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Home , Bernoulli polynomials, Clausen function, Dirichlet series, Gamma function, Polylogarithm, Riemann zeta function

New representations of the

Djurdje Cvijović1

1 Atomic Laboratory, Vinča Institute of Nuclear Sciences,

P.O. Box 522, 11001 Belgrade, Serbia.

Abstract. Maximon has recently given an excellent summary of the properties of the Euler

dilogarithm function and the frequently used generalizations of the dilogarithm, the most

important among them being the polylogarithm function Lis () z . The polylogarithm function appears in several fields of and in many physical problems. We, by making use of elementary arguments, deduce several new integral representations of the polylogarithm

Lis () z for any complex z for which z < 1. Two are valid for all complex s, whenever

Res > 1. The other two involve the and are valid in the important

special case where the parameter s is an positive integer. Our earlier established results on the

integral representations for the zeta function ζ (2n + 1) , nN∈ , follow directly as

corollaries of these representations.

Keywords: , integral representation, Riemann’s zeta function, Bernoulli

polynomials

2000 Mathematics Subject Classification. Primary 11M99; Secondary 33E20.

1 1. Introduction

Recently, Maximon [1] has given an excellent summary of the defining equations and

properties of the Euler dilogarithm function and the frequently used generalizations of the

dilogarithm, the most important among them being the polylogarithm function Lis () z . These

include integral representations, expansions, linear and quadratic transformations,

functional relations and numerical values for special arguments.

Motivated by this paper we have begun a systematic study of new integral

representations for Lis () z since it appears that only half a dozen of them can be found in the literature. Among known representations the following

zt∞ s−1 Li() z = dt (Re s > 0, z ∉ (1,)) ∞ , s ∫ t Γ−()sez0

and

ztzt∞ log ss−−111 log (1/ ) Li() z = dt = dt (Re s > 0), s ∫∫ Γ−Γ()sttz10 ( ) () s 1 − zt

are best known [2, pp. 30-31].

Here, by making use of fairly elementary arguments, we deduce several new integral

representations of the polylogarithm function for any complex z for which z < 1. Two are

valid for all complex s, whenever Res > 1 (see ). The other two given in Corollary

1 are valid in the important special case where the parameter s is an positive integer. Our

earlier published results [3] on the integral representations for the

ζ (2n + 1) , nN∈ , directly follow as corollaries of these representations (see Corollary 2).

2 2. The polylogarithm function

The polylogarithm function (also known as Jonquiére’s function), Lis () z , is defined

for any complex s and z as the of the

∞ zk Li() z z < s C ) s =∑ s ( 1; ∈ (1) k =1 k

which is absolutely convergent for all s and all z inside the unit disc in the complex z -

plane. Sometimes, Lis () z is referred to as the polylogarithm of order s and argument z, most frequently, however, it is simply called the polylogarithm.

The function Lis () z , for fixed s, has no poles or essential singularities anywhere in the finite complex z - plane, and, for fixed z, has no poles or essential singularities anywhere in the finite complex s - plane. It has only one essential singular point at s = ∞. For fixed s, the function has two branch points: z = 1 and z = ∞. The principal branch of the

polylogarithm is chosen to be that for which Lis () z is real for z real, 0 ≤ z ≤ 1, (when s is real) and is continuous except on the positive real axis, where a cut is made from z = 1 to ∞ such that the cut puts the real axis on the lower half plane of z. Thus, for fixed s the function

Lis () z is a single-valued function in the z-plane cut along the interval (1, ∞) where it is continuous from below.

In the important case where the parameter s is an integer, Lis () z will be denoted by

Lin () z (or Li−n () z when negative). Li0 () z and all Li−n () z , n = 1, 2, …, are rational

functions in z. Lin () z , n = 1, 2, …, is the polylogarithm of order n, (i.e. the n-th order

polylogarithm). The special case n = 1 is the ordinary Li1() z=− log(1 − z ), while

3 the cases n = 2, 3, 4, …, are classical polylogarithms known, respectively, as dilogarithm,

trilogarithm, quadrilogarithm, etc.

For more details and an extensive list of references in which polylogarithms appear in

physical and mathematical problems we refer the reader to Maximon [1]. The polylogarithm

Lin () z of order n = 1, 2, 3, … is thoroughly covered in Lewin’s standard text [4], while many

formulae involving Lis () z can be found in [2, pp. 30-31] and [5]. Berndt’s treatise [6] can

serve as an excellent introductory text on the polylogarithm (and numerous related functions)

and as an encyclopaedic source.

