Integrate \(\int_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x\)

How to evaluate
\begin{equation}
\int\limits_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x
\end{equation}
was a question on Mathematics Stack Exchange. Here is another solution method.

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}

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