TITLE: How to evaluate the integral $I = \int_o^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}dx$?
QUESTION [8 upvotes]: $$I = \int_{0}^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}\,dx$$.
I tried to solve by substituting $t = 2\pi\sqrt x \implies t^2 = 4{\pi}^2x \implies tdt = 2\pi^2dx$
$$I = \frac1{8\pi^4}\int_0^{\infty}\frac{t^3}{\sqrt{e^t-1}}dt.$$
But now I'm unable to solve from here.

REPLY [3 votes]: Let $\;x=\dfrac{y^2}{\pi^2},\;$ then
$$I=\int\limits_0^\infty\dfrac{x\,\text dx}{\sqrt{e^{2\pi\sqrt x}-1}}
=\dfrac2{\pi^4}\int\limits_0^\infty\dfrac{y^3\,\text dy}{\sqrt{e^{2y}-1}}
=\dfrac2{\pi^4}\int\limits_0^\infty\dfrac{y^3 e^{-y}\,\text dy}{\sqrt{1-e^{-2y}}}
=-\dfrac2{\pi^4}\int\limits_0^\infty y^3\,\text d\arcsin e^{-y}$$
$$\overset{\text{IBP}}{=\!=}\;\dfrac6{\pi^4}\int\limits_0^\infty y^2 \arcsin e^{-y}\,\text dy
=\dfrac6{\pi^4}\int\limits_0^1 \ln^2 z\, \dfrac{\arcsin z}z\,\text dz
=\dfrac{1}{8\pi^3}\big(12\zeta(3)+\ln^3 4\ +\pi^2\ln4\big).$$