Sums of generalized harmonic series for kids from five to fifteen

8/9/2019 Sums of generalized harmonic series for kids from five to fifteen

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a r X i v : 1 0 0 3 . 3 6 0 2 v 1 [ m a t h . C A ] 1 6 M a r 2 0 1 0 Sums of generalized harmonic series for kids

from ve to fteen

Z. K. SilagadzeBudker Institute of Nuclear Physics and

Novosibirsk State University, 630 090, Novosibirsk, Russia

Abstract

We reexamine remarkable connection, rst discovered by Beuk-ers, Kolk and Calabi, between (2n ), the value of the Riemann zeta-function at even positive integer, and the volume of some 2 n -dimensio-nal polytope. It can be shown that this volume equals to the traceof some compact self-adjoint operator. We provide an explicit ex-pression for the kernel of this operator in terms of Euler polynomials.This explicit expression makes it easy to calculate the volume of thepolytope and hence (2n ). In the case of odd positive integers, theexpression for the kernel enables to rediscover an integral representa-tion for (2n + 1), obtained originally by different method by Cvijovicand Klinowski. Finally, we indicate that the origin of the Beukers-Kolk-Calabis miraculous change of variables in the multidimensionalintegral, which is at the heart of all of this business, can be traceddown to the amoeba associated with the certain Laurent polynomial.

1 Introduction

In a nice little book [1] Vladimir Arnold has collected 77 mathematical prob-lems for Kids from 5 to 15 to stimulate the development of a culture of

critical thinking in pupils. Problem 51 in this book asks to calculate the sumof inverse squares of the positive integers and prove the Eulers celebratedformula

n =1

1n2

=2

6. (1)

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Well, there are many ways to do this (see, for example, [2, 3, 4, 5, 6, 7]

and references therein), some maybe even accessible for kids under fteen.However, in this note we concentrate on the approach of Beukers, Kolk andCalabi [8], further elaborated by Elkies in [9]. This approach incorporatespleasant features which all the kids (and even some adults) adore: simplicity,magic and the depth that allows to go beyond the particular case ( 1). Thesimplicity, however, is not everywhere explicit in [8] and [9], while the magiclongs for explanation after the rst admiration fades away. Below we will tryto enhance the simplicity of the approach and somewhat uncover the secretof magic.

The paper is organized as follows. In the rst two sections we reconsiderthe evaluation of (2) and (3) in order technical details of the general casenot to obscure the simple underlying ideas. Then we elaborate the generalcase and provide the main result of this work, the formula for the kernelwhich allows to simplify considerably the evaluation of (2n) from [8, 9]and re-derive Cvijovic and Klinowskis integral representation [10] for (2n +1). Finally, we ponder over the mysterious relations between the sums of generalized harmonic series and amoebas, rst indicated by Passare in [ 7].This relation enables to somewhat uncover the origin of the Beukers-Kolk-Calabis highly non-trivial change of variables.

2 Evaluation of (2)Recall the denition of the Riemann zeta function

(s) =

n =1

1ns

. (2)

The sum ( 1) is just (2) which we will now evaluate following the methodof Beukers, Kolk and Calabi [8]. Our starting point will be the dilogarithmfunction

Li 2(x) =

n =1

xn

n2. (3)

Clearly, Li 2(0) = 0 and Li 2(1) = (2). Differentiating ( 3), we get

xd

dxLi 2(x) =

n =1

xn

n= ln(1 x),

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where is the image of the unit square under the transformation ( x, y)

(u, v). It is easy to realize that is the isosceles right triangle = {(u, v) :u 0, v 0, u + v / 2} and, therefore,

(2) =43

12

2

2=

2

6. (10)

Beautiful even more so, as the same method of proof extends to thecomputation of (2k) in terms of a 2k-dimensional integral, for all k 1[11]. However, before considering the general case, we check whether thetrick works for (3).

3 Evaluation of (3)In the case of (3), we begin with trilogarithm

Li 3(x) =

n =1

xn

n3, (11)

and using

xd

dxLi 3(x) = Li2(x) =

x

0

ln(1 y)y

dy,

we get

(3) = Li 3(1) = 1

0

dxx

x

0

ln(1 y)y

dy. (12)

But

1x

x

0

ln(1 y)y

dy = 1

0

ln(1 xz)xz

dz =1

0

dz1

0

dy1 xyz

,

and nally (3) = Li 3(1) =

3

dxdydz1 xyz

, (13)

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where 3 = {(x,y,z ) : 0 x 1, 0 y 1, 0 z 1} is the unit cube.

