An integral representation of the Riemann zeta function

Question

I am referring to the equality in equation $3.29$ (page 12) and $4.20$ (page 17) in this paper.

I am unable to recognize where this comes from or what is the general expression for values other than $3$.

I checked at some online reviews like this - http://www.math.utah.edu/~milicic/zeta.pdf but nothing seems to match.

It would be great if someone can help.

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3.29

Use the integral representations of $\zeta(s)\Gamma(s)$ for $s=3$. — Dietrich Burde, Commented Aug 28, 2013 at 21:28
The equation is incorrect as stated; you want $\zeta$, not $\xi$. But be that as it may, asking us where it comes from without telling us what you already know is a really good way to get your question closed. — Steven Landsburg, Commented Aug 28, 2013 at 21:38
"I came across" is a poor introduction. Where did you see this? Context? Of course some (possibly corrected) version can be adduced from known things, etc., but ... — paul garrett, Commented Aug 28, 2013 at 23:27
Why not actually put the equation here in the post? Asking people to go dig through a linked PDF is perhaps asking too much. — Kim Morrison, Commented Sep 3, 2013 at 0:32

Carlo Beenakker · Accepted Answer · 2013-09-03 12:29:33Z

3

you ask for the "general expression" for values other than $q=3$:

$$\int_{0}^{\infty}d\lambda\frac{\lambda^{q/2-1}}{1+e^{2 \pi \sqrt{\lambda}}}=2^{1-2q}(2^q-2)\pi^{-q}\Gamma(q) \zeta(q),\text{ for Re }q>0$$

$$\int_{0}^{\infty}d\lambda\frac{\lambda^{q-1}{\rm coth}(\pi\lambda)}{1+e^{2 \pi \lambda}}=(2\pi)^{-q}\Gamma(q) \zeta(q),\text{ for Re }q>1$$

answered Sep 3, 2013 at 12:20

Carlo Beenakker

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$\begingroup$ Can you give a reference to its derivation? I hope there is a way to relate these identities to other better known representations? $\endgroup$
– user6818
Commented Sep 3, 2013 at 14:50
$\begingroup$ functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/07/01 $\endgroup$
– Carlo Beenakker
Commented Sep 3, 2013 at 14:57
$\begingroup$ Thats just a list of formulas :( Isn't there any first principles way of getting these identities? Like from the power-series definition itself? It looks like an impossible task to just know or even guess these identities when they come up in a calculation! [...i wonder how the authors of this paper saw this while doing this!..] $\endgroup$
– user6818
Commented Sep 3, 2013 at 17:42

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