2. Statement of the results

In preparation for the statement of the results we need to make several definitions.

The Riemann zeta function and the , ζ ()s and ζ (,sa ), are, respectively, defined by means of the series [7, p. 807, Eqs. 23.2.1, 23.2.19 and 23.2.20]:

⎧ ∞∞11 1 = (Res > 1) ⎪∑∑ss− s ⎪ kk==10kk12−+ (21) ζ ()s = ⎨ (2) 1(1)∞ − k −1 ⎪ (Res > 0; s ≠ 1) ⎪ 1−ss∑ ⎩12− k =1 k

and [2, pp. 24-27]

∞ 1 (,)sa Res > 1, a 0,-1,-2,...) . (3) ζ =(≠∑ s k =0 ()ka+

Both are analytic over the whole , except at s =1, where they have a simple pole.

4 Next, for any real x and any complex s with Res > 1, we define

∞ sin(kx ) Sx()= , (4a) s ∑ s k =1 k and

∞ cos(kx ) Cx() . (4b) s = ∑ s k =1 k

We also use the Bernoulli polynomials of degree n in x, Bn ()x , defined for each nonnegative integer n by means of their [7, p. 804, Eq. 23.1.1]

tetx ∞ t n Bx() (t < 2) t = ∑ n π , (5a) en−1!n=0

and given explicitly in terms of the Bernoulli numbers BBnn= (0) by [5, p. 765]

n ⎛⎞n nk − B ()xB= ⎜⎟ x. (5b) n∑ ⎝⎠k k k =0

Our results are as follows.

Theorem. Assume that s and z are complex numbers, let Lis () z be the polylogarithm function

and let Sxs () and Cxs () be defined as in Eqs. (4a) and (4b), respectively. If Res > 1 and z < 1, then

5 12δ ztsin(2)π Liz()= S (2)π t dt, (6a) ss∫ 2 δπ0 12cos(2)−+ztz

11δ − z2 Liz()= (2)π t dt= (6b) ss∫ 2 δ0 12cos(2)−+ztzπ

1 δ 2cos(2)ztz()π − = Ct(2π ) dt. (6c) ∫ s 2 δ0 12cos(2)−+ztzπ

Here, and throughout the text, we set either δ =1 or δ =1/2.

Remark 1. Observe that our integral representations in the above Theorem essentially involve either the Hurwitz zeta function or the generalized . Indeed,

between the functions Sxs () and Cxs () and the Hurwitz zeta function defined as in (3) there exists the relationship [8, p. 726, Entry 5.4.2.1]

s ⎧⎫Sts (2π ) (2π ) ⎧⎫csc(π s / 2) ⎨⎬=− ⎨ ⎬⎣⎡ζζ()(1,st∓ 1,1− s− t)⎦⎤ ⎩⎭Cts (2ππ )4()Γ s ⎩⎭ sec(s / 2)

⎧⎫sn≠ 2 (0≤≤ts 1; Re > 1; ⎨⎬,) n∈N, ⎩⎭sn≠+21

where Γ()s is the familiar function. In the cases when this relationship does not hold

Sxs ()and Cxs () are used to define the generalized Clausen function [4, p. 281, Eq. 3]

⎧⎫Cxn ( ) if n is odd Cln () x =∈⎨⎬()n N . ⎩⎭Sxn ()ifiseven n

What is most important, however, is that Sx21n− ()and Cx2n () (n∈ N) are expressible in terms of the Bernoulli polynomials and this makes possible our corollaries.

6 Corollary 1. Assume that n is a positive integer, let Lin () z be the polylogarithm function and

let Bn ()x be the Bernoulli polynomials of degree n in x given by Eqs. (5). Then, provided

that z < 1, we have

1(2)ππ21n− δ 2sin(2)zt Liz()=− (1)n Bt() dt, (7a) 21nn−−∫ 21 2 δπ2(2nzt−−+ 1)!0 1 2 cos(2 ) z

1(2)π 22n δ 1− z Liz()=− (1)n−1 Bt() dt= (7b) 22nn∫ 2 δπ2(2nzt )!0 1−+ 2 cos(2 ) z

1(2)π 2n δ 2cos(2)ztz( π − ) =−(1)n−1 B ()tdt. (7c) ∫ 2n 2 δ2(2nzt )!0 1−+ 2 cos(2π ) z

Corollary 2. Assume that n is a positive integer, let ζ ()s be the Riemann zeta function

defined as in (2) and let Bn ()x be the Bernoulli polynomials of degree n in x given by Eqs.