By the similar trick as before, we can transform ( 13) into the integral

(3) =87

3

dxdydz1 x2y2z2

, (14)

and here the analogy with the previous case ends, unfortunately, because thegeneralization of the Beukers-Kolk-Calabi change of variables does not leadin this case to the simple integral. However, it is interesting to note that thehyperbolic version of this change of variables

x =sinh u

cosh v, y =

sinh v

cosh w, z =

sinh w

cosh u(15)

does indeed produce an interesting result

(3) =87

U 3

dudvdw =87

Vol( U 3), (16)

where U 3 is a complicated 3-dimensional shape dened by the inequalities

u 0, v 0, w 0, sinh u cosh v, sinh v cosh w, sinh w cosh u.

Unfortunately, unlike the previous case, there is no obvious simple way tocalculate the volume of U 3.

However, there is a second way to convert the integral ( 12) for (3) inwhich the Beukers-Kolk-Calabi change of variables still plays a helpful role.We begin with the identity

(3) = 1

0

dxx

x

0

ln(1 y)y

dy = D

ln(1 y)xy

dxdy, (17)

where the domain of the 2-dimensional integration is the triangle D ={(x, y) : x 0, y 0, y x}. Interchanging the order of x and y inte-grations in the evaluation of the 2-dimensional integral ( 17), we get

(3) = 1

0

ln(1 y)y

dy1

y

dxx

,

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which can be transformed further as follows

(3) =1

0

ln(1 y) ln yy

dy = 1

0

ln y dy1

0

dx1 xy

= ln y1 xy dxdy,or in a more symmetrical form

(3) = 12 ln (xy)1 xy dxdy. (18)

Note that

2xy ln (xy)1 x2y2 dxdy = 14 ln (x2y2)1 x2y2 dx2 dy2 = 14 ln (xy)1 xy dxdy.Therefore, we can modify ( 5) and (6) accordingly and using them transform(18) into

(3) = 47 ln (xy)1 x2y2 dxdy. (19)

At this point we can use Beukers-Kolk-Calabi change of variables ( 8) in (19)and as a result we get

(3) = 47

ln (tan u tan v) dudv = 87

ln(tan u) dudv. (20)

But this equation indicates that

(3) = 87

/ 2

0

du ln (tan u)

/ 2 u

0

dv = 87

/ 2

0

2

u ln (tan u) du,

which after substitution x = 2 u becomes

(3) = 87

/ 2

0

x ln (cot x) dx =87

/ 2

0

x ln (tan x) dx. (21)

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But

/ 2

0

ln(tan x) dx = 0

/ 2

ln (cot u) du = / 2

0

ln(tan u) du = 0 ,

which enables to rewrite ( 21) as follows

(3) =87

/ 2

0

x 4

ln (tan x) dx =87

/ 2

0

ln (tan x)d

dxx2

2

4

x dx,

and after integration by parts and rescaling x x/ 2 we end with

(3) =17

0

x( x)sin x

dx. (22)

This is certainly an interesting result. Note that until quite recently veryfew denite integrals of this kind, involving cosecant or secant functions,were known and present in standard tables of integrals [ 12, 13, 14]. In fact(22) is a special case of the more general result [10] which we are going nowto establish.

4 The general case of (2n)The evaluation of (2) can be straightforwardly generalized. The polyloga-rithm function

Li s (x) =

n =1

xn

n s(23)

obeys

xd

dxLi s (x) = Li s 1(x),

and hence

Li s (x) =x

0

Li s 1(y)y

dy. (24)

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Repeated application of this identity allows to write

(n) = Li n (1) =1

0

dx1x1

x 1

0

dx2x2

. . .

x n 2

0

dxn 1xn 1

[ ln(1 xn 1)] . (25)

