how to calculate $\int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\mathrm{d}x$ I was trying to solve another integral when then I reached this, I've no idea of how to select the contour for the integration.
 A: Sub $x=\log{(1+y^2)}$; then the integral is equal to
$$\int_{-\infty}^{\infty} dy \frac{\log{(1+y^2)}}{1+y^2} $$
I will illustrate how to use complex analysis to evaluate this integral.  Consider the following contour integral:
$$\oint_C dz \frac{\log{(1+z^2)}}{1+z^2} $$
where $C$ is the following contour:

i.e., a semicircular contour of radius $R$ with a detour around the branch point at $z=i$ of radius $\epsilon$.  The contour integral is equal to
$$\int_{-R}^R dx \frac{\log{(1+x^2)}}{1+x^2} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{\log{(1+R^2 e^{i 2 \theta})}}{1+R^2 e^{i 2 \theta}} \\ + i \int_R^{1+\epsilon} dy \frac{\log{(y^2-1)}+i \pi}{1-y^2} + i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \frac{\log{[1+(i+\epsilon e^{i \phi})^2]}}{1+(i+\epsilon e^{i \phi})^2} \\ + i \int_{1+\epsilon}^R dy \frac{\log{(y^2-1)}-i \pi}{1-y^2} + i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{\log{(1+R^2 e^{i 2 \theta})}}{1+R^2 e^{i 2 \theta}}$$
Note that the third and fifth integrals are on opposite sides of the branch cut along the imaginary axis above $z=i$.  Also note the limits on the fourth integral: the upper limit is less than the lower limit because the contour traverses clockwise locally about the branch point $z=i$.
We consider the limits as $R \to \infty$ and $\epsilon \to 0$.  In these limits, the second and sixth integrals vanish.  Rearranging things a bit, we get for the contour integral
$$\int_{-\infty}^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} - i (-i 2 \pi) \int_{1+\epsilon}^{\infty} \frac{dy}{y^2-1} + \frac12 \int_{\pi/2}^{-3 \pi/2} d\phi \, \left [\log{(i 2 \epsilon)} + i \phi \right ] $$
Note that, while there appears to be singular behavior as $\epsilon \to 0$, that singular behavior will cancel out as we will see.
By Cauchy's theorem, the contour integral is zero.  Doing out the second and third integrals, we find that
$$\int_{-\infty}^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} - \pi \left [\log{\left (\frac{y-1}{y+1} \right )} \right ]_{1+\epsilon}^{\infty} - \pi \log{(i 2 \epsilon)} + i \frac14 (2 \pi^2) = 0$$
Simplifying, and taking $\log{i} = i \pi/2$, we get
$$\int_{-\infty}^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} + \pi \log{\epsilon} - \pi \log{2} - i \frac{\pi^2}{2} - \pi \log{2} - \pi \log{\epsilon} + i \frac{\pi^2}{2} = 0$$
Thus...

