

)as x →−∞,

and

Fs(x) = O

(1

x

)as x →−∞.

Now apply the formula for the cosine of a difference to compute∫ s2

−∞

∫ s3

−∞

cos(s23 − s2

4) ds4 ds3 =

∫ s2

−∞

(cos(s2

3)Fc(s3)+ sin(s23)Fs(s3)

)ds3

=Fc(s2)

2+ Fs(s2)

2

2.




Now use integration by parts. (Here we write F2(u) for (F(u))2.)

I =1

2lim

R→∞

∫ R

−∞

∫ s1

−∞

(cos(s2

1) cos(s22)+ sin(s2

1) sin(s22)) (

F2c (s2)+ F2

s (s2))

ds2 ds1

=1

2lim

R→∞

[Fc(R)

∫ R

−∞

cos(x2)(F2

c (x)+ F2s (x)

)dx

+ Fs(R)∫ R

−∞

sin(x2))(F2

c (x)+ F2s (x)

)dx

−

∫ R

−∞

(cos(x2)Fc(R)+ sin(x2)Fs(x)

)(F2

c (x)+ F2s (x)

)dx

]

=

√π

8

∫∞

−∞

(cos(x2)+ sin(x2)

)(F2

c (x)+ F2s (x)

)dx −

1

8

(F2

c (x)+ F2s (x)

)2∣∣∣∣∞−∞

=

√π

8

∫∞

−∞

(cos(x2)+ sin(x2)

)(F2

c (x)+ F2s (x)

)dx −

π2

8.

Here, in going from the second line to the third, we have used the asymptotics aboveto conclude that

cos(x2)(F2

c (x)+ F2s (x)

)= π cos(x2)+

√π

2

sin(2x2)− 1− cos(2x2)

2x+ O

(1

x3

)as x →+∞. The oscillatory terms in this sum and the error term are integrable; hencefrom this formula and the analogous one for the sine, we get∫ R

−∞

cos(x2)(F2

c (x)+ F2s (x)

)dx = −

√π

2√

2log R + O(1)

and ∫ R

−∞

sin(x2)(F2

c (x)+ F2s (x)

)dx =

√π

2√

2log R + O(1)

as R → +∞. All other terms converge, and these log R terms cancel in the limit, soin particular the integral in the third line exists.

To evaluate this integral, note that∫∞

−∞

(cos(x2)+ sin(x2)

)(F2

c (x)+ F2s (x)

)dx

= Re

[(1+ i)

∫∞

−∞

e−i x2(F2

c (x)+ F2s (x)

)dx

]= lim

a→iRe

[√

2a∫∞

−∞

e−ax2(F2

c (x)+ F2s (x)

)dx

],

where the final limit is taken over a with Re(a) > 0. To justify exchanging the integraland the limit, note that by explicit computation

lima→i

∫∞

0e−ax2

dx =∫∞

0e−i x2

dx;

lima→i

∫∞

1

e−ax2+ibx2

xdx =

∫∞

1

e−i x2+ibx2

xdx




for b = 0, −2, or 2. (For the first limit, this is the standard argument for evaluatingFc(∞) = Fs(∞) =

√π/2. The second is similar.)

These limits correspond to the leading terms of the asymptotics of the integrand.The error terms tend to zero fast enough (like x−3 at +∞ and x−2 at −∞) that theDominated Convergence Theorem applies. Thus we can exchange the limit and theintegral. Noting that

F2c (x)+ F2

s (x) =∫∞

0ei(x−s)2ds

∫∞

0e−i(x−t)2dt,

we see that the required integral is a standard sort of Gaussian integral and

√2a∫∞

−∞

e−ax2(Fc(x)

2+ Fs(x)

2)

dx

=√

2a∫∞

−∞

∫∞

0

∫∞

0e−ax2

+i(x−s)2−i(x−t)2ds dt dx

=

√2π

4

[π − i log

(ai − 1

ai + 1

)].

As Re(a) tends to zero, the argument of ai−1ai+1 tends to π/2 (though the magnitude

grows without bound). Hence the real part of the expression tends to∫∞

−∞

(cos(x2)+ sin(x2)

)(F2

c (x)+ F2s (x)

)dx =

3π 3/2

4√

2,

and we get

I =1

2

√π

2·

3π3/2

4√

2−π 2

8=π2

16.

Editorial comment. The (relatively long) solution published here avoids the hand-waving arguments found in some of the other solutions. The proposer notes that this in-tegral is connected to the quantum-mechanical Landau–Zener problem. He conjecturesthat if we do this with 2n nested integrals rather than 4, we get the value 2(π/4)n/n!.

The proposer’s solution to this problem unfortunately appeared on the web in anessay http://arxiv.org/abs/1201.1975, in time for would-be solvers to haveread it.

Also solved by T. Amdeberhan & A. Straub, M. L. Glasser, J. A. Grzesik, M. Omarjee (France), J. G. Sim-monds, and the proposer.

Limits and Derivatives (Correction)

11603 [2011, 847]. Proposed by Alfonso Villani, Universita di Catania, Catania, Italy.In the May 2013 issue solution by Iosif Pinelis, on p. 476 line 2, one step of the solutionwe published was a garbled version of what we received. It should have read

| f (x)| ≥ | f (xk)| −

∣∣∣∣∫ x

xk

f ′(u) du

∣∣∣∣ > 2ε − ε = ε.




amer.math.monthly.120.07.660 (1)

Documents

Transcript of amer.math.monthly.120.07.660 (1)