After rescaling

x1 = y1, x2 = x1y2, x3 = x2y3, . . . , x n 1 = xn 2yn 1 = y1y2 yn 1,

and using

1

0 dyn1 y1y2 yn = 1y1y2 yn 1 ln(1 y1y2 yn 1),we get

(n) = n

dy1 dy2 dyn1 y1y2 yn

, (26)

where n is n-dimensional unit hypercube. The analogs of ( 5) and (6) are

n

dx1 dxn1 x1 xn

+ n

dx1 dxn1 + x1 xn

= 2 n

dx1 dxn1 x21 x2n

and

n

dx1 dxn1 x1 xn

n

dx1 dxn1 + x1 xn

=1

2n 1 n

dx1 dxn1 x1 xn

,

from which it follows that ( 26) is equivalent to

(n) =2n 1

2n n

dx1 dxn1 x21 x2n

. (27)

If we now make a change of variables that generalizes ( 8), namely

x1 =sin u1cos u2

, x2 =sin u2cos u3

, . . . , x n 1 =sin un 1cos un

, xn =sin uncos u1

. (28)

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we, in general, encounter a problem because the Jacobian of ( 28) is [8, 9]

(x1, . . . , x n ) (u1, . . . , u n )

= 1 ( 1)n x21x22 x

2n ,

and, therefore, only for even n we will get a simple integral. For thehyperbolic version of (28),

x1 =sinh v1cosh v2

, x2 =sinh v2cosh v3

, . . . , x n 1 =sinh vn 1cosh vn

, xn =sinh vncosh v1

, (29)

the Jacobian has a right form

(x1 , . . . , x n ) (v1 , . . . , vn )= 1 x21x22 x2n ,

and we get

(n) =2n

2n 1 U n

dv1 dvn =2n

2n 1Voln (U n ). (30)

However, the gure U n has a complicated shape and it is not altogetherclear how to calculate its n-dimensional volume Voln (U n ). Therefore, fora moment, we concentrate on the even values of n for which (28) works

perfectly well and leads to

(2n) =22n

22n 1 2 n

du1 dun =22n

22n 1Vol2n ( 2n ), (31)

where n is a n-dimensional polytope dened through the inequalities

n = (u1, . . . , u n ) : u i 0, ui + ui+1 2

. (32)

It is assumed in ( 32) that u i are indexed cyclically (mod n) and therefore

un +1 = u1.There exists an elegant method due to Elkies [ 9] how to calculate then-volume of n (earlier calculations of this type can be found in [ 15]). Ob-viously

Voln ( n ) =2

nVoln (n ), (33)

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where Vol n (n ) is the n-dimensional volume of the rescaled polytope

n = {(u1, . . . , u n ) : ui 0, ui + u i+1 1} . (34)If we introduce the characteristic function K 1(u, v) of the isosceles right tri-angle {(u, v) : u, v 0, u + v 1} that is 1 inside the triangle and 0 outsideof it, then [9]

Voln (n ) =1

0

. . .1

0

n

i=1

K 1(u i , u i+1 ) du1 . . . du n =1

0

du1

1

0

du2 K 1(u1, u2) . . .

1

0dun 1 K 1(un 2, un 1)

1

0dun K 1(un 1, un ) K 1(un , u1). (35)

Let us note that K 1(u, v) can be interpreted [9] as the kernel of the linearoperator T on the Hilbert space L2(0, 1), dened as follows

(T f )(u) =1

0

K 1(u, v)f (v) dv =1 u

0

f (v) dv. (36)

Then ( 35) shows that Vol n (n ) equals just to the trace of the operator T n :

Voln (n ) =1

0K n (u1, u1) du1, (37)

whose kernel K n (u, v) obeys the recurrence relation

K n (u, v) =1

0

K 1(u, u 1) K n 1(u1, v) du1. (38)

Surprisingly, we can nd a simple enough solution of this recurrence relation.Namely,

K 2n (u, v) = ( 1)n22n 2

(2n 1)!

E 2n 1u + v

2+ E 2n 1

u v2

(u v)+

E 2n 1u + v

2+ E 2n 1

v u2

(v u) , (39)

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and

K 2n +1 (u, v) = ( 1)n 22n 1

(2n)!

E 2n1 u + v

2+ E 2n

1 u v2

(1 u v)+

E 2n1 u + v

2 E 2n

u + v 12

(u + v 1) . (40)

In these formulas E n (x) are the Euler polynomials [16] and (x) is the Heav-iside step function

(x) =

1, if x > 0,

12 , if x = 0 ,

0, if x < 0.

