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.

(Images added by J.O'Rourke)

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    $\begingroup$ Where did you come across this? $\endgroup$ – Steven Landsburg Aug 28 '13 at 20:51
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    $\begingroup$ Use the integral representations of $\zeta(s)\Gamma(s)$ for $s=3$. $\endgroup$ – Dietrich Burde Aug 28 '13 at 21:28
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    $\begingroup$ 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. $\endgroup$ – Steven Landsburg Aug 28 '13 at 21:38
  • $\begingroup$ "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 ... $\endgroup$ – paul garrett Aug 28 '13 at 23:27
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    $\begingroup$ Why not actually put the equation here in the post? Asking people to go dig through a linked PDF is perhaps asking too much. $\endgroup$ – Scott Morrison Sep 3 '13 at 0:32

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

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

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

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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 Sep 3 '13 at 14:50
  • $\begingroup$ functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/07/01 $\endgroup$ – Carlo Beenakker Sep 3 '13 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 Sep 3 '13 at 17:42

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