(5). We then have:

1(2)π 21n+ δ ζ(2nB+=− 1) ( 1)n−1 (t )cot(πt )dt, (8a) ∫ 21n+ δ 2(2n + 1)! 0

21(2)22nnπ +1δ ζπ(2nB+=− 1) ( 1)n−1 (t )tan(t )dt. (8b) 2n ∫ 21n+ 212(21)!−+δ n 0

Remark 2. The integral for ζ (2n + 1) in (8a) when δ = 1 is well known [7, p. 807, Eq.

23.2.17] while the other in Corollary 2 were recently deduced [3, Theorem 1].

However, we have failed to find the integral representations given in Theorem and Corollary

1 in the literature.

7 Remark 3. The integral representations given in (6) and (7) hold for z < 1. However, our

results may be extended to any complex z, z > 1, by means of the inversion formula [2, pp.

30-31; 5, pp. 762-763]

s (2π ) isππ/2 ⎛⎞⎛log(−z ) 1is 1 ⎞ Liss() z= eζ ⎜⎟⎜1− s , + −e Li ⎟ z ∉(0,1), Γ()siz⎝⎠⎝2π 2 ⎠

where ζ (,sa ) is the Hurwitz zeta function (3) or, by means of the following particularly simple inversion formula

n n−1 ⎛⎞1l()2π i ⎛⎞ogz Linn() z= (1) − Li ⎜⎟ − B n⎜⎟ n = 0,1,2,,… ⎝⎠zn!2⎝⎠π i

valid for the n-th order polylogarithm.

Remark 4. We remark that the existence of the integrals in (8) is assured since the integrands

on [0, α], 0 < α < 1, have only removable singularities. This can be demonstrated easily by

making use of some basic properties of Bn ()x . For instance, knowing that the odd-indexed

Bernoulli numbers Bn , apart from B1 = −1/2, are zero [7, p. 805, Eq. 23.1.19], we have

Bt21nn+ () (2 nBt+ 1)2 () 12− n limBt21n+ ( ) tan(π t )== lim lim =−+(1 2 )(2nB 1)2n /π ttt→→→1/2 1/2 cos(πππtt )1/2 − sin( )

1−n since Bn(1 / 2)=− (2 1) Bn [7, p. 805, Eq. 23.1.21]

3. Proof of the results

In what follows we shall need the following lemma and we provide two proofs for it.

The second proof was suggested by one of the anonymous referees and it makes use of well-

known and .

8 Lemma. Assume that n is a positive integer and that δ = 1 and δ = ½. Then we have:

δ ⎧⎫cos(nt 2π ) 2sin(2)ztπ ⎧0 ⎫ dt = , (9a) ∫ ⎨⎬ 2 ⎨ n ⎬ 0 ⎩⎭sin(nt 2π ) 12cos(2)−+ztzπ ⎩⎭δ z

δ ⎧⎫cos(nt 2π ) 1− z2 ⎧δ zn ⎫ dt = . (9b) ∫ ⎨⎬ 2 ⎨ ⎬ 0 ⎩⎭sin(nt 2π ) 12cos(2)−+ztzπ ⎩⎭0

First Proof of Lemma. It is clear that the integrals in (9) can be rewritten as follows:

2δπ ⎧⎫cos(nt ) 2sinzt ⎧0 ⎫ dt = , (10a) ∫ ⎨⎬ 2 ⎨ n ⎬ 0 ⎩⎭sin(nt ) 12cos−+ztz ⎩⎭2δπ z

2δπ ⎧⎫cos(nt ) 1− z2 ⎧2δπ zn ⎫ dt = . (10b) ∫ ⎨⎬ 2 ⎨ ⎬ 0 ⎩⎭sin(nt ) 12cos−+ztz ⎩⎭0

We first establish the case δ = 1 in (10). For any positive integer n and arbitrary complex z consider the integrals I 1 and I 2 with parameters given by

