mathematics Article A Probabilistic Proof for Representations of the Riemann Zeta Function Jiamei Liu, Yuxia Huang and Chuancun Yin * School of Statistics, Qufu Normal University, Shandong 273165, China; [email protected] (J.L.); [email protected] (Y.H.) * Correspondence: [email protected] Received: 4 April 2019; Accepted: 19 April 2019; Published: 24 April 2019 Abstract: In this paper, we present a different proof of the well known recurrence formula for the Riemann zeta function at positive even integers, the integral representations of the Riemann zeta function at positive integers and at fractional points by means of a probabilistic approach. Keywords: Bernoulli numbers; (half-) logistic distribution; integral representation; probabilistic approach; Riemann zeta function 1. Introduction The well known Riemann zeta function ζ is defined by 8 P 1 > s , if (s) > 1, <> n1=1 n ζ(s) = n+1 < > 1 P ( 1) : − s , if (s) > 0, s 1, 1 21 s n1=1 n , − − < which can be continued meromorphically to the whole complex s-plane, except for a simple pole at s = 1, see [1–3] for details. Finding recurrence formulas and integral representations of the zeta function zeta function has become an important issue in complex analysis and number theory. One of the famous formulas is the following recursion formula for positive even integers 2n 1 n+1 2 − 2n ζ(2n) = ( 1) π B n, n N , (1) − (2n)! 2 2 0 where N = N 0 and Bn is the nth Bernoulli number. Here N is the set of positive integers. Several 0 [ f g new proofs to (1) can be found in [4–7]. A new parameterized series representation of zeta function is derived in [8]. However, no similar closed-form representation of ζ(s) at odd integers or fractional points can be found in literature. The Riemann zeta function for positive odd integer arguments can be expressed by series and integrals. One possible integral expression is established by [9] as follows 2n+1 Z 1 n+1 (2π) ζ(2n + 1) = ( 1) B2n+1(u) cot(πu)du , n N , (2) − 2(2n + 1)! 0 2 where Bn(x) are Bernoulli polynomials defined by the generating function [10] tetx X1 tn = Bn(x) , t < 2π . et 1 n! j j − n=0 Mathematics 2019, 7, 369; doi:10.3390/math7040369 www.mdpi.com/journal/mathematics Mathematics 2019, 7, 369 2 of 7 The Bernoulli numbers Bn = Bn(0) are well-tabulated (see, for example, [3]): 1 1 1 1 B = 1, B = , B = , B = , B = , B + = 0 (n = 1, 2, ), . 0 1 −2 2 6 4 −30 6 42 2n 1 ··· ··· More lists of Bernoulli numbers and their estimation can be found in the recent work by Qi [11]. The zeta function ζ(s) has many integral representations, one of which is the following [12] (P.172) (note that there is an extra 2 in (51) of [12] (P.172): 1 s 1 Z s t (1 2 − )− 1 t e ζ(s) = − dt , (s) > 0 . (3) 2 G(s + 1) 0 (et + 1) < The aim of this note is to present a new proof of (1) for ζ(2n) and deduce the integral representations for ζ(n) and ζ n 1 . The proofs are based on the characteristic function and the moment generating − 2 function of logistic, half-logistic and elliptical symmetric logistic distributions in probability theory and mathematical statistics. 2. The Main Results and Their Proofs In this section we present a new proof to the following results by using a probabilistic method. To the best of our knowledge, the result (5) is new. Proposition 1. For Riemann’s zeta function ζ, we have 2n 1 n 1 2 − 2n ζ(2n) = ( 1) − π B n , n N , (4) − (2n)! 2 2 0 n 1 R (ln y) − 2 2n 1 dy 1 1 (1+y)2 = ζ n 2n 3 , n N , (5) − 2 − 2 pπ(2n 1)!! 