After they are guessed, it is quite straightforward to prove ( 39) and (40) byinduction using the recurrence relation ( 38) and the following properties of the Euler polynomials

ddx

E n (x) = nE n 1(x), E n (1 x) = ( 1)n E n (x). (41)

In particular, after rather lengthy but straightforward integration we get1 u

0

K 2n +1 (u1, v)du1 = K 2n +2 (u, v) X,

where

X = ( 1)n +122n

(2n + 1)!E 2n +1

1 + v2

+ E 2n +11 v

2.

But1 v2 = 1 1 + v2

and the second identity of ( 41) then implies that X = 0.Therefore the only relevant question is how ( 39) and (40) were guessed.

Maybe the best way to explain the method used is to refer to the problem

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13 from the aforementioned book [ 1]. To demonstrate the cardinal differ-

ence between the ways problems are posed and solved by physicists and bymathematicians, Arnold provides the following problem for children:On a bookshelf there are two volumes of Pushkin s poetry. The thickness

of the pages of each volume is 2 cm and that of each cover 2 mm. A wormholes through from the rst page of the rst volume to the last page of thesecond, along the normal direction to the pages. What distance did it cover?

Usually kids have no problems to nd the unexpected correct answer, 4mm, in contrast to adults. For example, the editors of the highly respectablephysics journal initially corrected the text of the problem itself into: fromthe last page of rst volume to the rst page of the second to match theanswer given by Arnold [1, 17]. The secret of kids lies in the experimentalmethod used by them: they simple go to the shelf and see how the rst pageof the rst volume and the last page of the second are situated with respectto each other.

The method that led to ( 39) and (40) was exactly of this kind: we simplycalculated a number of explicit expressions for K n (u, v) using (38) and triedto locate regularities in this expressions.

Having (39) at our disposal, it is easy to calculate the integral in ( 37).Namely, because

K 2n (u, u ) = ( 1)n22n 2

(2n 1)![E 2n 1(u) + E 2n 1(0)] , (42)

andE 2n 1(u) =

12n

ddu

E 2n (u), (43)

we get

Vol2n (2n ) =1

0

K n (u, u ) du = ( 1)n22n 2

(2n 1)!E 2n 1(0), (44)

(note that E 2n (0) = E 2n (1) = 0.) But E 2n 1(0) can be expressed trough theBernoulli numbers

E 2n 1(0) = 2

2n(22n 1)B2n , (45)

and combining ( 31), (33), (44) and ( 45), we nally reproduce the celebratedformula

(2n) = ( 1)n +122n 1

(2n)!2n B2n . (46)

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5 The general case of (2n + 1)The evaluation of (3) can be also generalized straightforwardly. We have

(n) =1

0

Li n 1(x1)x1

dx1 =1

0

dx1x1

x 1

0

Lin 2(x2)x2

dx2 = D

Li n 2(x2)x1x2

dx1dx2.

Interchanging the order of integrations in the two-dimensional integral, weget

(n) =1

0Lin 2(x2)

x2dx2

1

x 2dx1

x1=

1

0ln (x2)Li n 2(x2)

x2dx2. (47)

Now we can repeatedly apply the recurrence relation ( 24), accompanied withLi1(x) = ln(1 x) at the last step, and transform ( 47) into

(n) =1

0

ln x1x1

dx1

x 1

0

dx2x2

. . .

x n 4

0

dxn 3xn 3

x n 3

0

ln(1 xn 2)xn 2

dxn 2,

which after rescaling

x2 = x1y2, x3 = x2y3 = x1y2y3, . . . , x n 2 = xn 3yn 2 = x1y2 yn 2,

takes the form

(n) =1

0

ln x1x1

dx1

1

0

dy2y2

. . .1

0

dyn 3yn 3

1

0

ln(1 x1y2 yn 2)yn 2

dyn 2. (48)

Then the relation1

0

dyn 11 x1y2 yn 1

= ln(1 x1y2 yn 2)

y1y2 yn 2

shows that ( 48) is equivalent to the ( n 1)-dimensional integral

(n) = n 1

ln x11 x1 xn 1

dx1 dxn 1. (49)

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As in the previous case, ( 49) can be further transformed into

(n) = 2n

2n 1 n 1

ln x11 x21 x2n 1

dx1 dxn 1,

or, in the more symmetrical way,

(n) = 2n

2n 11

n 1 n 1

ln (x1 xn 1)1 x21 x2n 1

dx1 dxn 1. (50)

Let us now assume that n is odd and apply the Beukers-Kolk-Calabi changeof variables (28) to the integral ( 50). As the result, we get

(2n + 1) = 1

2n22n +1

22n +1 1 2 n

ln[tan( u1) tan( u2n )] du1 du2n ,

which is the same as

(2n + 1) = 22n +1

22n +1 1 2 n

ln[tan( u1)] du1 du2n .

