A Probabilistic Proof for Representations of the Riemann Zeta Function

where N0 = N ∪ {0} and Bn 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


Introduction
The well known Riemann zeta function ζ is defined by which can be continued meromorphically to the whole complex s-plane, except for a simple pole at s = 1, see [1][2][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 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][5][6][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 where B n (x) are Bernoulli polynomials defined by the generating function [10] te The Bernoulli numbers B n = B n (0) are well-tabulated (see, for example, [3]): 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): 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 2 .The proofs are based on the characteristic function and the moment generating function of logistic, half-logistic and elliptical symmetric logistic distributions in probability theory and mathematical statistics.

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. and where B n 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) Then the moment generating function (mgf) of X is given by 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.Lemma 2. We assume that random variable X has the standard 1-dimensional elliptically symmetric logistic distribution with pdf where Then the characteristic function of X is given by where ζ is the Riemann zeta function.
Proof.Using the Taylor expansion f in ( 9) can be written as we only need to determine the even-order moments.For m ≥ 1 we get where we have used the fact that For any t ∈ (−∞, ∞), we get the characteristic function of X by performing the following calculations This ends the proof of Lemma 2.
Lemma 3. We assume that random variable X has the standard half-logistic distribution with the pdf Then the mgf of X is given by where ζ is the Riemann zeta function.
Proof.The mean of X is given by Using the expansion we get, for any positive integer n > 1, where we have used the fact lim This completes the proof of Lemma 3.
Using the series expansion (see e.g., [15]) where B 2k is the 2kth Bernoulli numbers, we have from which we deduce that This completes the proof of ( 4).Now we prove (5).Denoted by where f is defined by (7).Taking 2nth and (2n + 1)th derivatives of the two functions with respect to t, we get Note that H(t) = G(t) for any real t, and thus H (n) (t) = G (n) (t) for any real t and any positive integers n.In particular, H which concludes the proof of ( 5).Finally, we prove (6).Using (12) one has Taking the nth derivative of both sides of ( 14) with respect to t and then setting t = 0 yields the desired result.