$$\int_0^{\infty} dx \frac{x}{\sqrt{e^x-1}} = 2 \pi \log{2} $$

A: Let $t^2= e^x-1$. We have
$$2tdt = e^xdx = (1+t^2)dx \implies dx = \dfrac{2tdt}{1+t^2}$$
Hence, we have
$$I = \int_0^{\infty} \dfrac{xdx}{\sqrt{e^x-1}} = \int_0^{\infty} \dfrac{2t \log(1+t^2)dt}{(1+t^2)t} = 2\int_0^{\infty} \dfrac{\log(1+t^2)}{(1+t^2)}dt$$
Let
$$I(a) = \int_0^{\infty} \dfrac{\log(1+a^2t^2)}{1+t^2}dt \,\,\, (\clubsuit)$$
We need $2I(1)$. Differentiating $(\clubsuit)$, we obtain
$$I'(a) = \int_0^{\infty} \dfrac{2at^2}{(1+a^2t^2)(1+t^2)}dt = \dfrac{2a}{a^2-1}\left(\int_0^{\infty} \dfrac{dt}{1+t^2} - \int_0^{\infty} \dfrac{dt}{1+a^2t^2} \right)$$
Hence,
$$I'(a) = \dfrac{2a}{a^2-1}\left(\dfrac{\pi}2 - \dfrac{\pi}{2a}\right) = \dfrac{\pi}{(1+a)} \,\,\, (\spadesuit)$$
Further, we have $I(0) = 0$. Hence, integrating $(\spadesuit)$, we obtain
$$I(a) = \pi \log(1+a)$$
The desired integral is $2I(1) = \pi \log(2)$.
A: Let $z=\mathrm{e}^{x}-1$, so that we have
\begin{equation}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}(z+1)}{\sqrt{z}}\frac{1}{z+1} \mathrm{d} z
\end{equation}
Let us consider
\begin{equation}
I(a) = \int\limits_{0}^{\infty} \frac{(z+1)^{a}}{\sqrt{z}} \mathrm{d} z = \mathrm{B}\left(\frac{1}{2}, -\frac{1}{2}-a\right)
= \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(-\frac{1}{2}-a\right)}{\Gamma(-a)}
\end{equation}
so that
\begin{equation}
\lim_{a \to -1} \frac{\partial I(a)}{\partial a} = \int\limits_{0}^{\infty} \frac{\mathrm{ln}(z+1)}{\sqrt{z}}\frac{1}{z+1} \mathrm{d} z =
\int\limits_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x
\end{equation}
Then,
\begin{equation}
\frac{\partial I(a)}{\partial a} = \Gamma\left(\frac{1}{2}\right)\left[\frac{-\Gamma(-a)\Gamma\left(-\frac{1}{2}-a\right)\psi^{0}\left(-\frac{1}{2}-a\right) + \Gamma\left(-\frac{1}{2}-a\right)\Gamma(-a)\psi^{0}(-a)}{\Gamma(-a)\Gamma(-a)} \right]
\end{equation}
\begin{align}
\lim_{a \to -1} \frac{\partial I(a)}{\partial a} & = -\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}
\left[\psi^{0}\left(\frac{1}{2}\right) - \psi^{0}(1)\right] \\
& = -\pi[(-\gamma-\mathrm{ln}4) -(- \gamma)] \\
& = \pi\mathrm{ln}4 \\
& = \int\limits_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{t \equiv \expo{x} - 1 \implies x = \ln\pars{1 + t}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{x \over \root{\expo{x} - 1}}\,\dd x} =
\int_{0}^{\infty}
t^{-1/2}\,\,\,{\ln\pars{1 + t} \over 1 + t}\,\dd t
\\[5mm] = &\
\int_{0}^{\infty}
t^{\color{red}{1/2} - 1}\bracks{%
-\sum_{k = 0}^{\infty}H_{k}\,\pars{-t}^{k}}\,\dd t
\\[5mm] = &\
-\int_{0}^{\infty}
t^{\color{red}{1/2} - 1}
\bracks{%
\sum_{k = 0}^{\infty}
\color{red}{H_{k}\,\Gamma\pars{1 + k}}
{\pars{-t}^{k} \over k!}}\dd t
\end{align}
With the Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{x \over \root{\expo{x} - 1}}\,\dd x}
\\[5mm] = &\
-\ \underbrace{\Gamma\pars{\color{red}{1 \over 2}}}
_{\ds{\root{\pi}}}\
\underbrace{\quad H_{\color{red}{-1/2}}\quad}
_{\ds{\int_{0}^{1}{1- t^{\color{red}{-1/2}} \over 1 - t}
\,\dd t}}\
\Gamma\pars{1 \color{red}{- {1 \over 2}}}
\\[5mm] = &\
-\pi\int_{0}^{1}{1- t^{-1} \over 1 - t^{2}}\,2t\,\dd t =
2\pi\int_{0}^{1}{\dd t \over 1 + t}
\\[5mm] = &\
\bbx{2\pi\ln\pars{2}} \approx 4.3552 \\ &
\end{align}
