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; 18463757687@163.com (J.L.); yxhuang1216@163.com (Y.H.) * Correspondence: ccyin@qfnu.edu.cn 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 ζ(s)= ∑ ∞ n=1 1 n s , if ℜ(s) > 1, 1 1−2 1−s ∑ ∞ n=1 (−1) n+1 n s , if ℜ(s) > 0, s 1, 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 ζ(2n)=(−1) n+1 2 2n−1 (2n)! π 2n B 2n , n ∈ N 0 , (1) where N 0 = N ∪{0} and B n is the nth Bernoulli number. Here N is the set of positive integers. Several 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 ζ(2n + 1)=(−1) n+1 (2π) 2n+1 2(2n + 1)! 1 0 B 2n+1 (u) cot(πu)du , n ∈ N , (2) where B n (x) are Bernoulli polynomials defined by the generating function [10] te tx e t − 1 = ∞ n=0 B n (x) t n n! , |t| < 2π . Mathematics 2019, 7, 369; doi:10.3390/math7040369 www.mdpi.com/journal/mathematics