$$I = \int_{0}^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}\,dx$$.
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.
$$I = \int_0^{\infty}\frac{x}{\sqrt{e^{2\pi\sqrt x}-1}}dx$$
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 |
+---------+--------+
$$\begin{align*} f(s) & = \int_0^1 \frac{u^{s-1}}{\sqrt{\frac 1{u}-1}}du\\ & = \int_0^1 u^{s-\frac 1{2}}(1-u)^{-\frac1{2}}du\\ & = \beta \left(s+\frac 1{2}, \frac 1{2} \right)\\ & = \frac{\Gamma(s+\frac 1{2})\Gamma(\frac 1{2})}{\Gamma(s+1)}\\ & \implies \color{red}{f_1(s)} = \color{blue}{f(s)}\color{green}{\left(\psi_0\left(s+\frac 1{2}\right) - \psi_0(s+1)\right)} = \color{blue}{f}\color{green}{\times g}\\ \end{align*}$$
$$\begin{align*}f_1 = & f\times g\\ & \implies f_2 = f_1g + fg_1 = (fg)g+fg_1 = fg^2+fg_1\\ & \implies f_3 = f_1g^2+2fgg_1+f_1g_1+fg_2\\ & \implies f_3(s) = fg^3 + 3fgg_1 + fg_2 = \color{blue}{f(s)\left[g^3(s)+3g(s)g_1(s) + g_2(s)\right] = f_3(s)}\text{.................#2}\\ \end{align*}$$
let's find zeros of function $$\color{green}{f(s=0)} = \beta\left(\frac 1{2}, \frac 1{2}\right) = \frac {\sqrt\pi\sqrt\pi}{1} = \color{green}{\pi}$$
$$\begin{align*} \color{green}{g(s)} & = \left(\psi_0(s+ \frac 1{2}) - \psi_0(s+1)\right) \\ &\implies g(s=0) = \left(\color{red}{\psi_0(\frac 1{2})} - \psi(1)\right) \\ &= (\color{red}{(-\gamma-2\ln2)}+\gamma) = \color{green}{-2\ln(2)}\\ \end{align*}$$
$g_1(0)$ $$\begin{align*} \color{green}{g_1(s)} &= \left(\psi_1(s+ \frac 1{2}) - \psi_1(s+1)\right)\\ &\implies g_1(0) = \left(\color{red}{\psi_1(\frac 1{2})} - \psi_1(1)\right)\\ & = \left(\color{red}{\frac{\pi^2}{2}}-\frac{\pi^2}{6}\right) = \color{green}{\frac{\pi^2}{3}}\\ \end{align*}$$
$g_2(0)$
$$\begin{align*} \color{green}{g_2(s)} = &\left(\psi_2(s+ \frac 1{2}) - \psi_2(s+1)\right)\\ &\implies g_2(0) = \left(\color{red}{\psi_2( \frac 1{2})} - \psi_2(1)\right)\\ & = \color{red}{-14\zeta(3)} + 2\zeta(3) = \color{green}{-12\zeta(3)} \end{align*}$$
From above $$\begin{align*}I = & -\frac 1{8\pi^4} f_3(0) = -\frac 1{8\pi^4}\left(\pi\left((-2\ln 2)^3 + 3(-2\ln 2)\times \frac{\pi^2}{3} - 12\zeta(3)\right)\right)\\ & = \frac {\ln^3(2)}{\pi^3} + \frac{\ln(2)}{4\pi} + \frac {3\zeta(3)}{2\pi^3} \\ \end{align*}$$
Too long for a comment.
If we consider the problem of $$I_n=\int \frac{t^n}{\sqrt{e^t-1}}dt$$ for $n>1$ they all write as $$I_n=-2 t^n \sin ^{-1}\left(e^{-t/2}\right)-4ne^{-t/2} J_n$$ where $J_n$ are expressed in terms of generalized hypergeometric functions.
For example $$J_2=t \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};e^{-t} \right)+2 \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{ 2},\frac{3}{2};e^{-t}\right)$$ $$J_3=t^2 \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};e^{-t} \right)+4 t \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{ 2},\frac{3}{2};e^{-t}\right)+$$ $$8 \, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{ 2},\frac{3}{2},\frac{3}{2},\frac{3}{2};e^{-t}\right)$$ $$J_4=t^3 \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};e^{-t} \right)+6 t^2 \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{ 2},\frac{3}{2};e^{-t}\right)+$$ $$24 t \, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{ 2},\frac{3}{2},\frac{3}{2},\frac{3}{2};e^{-t}\right)+48 \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{ 2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};e^{-t}\right)$$
where we can notice simple patterns.
Now, for the definite integrals $$K_n=\frac 1 \pi\int_0^\infty \frac{t^n}{\sqrt{e^t-1}}dt$$ the term $2 t^n \sin ^{-1}\left(e^{-t/2}\right)$ disappears and we have for the first ones $$K_2=\frac{\pi ^2}{3}+4 \log ^2(2)$$ $$K_3=2 \pi ^2 \log (2)+8 \log ^3(2)+12 \zeta (3)$$ $$K_4=8 \pi ^2 \log ^2(2) +16 \log ^4(2)+96 \zeta (3) \log (2)+114 \zeta(4) $$
Let $\;x=\dfrac{y^2}{\pi^2},\;$ then
$$I=\int\limits_0^\infty\dfrac{x\,\text dx}{\sqrt{e^{2\pi\sqrt x}-1}}
=\dfrac2{\pi^4}\int\limits_0^\infty\dfrac{y^3\,\text dy}{\sqrt{e^{2y}-1}}
=\dfrac2{\pi^4}\int\limits_0^\infty\dfrac{y^3 e^{-y}\,\text dy}{\sqrt{1-e^{-2y}}}
=-\dfrac2{\pi^4}\int\limits_0^\infty y^3\,\text d\arcsin e^{-y}$$
$$\overset{\text{IBP}}{=\!=}\;\dfrac6{\pi^4}\int\limits_0^\infty y^2 \arcsin e^{-y}\,\text dy
=\dfrac6{\pi^4}\int\limits_0^1 \ln^2 z\, \dfrac{\arcsin z}z\,\text dz
=\dfrac{1}{8\pi^3}\big(12\zeta(3)+\ln^3 4\ +\pi^2\ln4\big).$$