2π 2sinzt nz, = edint t= I 1 ()∫ 2 0 12cos−+ztz

2π 2sinzt 2π 2sinzt = cos(nt ) dt + intsin( ) dt, (11a) ∫ 2 ∫ 2 0 12cos−+ztz 0 12cos−+ztz

and by

2π 1− z2 I nz, = edint t= 2 ()∫ 2 0 12cos−+ztz

9 2π 1− z2 2π 1− z2 = cos(nt ) dt + intsin( ) dt. (11b) ∫ 2 ∫ 2 0 12cos−+ztz 0 12cos−+ztz

In order to derive the integrals I 1 and I 2 we make use of and of residues. By setting

1 1 τ = eit , cost =+(τ 1/τ ) and sint =−()τ 1/τ 2 2i

where i =−1 , we arrive at

idτ n (1)ττ2 − ττn−12(1)− nz, = = dτ = I 1 ()∫ ∫ τ =1 ()(1/)τ −−zzττiτ =1 ()(1/)ττ−−zz

ττn−12(1)− = 2πi Res = 2πi z n , (12a) τ = z ()(1/ττ−−z z)

and

τ n+12(1)zd− τ τ n (1)z2 − nz, = = dτ = I 2 ()∫ ∫ τ =1 zz()(1/)τ −−ττ zi τ =1 iz()(1/)ττ−− z z

τ n (1)z2 − = 2πi Res = 2πiiz (− n ) = 2π z n . (12b) τ = z iz()(1/ττ−− z z)

Combining (11) and (12) together and equating the real and imaginary parts on both

sides gives the integrals in (10) where δ = 1. In this way we evaluate the integrals in (9)

where δ = 1. Observe that in both integrals in (12) the contour is the unit circle and is

traversed in the positive (counterclockwise) direction and the only singularities of the integrands that lie inside the contour are at τ = z.

10 Next, it can be easily shown that the case δ = 1/2 in (10) reduces to the above considered case δ = 1. Indeed, in view of the following well-known property [8, p. 272 and

273, Eqs. 2.1.2.20 and 2.1.2.21]

⎧0 fat (2−=− ) ft ( ) 2a ⎪ f ()tdt= ⎨ a ∫ 2ftdt ( ) f (2 a−= t ) ft ( ) 0 ⎪ ∫ ⎩ 0 it suffices to show that both integrands in (10) are such that f (2π − tft )= ( ).

This completes the proof of our lemma.

Second Proof of Lemma. Consider the Chebyshev polynomials of the first and second kind,

Txm () and Uxm () , defined by [7, p. 776, Eqs. 22.3.15 and 22.3.16]

Tm (cosθ )= cos(mθ ) and Umm (cosθ )= sin[(+ 1)θθ ]/ sin , and recall that their generating functions are given by [7, p. 783, Eqs. 22.9.9 and 22.9.10]

1− z2 ∞ 1 ∞ =+12Txz ()m , = Uxz( )m ( z < 1 ) . 2 ∑ m 2 ∑ m 12−+xz z m=1 12−+xz z m=0

Now, by applying the orthogonality relation of the we, for instance, have the sought formulas in (9b)

δδ⎧⎫cos(nt 2π ) 11−−zz222 π⎧⎫cos(nt ) 1 dt = dt ∫∫⎨⎬ 22⎨⎬ 00⎩⎭sin(nt 2π ) 1−+ 2ztz cos(2ππ ) 2 ⎩⎭sin(nt ) 1−+ 2ztz cos

1122δπ ⎧⎫cos(nt ) ⎛⎞∞ δπ ⎧⎫cos(nt ) =+12Ttzdt (cos)m =dt ∫∫⎨⎬⎜⎟∑ m ⎨⎬ 22ππ00⎩⎭sin(nt ) ⎝⎠m=1 ⎩⎭sin(nt )

∞ 2δπ n m 1 ⎛⎞⎧⎫cos(nt ) ⎧δ z ⎫ +=2c∑ zm∫ ⎜⎟⎨⎬os(t)dt ⎨⎬. m=1 2π 0 ⎝⎠⎩⎭sin(nt ) ⎩⎭0

The integrals given in (9a) follow in a similar manner.

