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so I'm trying to prove the functional equation of Lerch Zeta, through the Hankel contour and Residue theorem, did the following.

In the article "Note sur la function" by Mr. Mathias Lerch, a complex function is defined:

$$ R(w,x,s)=\sum_{k=0}^{\infty} \frac{e^{2\pi ikx}}{(w+k)^s} $$

Having the integral over Hermann Hankel's contour:

$$ K(w,x,s)=\int_H\frac{z^{s-1}e^{-wx}}{1-e^{2\pi ix-z}}dz $$

The relationship between R and K is as follows:

$$ R(w,x,s)=\frac{e^{-s\pi i}}{2\pi i}\Gamma(1-s)K(w,x,s) $$

But using the residue theorem on K we get:

$$ \frac{ie^{2\pi iwx}}{(2\pi)^{s}}K(w,x,s)= $$

$$ =e^{-\frac{\pi i}{2}(1-s)}\sum_{k=0}^{\infty} \frac{e^{-2\pi ikw}}{(x+k)^{1-s}} + e^{2\pi iw+\frac{\pi i}{2}(1-s)}\sum_{k=0}^{\infty} \frac{e^{2\pi ikw}}{(1-x+k)^{1-s}} $$

And then there's a moment in the article that I don't understand. There is something written in French. You can download the original here:

https://projecteuclid.org/journals/acta-mathematica/volume-11/issue-none/Note-sur-la-fonction-mathfrakKw-x-s--sumlimits_k-/10.1007/BF02612318.full

I don't understand this:

$$ (x+k)^{1-s}=[(x+k)^{1-s}] $$ and $$ (1-x+k)^{1-s}=e^{2\pi i(1-s)}[(1-x+k)^{1-s}] $$

I suspect that it is about the complex argument, Riemann surfaces and some kind of discontinuity, "branch cut". And while I know how it works with contour integration, I don't understand it here.

Why do we multiply by the exponent for the series with 1-x+k but not for the series with x+k? (please, step by step) It is important, because it changes final result which is functional equation.

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    $\begingroup$ Without going through the full paper, the first full paragraph on p. 23 reads, "We've only defined the function ${\mathfrak R}$ for real and positive values of $w$. Denote by $[u^{\sigma}]$ the quantity $e^{\sigma \log u}$, the imaginary part of $\log u$ being between the bounds $(-\pi, \dots, \pi)$, so that the function $[u^{\sigma}]$ is continuous and uniform on the space of points $u$ outside the cut $(-\infty, \dots 0)$, along which it's discontinuous." (Or something like that; my technical vocabulary here is a bit weak.) Short answer: There's a branch cut to define $u^{\sigma}$. $\endgroup$
    – anomaly
    Commented Jun 23 at 13:04
  • $\begingroup$ Even shorter answer: Because of the branch cut, we don't have $[u^{\sigma} u^{\sigma'}] = [u^{\sigma + \sigma'}]$. That discrepancy is the cause of the difference in those two equations. $\endgroup$
    – anomaly
    Commented Jun 23 at 13:07
  • $\begingroup$ Well it is still not clear for me. Why $$ arg(x+k)=0 $$ and $$ arg(1-x+k)=2\pi $$ ? $\endgroup$
    – Nightmare Integral
    Commented Jun 23 at 13:28
  • $\begingroup$ Not sure what the conventions on $x$ are without going through the whole paper, but $x + k$ and $1 - x + k$ have opposite signs on their imaginary parts ($k$ is an integer here), and the discussion at the bottom of p. 20 should give you the necessary argument. $\endgroup$
    – anomaly
    Commented Jun 23 at 13:34
  • $\begingroup$ I understand your way of thinking, but I still expect a less heuristic and more rigorous answer. Maybe someone else will know. $\endgroup$
    – Nightmare Integral
    Commented Jun 24 at 13:17

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