By rescaling variables, we can go from the polytope 2n to the polytope 2nin this 2n-dimensional integral and get

(2n + 1) = 22n +1

22n +1 12

2n 2 n

ln tan u12

du1 du2n . (51)

Using the kernel K 2n (u, v), we can reduce the evaluation of ( 51) to the eval-uation of the following one-dimensional integral

(2n + 1) = 22n

22n +1 1

1

0

ln tan2

u K 2n (u, u ) du. (52)

Butln tan

2

(1 u) = ln cot2

u = ln tan2

u ,

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which enables to rewrite ( 52) as

(2n +1) = 2n

22n +1 1

1

0

ln tan2

u [K 2n (u, u ) K 2n (1 u, 1 u)] du.

(53)However, from (42) and (43) we have (recall that E 2n 1(1 u) = E 2n 1(u))

K 2n (u, u ) K 2n (1 u, 1 u) = ( 1)n22n 1

(2n)!d

duE 2n (u),

and the straightforward integration by parts in ( 53) yields nally the result

(2n + 1) =( 1)n 2n +1

4 [1 2 (2n +1) ] (2n)!

1

0

E 2n (u)sin( u)

du. (54)

This is exactly the integral representation for (2n + 1) found in [10]. Ourearlier result ( 22) for (3) is just a special case of this more general formula.

6 concluding remarks: (2) and amoebasIt remains to clarify the origin of the Beukers-Kolk-Calabis highly non-trivial

miraculous change of variables ( 28). Maybe an interesting observation dueto Passare [ 7] that (2) is related to the amoeba of the polynomial 1 z1 z2gives a clue.

Amoebas are fascinating objects in complex geometry [ 18, 19]. They aredened as follows [20]. For a Laurent polynomial P (z1, . . . , zn ), let Z P denotethe zero locus of P (z1, . . . , zn ) in (C \{ 0})n dened by P (z1, . . . , zn ) = 0. Theamoeba A(P ) of the Laurent polynomial P (z1, . . . , zn ) is the image of thecomplex hypersurface Z P under the map

Log : (C \{ 0})n R n

dened through (z1, . . . , zn ) (ln |z1 |, . . . , ln |zn |).

Let us nd the amoeba of the following Laurent polynomial

P (z1, z2) = z1 z 11 i z2 z

12 . (55)

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Taking

z1 = eu

ei u

, z2 = ev

e i v

,we nd that the zero locus of the polynomial ( 55) is determined by conditions

cos u sinh u = sin v cosh v, sin u cosh u = cos v sinh v.

If we rewrite these conditions as follows

x =sinh vcosh u

=sin ucos v

, y =sinh ucosh v

=sin vcos u

, (56)

we immediately recognize the Beukers-Kolk-Calabi substitution ( 8) and itshyperbolic version with the only difference that in ( 8) we had 0 x, y 1.However, from (56) we get

cos2 u =1 x2

1 x2y2, cos2 v =

1 y2

1 x2y2, (57)

andcosh2 u =

1 + y2

1 x2y2, cosh2 v =

1 + x2

1 x2y2. (58)

It is clear from (57) and (58) that we should have

x2 1, y2 1.

Therefore, the amoeba A(P ) is given by relations

A(P ) = (u, v) : 1 sinh ucosh v

1, 1 sinh vcosh u

1 , (59)

and the hyperbolic version of the Beukers-Kolk-Calabi change of variables(8) transforms the unite square into the one-quarter of the amoeba ( 59).Then the analog of ( 9) indicates that (2) equals to the one-third of the areaof this amoeba.

As we see, the hyperbolic version of the Beukers-Kolk-Calabi change of

variables seems more fundamental and arises quite naturally in the context of the amoeba ( 59). Trigonometric version of it then is just an area-preservingtransition from the radial coordinates ( u, v) to the angular ones ( u , v).

