Connection between the polylogarithm and the Bernoulli polynomials.

Question

I have been studying the polylogarithm function and came across its relation with Bernoulli polynomials, as Wikipedia site asserts:

For positive integer polylogarithm orders $s$, the Hurwitz zeta function $\zeta(1−s, x)$ reduces to Bernoulli polynomials, $\zeta(1−n, x) = −B_n(x) / n$, and Jonquière's inversion formula for $n = 1, 2, 3, …$ becomes: $$Li_n(e^{2\pi ix}) + (-1)^n Li_n(e^{-2\pi ix}) = -\frac{(2\pi i)^n}{n!} B_n(x)$$ where again $0 \leq Re(x) < 1$ if $Im(x) \geq 0$, and $0 < Re(x) \leq 1$ if $Im(x) < 0$.

While I understand the formula itself and that the Hurwitz zeta function reduces to the Bernoulli polynomials for natural orders $s$, I am having trouble understanding how to obtain it. I have referred to the following sources:

Wikipedia - Polylogarithm

MathWorld - Jonquière's Relation

However, I still do not understand how to prove this formula. Could someone please explain, in detail, how is it obtained? Any insights or references to textbooks/research papers that provide a detailed derivation would be greatly appreciated.

Answer 1

See the Wikipedia page Hurwitz zeta function, which states the result under the name "Hurwitz's formula" as: $$ \zeta(1-s,a) = \frac{\Gamma(s)}{(2\pi)^s} \left( e^{-\pi i s/2} \sum_{n=1}^\infty \frac{e^{2\pi ina}}{n^s} + e^{\pi i s/2} \sum_{n=1}^\infty \frac{e^{-2\pi ina}}{n^s} \right) $$ This becomes the formula stated in the question, after taking $s$ to be a positive integer, renaming variables, and rearranging.

The page cites references to a number of books and articles where you can take your pick among several possible proofs:

Hurwitz's formula has a variety of different proofs. [See the references in Section 4 of Kanemitsu et al. 2017.] One proof uses the contour integration representation along with the residue theorem. [Apostol, Introduction to analytic number theory, Theorem 12.6][Whittaker and Watson, A Course Of Modern Analysis 4th ed., Section 13.15] A second proof uses a theta function identity, or equivalently Poisson summation. [Fine 1951] These proofs are analogous to the two proofs of the functional equation for the Riemann zeta function in Riemann's 1859 paper. Another proof of the Hurwitz formula uses Euler–Maclaurin summation to express the Hurwitz zeta function as an integral $$ \zeta(s,a) = s \int_{-a}^\infty \frac{\lfloor x \rfloor - x + \frac{1}{2}}{(x+a)^{s+1}} dx $$ ($−1 < Re(s) < 0$ and $0 < a \le 1$) and then expanding the numerator as a Fourier series. [Berndt 1972]

Answer 2

This can be obtained by contour integration. Here's the sketch. Consider $x \in [0, 1)$ and the following function: $$f(z) = \frac{\pi \csc(\pi z) e^{(2x-1)\pi i z}}{z^{n}}$$ By Cauchy theorem on a circle $\gamma_{N+\frac{1}{2}}$ and since $\csc(\pi z)$ has a first order pole in every integer point we have: $$\oint_{\gamma_{N+\frac{1}{2}}}f(z)dz = \sum\limits_{\substack{k=-N\\k\neq0}}^N \underset{z=k}{\rm Res}f(z) + \underset{z=0}{\rm Res}f(z)$$ It turns out that $\pi \csc{(\pi z)} e^{(2x-1)\pi i z}$ is bounded and $\frac{1}{z^{n}}\underset{N\to\infty}{\rightarrow} 0$ on this circle, what yields that $$\oint_{\gamma_{N+\frac{1}{2}}}f(z)dz \underset{N\to\infty}{\rightarrow} 0$$ It's not hard to see that for $k \in \mathbb{Z} \ (k \neq 0)$ $$\underset{z=k}{\rm Res}\frac{\pi \csc{(\pi z)} e^{(2x-1)\pi i z}}{z^{n}} = \lim\limits_{z\to k} \frac{\pi (z-k)e^{(2x-1)\pi i z}}{\sin{(\pi z)}z^{n}} = \frac{(-1)^k e^{(2x-1)\pi i k}}{k^{n}} = \frac{e^{2\pi i x k}}{k^n}$$ What means that $$\sum\limits_{\substack{k=-\infty\\{k \neq 0}}}^\infty \frac{e^{2\pi i x k}}{k^n} = -\underset{z=0}{\rm Res}\frac{\pi \csc(\pi z) e^{(2x-1)\pi i z}}{z^{n}}$$

Now we notice that $$\sum\limits_{\substack{k=-\infty\\{k \neq 0}}}^\infty \frac{e^{2\pi i x k}}{k^{n}} = \sum\limits_{k=1}^\infty \frac{e^{2\pi i x k}}{k^{n}} + (-1)^n\sum\limits_{k=1}^\infty \frac{e^{-2\pi i x k}}{k^{n}} = {\rm Li}_{n}(e^{2\pi i x}) + (-1)^n{\rm Li}_{n}(e^{-2\pi i x})$$ So we have $${\rm Li}_{n}(e^{2\pi i x}) + (-1)^n {\rm Li}_{n}(e^{-2\pi i x}) = -\underset{z=0}{\rm Res}\frac{\pi \csc(\pi z) e^{(2x-1)\pi i z}}{z^{n}}$$ Now we notice that $\frac{\pi \csc(\pi z) e^{(2x-1)\pi i z}}{z^{n}}$ has a pole of order $n+1$ in $z = 0$, so the residue is equal to the n-th coefficient in Taylor expansion of $\pi z \csc(\pi z) e^{(2x-1)\pi i z}$. That's where Bernoulli polynomials come from. By defenition $$\frac{te^{xt}}{e^t-1} = \sum\limits_{k=0}^\infty B_k(x)\frac{t^k}{k!}$$ Substituting $2\pi iz$ for $t$ we have $$\pi z\csc{(\pi z)}e^{(2x-1)\pi i z} = \sum\limits_{k=0}^\infty\frac{(2\pi i)^kB_{k}(x)}{k!}z^{k}$$ And there we have it! $${\rm Li}_{n}(e^{2\pi i x}) + (-1)^n {\rm Li}_{n}(e^{-2\pi i x}) = -\frac{(2\pi i)^nB_n(x)}{n!}$$ Now we can extend this result for non real values of x. Since ${\rm Li}_{n}(e^{2\pi i x})$ is analytical on a strip $\Re(x) \in (0, 1)$ and the identity is proved for $x \in (0, 1)$ by the Identity Theorem we can extend it for $\Re(x) \in (0, 1)$.