1 2− 2 − − and 1 n 1 Z n x (1 2 − )− 1 x e ζ(n) = − − dx , n N, n > 1 , (6) n! x 2 2 0 (1 + e− ) where Bn is the nth Bernoulli number. To prove the proposition, we need the following three lemmas. Lemma 1. We assume that random variable X has the standard logistic distribution with the probability density function (pdf) exp( x) f (x) = − , < x < . (7) (1 + exp( x))2 −∞ 1 − Then the moment generating function (mgf) of X is given by πt E[exp(tX)] = , t < 1 . (8) sin(πt) j j Proof. See Johnson et al. [13] (Equation (23.10)). As far as we are aware, the formulas for characteristic and moment generating functions given in the following two lemmas are new. Mathematics 2019, 7, 369 3 of 7 Lemma 2. We assume that random variable X has the standard 1-dimensional elliptically symmetric logistic distribution with pdf exp( x2) f (x) = c − , < x < , (9) (1 + exp( x2))2 −∞ 1 − where 0Z t 1 1 B 1 1 e− C− c = B t− 2 dtC . @B t 2 AC 0 (1 + e− ) Then the characteristic function of X is given by 2n X1 n c pπ t 2n 3 1 E[exp(itX)] = 1 + ( 1) 1 2− 2− ζ n , (10) − 22n+1 n! − − 2 n=1 where ζ is the Riemann zeta function. Proof. Using the Taylor expansion X 2 2 1 k 1 2 (1 + exp( x ))− = ( 1) − k exp( (k 1)x ) , x , 0 , − − − − k=1 f in (9) can be written as X 1 k 1 2 f (x) = c ( 1) − k exp( kx ) , x , 0 . − − k=1 Noting that f ( x) = f (x), < x < , we only need to determine the even-order moments. − − 1 1 For m 1 we get ≥ R R P k 1 E(X2m) = 1 x2m f (x)dx = c 1 x2m 1 ( ) k ( kx2)dx 2 0 2 0 1 − exp k=1 − − P1 k 1 pπ (2m)! 2m 1 = c ( 1) − k 2− 22m m! − k=1 − pπc (2m)! 2m 3 1 = 1 2− 2− ζ m , 22m m! − − 2 where we have used the fact that Z 1 2 2k pπ 1 3 2k 1 2k+1 exp( bx )x dx = − b− 2 . 0 − 2 2 2 ··· 2 For any t ( , ), we get the characteristic function of X by performing the following calculations 2 −1 1 " # 2n 2n [ ( )] = + P1 ( )n t2nX2n = + P1 ( )n t E(X ) E exp itX E 1 1 (2n)! 1 1 (2n)! n=1 − n=1 − P1 n c pπ t2n 2n 3 1 = 1 + ( 1) 1 2 2− n . 22n n! − ζ 2 n=1 − − − This ends the proof of Lemma 2. Lemma 3. We assume that random variable X has the standard half-logistic distribution with the pdf 2 exp( x) f (x) = − , x > 0. (11) (1 + exp( x))2 − Mathematics 2019, 7, 369 4 of 7 Then the mgf of X is given by X 1 1 n n E[exp(tX)] = 1 + 2 (1 2 − )ζ(n)t , t < 1, (12) − j j n=1 where ζ is the Riemann zeta function. Proof. The mean of X is given by Z x 1 xe E(X) = 2 − dx = 2 ln 2. 2 0 (1 + exp( x)) − Using the expansion P1 k 1 kx f (x) = 2 ( 1) − ke− k=1 − P1 (2k 1)x P1 2kx = 2 (2k 1)e− − 2 2ke− , x > 0, k=1 − − k=1 we get, for any positive integer n > 1, R E(Xn) = 1 xn f (x)dx 2 0 P R P R = 1 ( k ) 1 xne (2k 1)xdx 1 k 1 xne 2kxdx 2 2 1 0 − − 2 2 0 − k=1 − − k=1 P1 1 P1 1 = 2n! n 2n! n (2k 1) (2k) k=1 − − k=1 = 2n!(1 21 n)ζ(n). − − Then we have k 1 k P1 E(X ) P1 2k!ζ(k)(1 2 − ) ( tX) = + k = + + − k E e 1 k! t 1 2t ln 2 k! t k=1 k=2 P1 1 k k = 1 + 2t ln 2 + 2 (1 2 − )ζ(k)t , t < 1, k=2 − j j where we have used the fact lim(s 1)ζ(s) = 1. s 1 − ! This completes the proof of Lemma 3. Proof of Proposition 1. The mgf of the standard logistic distribution can be written as 2n 1 X1 (2 − 1)ζ(2n) E(etX) = 1 + − t2n, (13) 2(n 1) n=1 2 − see, for example, [14]. Comparing (7) and (10) yields g(t) = h(t), t < 1, where j j 2n 1 X1 (2 − 1)ζ(2n) g(t) = 1 + − t2n, 2(n 1) n=1 2 − and πt h(t) = . sin(πt) Mathematics 2019, 7, 369 5 of 7 Using the series expansion (see e.g., [15]) 2k πt X1 k 1 2 2 2k = ( 1) − − B (πt) , t < 1, sin(πt) − (2k)! 2k j j k=0 where B2k is the 2kth Bernoulli numbers, we have X 2n 1 X 2n 1 (2 − 1)ζ(2n) 2n 1 n 1 2 2 2n − t = ( 1) − − B n(πt) , t < 1, 2(n 1) − (2n)! 2 j j n=1 2 − n=1 from which we deduce that 2n 1 n 1 2 − 2n ζ(2n) = ( 1) − π B n. − (2n)! 2 This completes the proof of (4). Now we prove (5). Denoted by Z 1 H(t) = exp(itx) f (x)dx, −∞ and 2k X1 k c pπ t 2k 3 1 G(t) = 1 + ( 1) 1 2− 2− ζ k , − 22k+1 k! − − 2 k=1 where f is defined by (7).