One more amoeba related to (2) was found in [7]. Although the corre-sponding amoeba A(1 z1 z2) looks different from the amoeba ( 59), they

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do have the same area. The trigonometric change of variables used by Pas-

sare in [7] is also different from (8) but also leads to simple calculation of thearea of A(1 z1 z2) and hence (2). Of course it will be very interestingto generalize this mysterious relations between (n) and amoebas for n > 2and nally disentangle the mystery. Im afraid, however, that this game isalready not for kids under fteen.

References

[1] V. I. Arnold, Problems for Kids from 5 to 15 (MCNMO, Moscow, 2004)(in Russian).

[2] A. M. Yaglom and I. M. Yaglom, Elementary Presentation of Nonele-mentary Problems , problem 143 (Gostekhizdat, Moscow, 1954) (in Rus-sian).

[3] D. Kalman, Six Ways to Sum a Series , College Math. J. 24 , 402-421(1993).

[4] K. P. Kokhas, Sum of inverse squares , Mathematicheskoe Prosveshenie8 , 142-163 (2004). (in Russian). See also informal notes by R. Chapman,Evaluating (2), http://www.secamlocal.ex.ac.uk/ rjc/etc/zeta2.pdf

[5] J. Hofbauer, A Simple Proof of 1 +1

22 +1

32 + = 2

6 and Related Identities , Am. Math. Mon. 109 , 196-200 (2002).

[6] N. Lord, Yet Another Proof That 1n 2 =16

2, Math. Gazette 86 , 477-479 (2002).

[7] M. Passare, How to Compute 1n 2 by Solving Triangles , Am. Math.Mon. 115 , 745-752 (2008).

[8] F. Beukers, J. A. C. Kolk and E. Calabi, Sums of generalized harmonicseries and volumes , Nieuw Archief voor Wiskunde, fourth series, 11 ,217-224 (1993).

[9] N. D. Elkies, On the Sums

k= (4k + 1) n , Am. Math. Mon. 110 , 561-

573 (2003); Corr. ibid. 111 , 456 (2004).

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http://www.secamlocal.ex.ac.uk/~rjc/etc/zeta2.pdfhttp://www.secamlocal.ex.ac.uk/~rjc/etc/zeta2.pdfhttp://www.secamlocal.ex.ac.uk/~rjc/etc/zeta2.pdfhttp://www.secamlocal.ex.ac.uk/~rjc/etc/zeta2.pdf


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[10] D. Cvijovic and J. Klinowski, Integral Representations of the Riemann

Zeta Function for Odd-Integer Arguments , J. Comput. Appl. Math. 142 ,435-439 (2002).

[11] M. Aigner and G. M. Ziegler, Proofs from THE BOOK , p. 38 (Springer-Verlag, Berlin Heidelberg, third edition, 2004).

[12] D. Cvijovic and H. M. Srivastava, Closed-form evaluations of certain denite integrals by employing the Cauchy integral theorem , Numer. Al-gor. 49 , 129-141 (2008).

[13] D. Cvijovic, Closed-form evaluation of some families of cotangent and cosecant integrals , Integral Transform. Spec. Funct. 19 , 147-155 (2008).

[14] D. Cvijovic, Closed-form evaluation of some families of denite tan-gent and secant integrals , Integral Transform. Spec. Funct. 18 , 569-579(2007).

[15] J. Kubilius, Estimating the second central moment for strongly addi-tive arithmetic functions , Lithuanian Mathematical Journal 23 , 61-69(1983).

[16] See, for example, B. K. Karande and N. K Thakare, On the unication of Bernoulli and Euler polynomials , Indian J. Pure Appl. Math. 6 , 98-

107 (1975); G. Bretti and P. E. Ricci, Euler polynomials and the related quadrature rule , Georgian Math. J. 8 , 447-453 (2001).

[17] V. I. Arnold, Mathematics and Physics , in G. Boniolo, P. Budinich andM. Trobok (Eds.) The Role of Mathematics in Physical Sciences , pp.225-233 (Springer, Dordrecht, 2005).

[18] O. Viro, WHAT IS an amoeba? , Notices Amer. Math. Soc. 49 , 916-917(2002).

[19] M. Passare and A. Tsikh, Amoebas: their spines and their contours ,Contemp. Math. 377 , 275-288 (2005).

[20] I. M. Gelfand, M. M. Kapranov and A. V. Zelevinsky, Discriminants,resultants, and multidimensional determinants , (Birkhauser, Boston,1994).

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