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Given the known identity also known as "Hurwitz formula" or "Functional equation of Hurwitz zeta function", $$ \zeta(1 - s, x) = \frac{\Gamma(s)}{(2\pi)^s} \left( e^{- \frac{\pi i s}{2}} \sum_{n=1}^{\infty} \frac{e^{2\pi inx}}{n^s} + e^{\frac{\pi i s}{2}} \sum_{n=1}^{\infty} \frac{e^{-2\pi inx}}{n^s} \right) $$ (valid for $\Re(s) > 1$ and $0 < x ≤ 1$.) which reduces to Riemann zeta function's functional equation, how does the right-hand side behave when looping around the origin $j$ times, i.e., when $x \mapsto x + j$ for some integer $j$?

The branch cut of the polylogarithm $\sum_{n=1}^{\infty} \frac{e^{\pm 2\pi inx}}{n^s}$ increases as $x$ crosses certain values (as discussed in https://math.stackexchange.com/a/4605477/622884). Specifically, how can the domain of $x$ be extended beyond the usual constraint $0 < \Re(x) \leq 1$?

Dr Potato
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  • Formula (3) at Wolfram MathWorld: Hurwitz Zeta Function indicates $$\zeta(s,a)=\frac{1}{s-1} \sum\limits_{n=0}^{\infty} \frac{1}{n+1} \sum\limits_{k=0}^n (-1)^k \binom{n}{k} (a+k)^{1-s},,\quad a>-1$$ gives an analytic continuation to the entire complex plane except $s=1$. – Steven Clark Oct 07 '24 at 23:09
  • I do not catch how is this related to the Hurwitz formula $\zeta(1-s,a)=\frac{\Gamma(s)}{(2\pi)^s} ( \sum \cdot + \sum \cdot ) $ extending the domain of the variable $a$ outside the strip mentioned in the question. Indeed Hurwitz zeta function can be extended to broader sets of complex numers either in the $s$ or in the $a$ complex planes, but the question is about the Hurwitz formula precisely, which reduces to the relation between polylogarithm and Bernoulli polynomials, described in a linked question. – Dr Potato Oct 08 '24 at 00:30
  • I believe the Hurwitz formula is valid for $0<x\le 1$ (which you initially indicated), not $0<\Re(x)\le 1$ (which you later indicated). The formula I gave extends the Hurwitz zeta function to $a>-1$ as well as $s\in\mathbb{C}\land s\ne 1$. – Steven Clark Oct 08 '24 at 17:01
  • I don't see how you'll make any progress with the Hurwitz formula since $$\frac{\Gamma(s)}{(2 \pi )^s} \left(e^{\frac{\pi i s}{2}} \sum\limits_{n=1}^{\infty} \frac{e^{-2 \pi i n x}}{n^s}+e^{-\frac{\pi i s}{2}} \sum\limits_{n=1}^{\infty} \frac{e^{2 \pi i n x}}{n^s}\right)=\frac{\Gamma(s)}{(2 \pi )^s} \left(e^{\frac{\pi i s}{2}} \text{Li}_s\left(e^{-2 i \pi x}\right)+e^{-\frac{\pi i s}{2}} \text{Li}_s\left(e^{2 i \pi x}\right)\right)$$ and $$e^{-2 i \pi x}=e^{-2 i \pi (x \bmod 1)}\land e^{2 i \pi x}=e^{2 i \pi (x \bmod 1)},\quad x\in\mathbb{R}.$$ – Steven Clark Oct 08 '24 at 17:01
  • Thank you for your clarification. For $0<s$ naturals, as an example, if the parameter $z$ of $Li_s(z)$ verifies $1<|z|$, then every loop around the origin of $z=e^{2\pi i x}$ crosses the branch cut of the polylogarithm (in particular for $s=1$ it is the actual logarithm $Li_1(z)=\log(1-z)$, a multivalued function) which lieas along $[1,\infty)$ in the $z$-complex plane, and although $e^{2\pi i x}=e^{2\pi i (x \mod 1)}$ as you pointed out correctly, parametrizing curves crossing the branch cut throws multiple values for the same domain point for such multivalued functions :) – Dr Potato Oct 08 '24 at 17:48
  • ....for the $0<\Re(x)\leq 1$ instead of $0<x\leq 1$ part, let me quote the special case when $s$ is a natural number:

    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 $Im(x) < 0$. This was taken from the Wikipedia site on the Polylogarithm.

     – Dr Potato Oct 08 '24 at 21:21

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