How to evaluate the integral $I = \int_o^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}dx$? $$I = \int_{0}^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}\,dx$$.
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.
 A: Let $\;x=\dfrac{y^2}{\pi^2},\;$ then
$$I=\int\limits_0^\infty\dfrac{x\,\text dx}{\sqrt{e^{2\pi\sqrt x}-1}}
=\dfrac2{\pi^4}\int\limits_0^\infty\dfrac{y^3\,\text dy}{\sqrt{e^{2y}-1}}
=\dfrac2{\pi^4}\int\limits_0^\infty\dfrac{y^3 e^{-y}\,\text dy}{\sqrt{1-e^{-2y}}}
=-\dfrac2{\pi^4}\int\limits_0^\infty y^3\,\text d\arcsin e^{-y}$$
$$\overset{\text{IBP}}{=\!=}\;\dfrac6{\pi^4}\int\limits_0^\infty y^2 \arcsin e^{-y}\,\text dy
=\dfrac6{\pi^4}\int\limits_0^1 \ln^2 z\, \dfrac{\arcsin z}z\,\text dz
=\dfrac{1}{8\pi^3}\big(12\zeta(3)+\ln^3 4\ +\pi^2\ln4\big).$$

A: $$I =  \int_0^{\infty}\frac{x}{\sqrt{e^{2\pi\sqrt x}-1}}dx$$
Let
$$x = \frac{t^2}{4\pi^2} \implies dx = \frac{tdt}{2\pi^2}$$
Thus,
$$I =\frac1{8\pi^4} \int_0^{\infty}\frac{t^3}{\sqrt{e^t-1}}dt$$
Let's consider a more general integral.
$$f(s) = \int_0^{\infty} \frac{e^{-st}}{\sqrt{e^t-1}}dt$$
$$\frac {df(s)}{ds^3}=f_3(s) = -\int_0^{\infty}\frac{t^3e^{-st}}{\sqrt{e^t-1}}dt$$
$$\color{blue}{I = -\frac 1{8\pi^4}\left(\frac {df(s)}{ds^3}\right)_{s=0}}\text{...................#1}$$
Now,
$$u =e^{-t} \implies \ln u = -t \implies -\frac{du}{u} = dt$$
              +---------+--------+
       t -->> |    0    |   ∞    |
              +---------+--------+
       u -->> |    1    |   0    |
              +---------+--------+

$$\begin{align*}
f(s) 
& = \int_0^1 \frac{u^{s-1}}{\sqrt{\frac 1{u}-1}}du\\
& = \int_0^1 u^{s-\frac 1{2}}(1-u)^{-\frac1{2}}du\\
& = \beta \left(s+\frac 1{2}, \frac 1{2} \right)\\
& = \frac{\Gamma(s+\frac 1{2})\Gamma(\frac 1{2})}{\Gamma(s+1)}\\
& \implies \color{red}{f_1(s)} = \color{blue}{f(s)}\color{green}{\left(\psi_0\left(s+\frac 1{2}\right) - \psi_0(s+1)\right)} = \color{blue}{f}\color{green}{\times g}\\
\end{align*}$$

*

*What is $\psi$-function

*Derivative of $\beta$-function
$$\begin{align*}f_1 =
& f\times g\\
& \implies f_2 = f_1g + fg_1 = (fg)g+fg_1 = fg^2+fg_1\\
& \implies f_3 = f_1g^2+2fgg_1+f_1g_1+fg_2\\ 
& \implies f_3(s) = fg^3 + 3fgg_1 + fg_2 = \color{blue}{f(s)\left[g^3(s)+3g(s)g_1(s) + g_2(s)\right] = f_3(s)}\text{.................#2}\\
\end{align*}$$
let's find zeros of function
$$\color{green}{f(s=0)} = \beta\left(\frac 1{2}, \frac 1{2}\right) = \frac {\sqrt\pi\sqrt\pi}{1} = \color{green}{\pi}$$

*

*$g(0)$
$$\begin{align*} 
\color{green}{g(s)}  
& = \left(\psi_0(s+ \frac 1{2}) - \psi_0(s+1)\right) \\
&\implies g(s=0) = \left(\color{red}{\psi_0(\frac 1{2})} - \psi(1)\right) \\
&= (\color{red}{(-\gamma-2\ln2)}+\gamma) = \color{green}{-2\ln(2)}\\
\end{align*}$$

*

*$g_1(0)$
$$\begin{align*} \color{green}{g_1(s)} 
&= \left(\psi_1(s+ \frac 1{2}) - \psi_1(s+1)\right)\\
&\implies g_1(0) = \left(\color{red}{\psi_1(\frac 1{2})} - \psi_1(1)\right)\\
& = \left(\color{red}{\frac{\pi^2}{2}}-\frac{\pi^2}{6}\right) = \color{green}{\frac{\pi^2}{3}}\\
\end{align*}$$


*$g_2(0)$
$$\begin{align*}
\color{green}{g_2(s)} = 
&\left(\psi_2(s+ \frac 1{2}) - \psi_2(s+1)\right)\\
&\implies g_2(0) = \left(\color{red}{\psi_2( \frac 1{2})} - \psi_2(1)\right)\\
& = \color{red}{-14\zeta(3)} + 2\zeta(3) = \color{green}{-12\zeta(3)}
\end{align*}$$
From above
$$\begin{align*}I =
& -\frac 1{8\pi^4} f_3(0) = -\frac 1{8\pi^4}\left(\pi\left((-2\ln 2)^3 + 3(-2\ln 2)\times \frac{\pi^2}{3} 
 - 12\zeta(3)\right)\right)\\
& =  \frac {\ln^3(2)}{\pi^3} + \frac{\ln(2)}{4\pi} + \frac {3\zeta(3)}{2\pi^3}   \\
\end{align*}$$