11 Proof of Theorem. The proof of Theorem rests on the above Lemma. From (9a) we obtain the formula

1 δ 2zt sin(2π ) zktk = sin( 2πδ ) dt = 1, 1/2; k = 1, 2, ... , . ∫ 2 δπ0 1−+ 2ztz cos(2 )

Dividing the formula by ks and summing over k, we have

∞∞zktztk 1sin(2)δ ππ 2sin(2) =>dt Re s 1, ∑∑ss∫ 2 kk==11kkztzδπ0 1−+ 2 cos(2 )

so that

∞∞zktztk 1δ sin( 2ππ ) 2 sin(2 ) =>dt Re s 1. (13) ∑∑ss∫ 2 kk==11kkztzδπ0 1−+ 2 cos(2 )

Now, the required integral formula in (6a) directly follows from the last expression in

light of the definitions of Lis () z in (1) and Sxs () in (4a). It should be noted that inverting the order of summation and integration on the right-side of (13) is justified by of the series involved.

Starting from (9b) the formula in (6b) is derived in precisely the same way. In order to prove (6c) note that

1− z2 2cos(2)z ( πtz− ) =+1 , 12cos(2)−+−+z ππtz22 12cos(2) z tz

thus we have

12 δδ1− z2 2ztz( cos(2π ) − ) Ct(2ππ ) dtCt= (2 ) dt, ∫∫ss22 001−+ 2ztz cos(2ππ ) 1 −+ 2ztz cos(2 ) since

δδ∞ 1 C(2ππ t ) dt == cos 2k t dt 0. ∫∫s ∑ s () 00k =1 k

This proves the theorem.

Proof of Corollary 1. We need the Fourier expansions for the Bernoulli polynomials Bn ()x

[7, p. 805, Eqs. 23.1.17]

2(2nk− 1)!∞ sin(2π x ) Bx() (1)n 21n− =− 21nn−−∑ 21 (2π ) k=1 k

where 0≤≤x 1 for n = 2, 3, …, 0 < x < 1 for n = 1, and [7, p. 805, Eqs. 23.1.18]

2(2nk )!∞ cos(2π x ) Bx() (1)n−1 2n =− 22nn∑ (2π ) k =1 k

where 0≤≤x 1 for n = 1, 2, … . These Fourier expansions can be rewritten as follows

(2π )21n− Sx()=− (1)n Bx(), 21nn−−2(2n − 1)! 21

(2π )2n Cx()=− (1)n−1 Bx (), 22nn2(2n )!

in terms of the functions Sxs () and Cxs () defined in (4a) and (4b).

Finally, the integral formulae for n-th polylogarithms proposed in (7) are obtained from the above expansions in conjunction with Theorem.

13 Proof of Corollary 2. First, note that for z = 1 and z = -1 the polylogarithm reduces to the

Riemann zeta function

Lis (1)= ζ ( s ) ( Res > 1) , (14a)

1−s Lis (−= 1) (2 − 1)ζ (s ) (s ≠ 1) . (14b)

Secondly, it is not difficult to show that

2sin(2)ztπ lim = cot(πt ) (15a) z→1 12cos(2)−+ztzπ 2

and

2sin(2)ztπ lim = tan(πt ) . (15b) z→−112cos(2)−+ztzπ 2

Finally, we deduce the integral formulae in (8) starting from (7a) and by employing the expressions in (14) and in (15).

14 References

[1] L. C. Maximon, The dilogarithm function for complex argument, Proc. R. Soc. Lond. A

459 (2003) 2807-2819.

[2] A. Erdélyi, W. Magnus, F. Oberhettinger and F. G. Tricomi, Higher transcendental functions, volume I, McGraw-Hill, New York, 1953.

[3] D. Cvijović and J. Klinowski, Integral representations of the Riemann zeta function for odd-integer arguments, J. Comp. App. Math. 142 (2002) 435-439.

[4] L. Lewin, Polylogarithms and Associated Functions, North-Holland, New York, 1981.

[5] A. P. Prudnikov, Yu. A. Brychkov and O. I. Marichev, Integrals and Series, volume III:

More , Gordon and Breach, New York, 1990.

[6] B. C. Berndt, Ramanujan's Notebooks, Part I, Springer-Verlag, New York, 1985.

[7] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions with Formulas,

Graphs, and Mathematical Tables, Dover, New York, 1972.

[8] A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, Integrals and series, volume I:

Elementary Functions, Gordon and Breach, New York, 1986.

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