Prove that $2\int_0^\infty \frac{e^x-x-1}{x(e^{2x}-1)} \, \mathrm{d}x =\ln(\pi)-\gamma $ Let $\gamma$ be the Euler-Mascheroni constant.
I'm trying to prove that

$$2\int_0^\infty \frac{e^x-x-1}{x(e^{2x}-1)} \, \mathrm{d}x =\ln(\pi)-\gamma $$


I tried introducing a parameter to the exponent in the numerator and then differentiating under the integral sign. But doing so seems to result in an integral that doesn't converge.
 A: By Frullani's theorem we have:
$$ \int_{0}^{+\infty}\frac{e^x-1-x}{x}\,e^{-2m x}\,dx = -\frac{1}{2m}+\log\left(\frac{2m}{2m-1}\right)\tag{1}$$
hence it is straightforward to prove the claim by summing $(1)$ over $m\geq 1$, then exploiting:
$$ \gamma = \sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right] \tag{2}$$
that is just the usual definition of the Euler-Mascheroni constant, together with:
$$ \sum_{n\geq 1}(-1)^n \log\left(1+\frac{1}{n}\right) = -\log\left(\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\ldots\right) = -\log\frac{\pi}{2} \tag{3}$$
that follows from Wallis' product.
A: $$ \begin{align} 2\int_{0}^{\infty} \frac{e^x-x-1}{x(e^{2x}-1)} \, dx &= 2 \int_{0}^{\infty} \frac{1}{x(e^{2x}-1)} \sum_{n=2}^{\infty} \frac{x^{n}}{n!}\\ & = 2 \sum_{n=2}^{\infty}\frac{1}{n!} \int_{0}^{\infty} \frac{x^{n-1}}{e^{2x}-1} \, dx \\ &= 2 \sum_{n=2}^{\infty}\frac{1}{n!2^{n}} \int_{0}^{\infty} \frac{u^{n-1}}{e^{u}-1} \, du \\ &= 2 \sum_{n=2}^{\infty} \frac{\zeta(n)}{n 2^{n}} \tag{1} \\ &= 2 \sum_{n=2}^{\infty} \frac{1}{n 2^{n}} \sum_{k=1}^{\infty} \frac{1}{k^{n}} \\ &= 2 \sum_{k=1}^{\infty} \sum_{n=2}^{\infty} \frac{1}{n(2k)^{n}} \\ & =2  \sum_{k=1}^{\infty} \left[ \log \left(\frac{2k}{2k-1} \right) - \frac{1}{2k}\right] \\ & =2 \lim_{N \to \infty} \left(\log \left(\frac{(2N)!!}{(2N-1)!!} \right) - \frac{H_{N}}{2} \right) \\ &= 2 \lim_{N \to \infty} \left(\log \left(\frac{2^{2N}(N!)^2}{(2N)!} \right) - \frac{H_{N}}{2} \right) \tag{2} \\ &= 2 \lim_{N \to \infty} \left(\log \left(\frac{2^{2N}(2\pi N) \left(\frac{N}{e} \right)^{2N}}{\sqrt{2 \pi(2N)} \left(\frac{2N}{e} \right)^{2N}} \right) - \frac{H_{N}}{2} \right) \tag{3} \\ &= \lim_{N \to \infty} \left(\log(\pi) + \log(N) - H_{N} \right)  \\  &= \log(\pi) - \gamma \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Riemann_zeta_function#Definition
$(2)$ http://mathworld.wolfram.com/DoubleFactorial.html
$(3)$ https://en.wikipedia.org/wiki/Stirling's_approximation
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\color{#f00}{%
2\int_{0}^{\infty}{\expo{x} - x - 1 \over x\pars{\expo{2x} - 1}}\,\dd x} =
2\int_{0}^{\infty}{\pars{\expo{x} - 1}/x  - 1 \over \expo{2x} - 1}\,\dd x
\end{align}

Note that
  $\ds{{\expo{x} - 1 \over x} - 1 = \int_{0}^{1}\pars{\expo{xt} - 1}\,\dd t}$ such that

\begin{align}
&\color{#f00}{%
2\int_{0}^{\infty}{\expo{x} - x - 1 \over x\pars{\expo{2x} - 1}}\,\dd x} =
2\int_{0}^{1}\int_{0}^{\infty}{\expo{tx} - 1 \over \expo{2x} - 1}\,\dd x\,\dd t \\[4mm] \stackrel{\expo{-2x}\,\,\, \equiv\ y}{=}\
 &\
-\int_{0}^{1}\int_{0}^{1}{1 - y^{-t/2} \over 1 - y}\,\dd x\,\dd t =
-\int_{0}^{1}\bracks{\Psi\pars{1 - {t \over 2}} + \gamma}\,\dd t =
2\ln\pars{\Gamma\pars{1/2} \over \Gamma\pars{1}} - \gamma
\\[4mm] = &\
\color{#f00}{\ln\pars{\pi} - \gamma}
\end{align}

because $\ds{\quad\Gamma\pars{\half} = \pi^{1/2}\quad}$ and
  $\ds{\quad\Gamma\pars{1} = 1}$.

A: Hint. One may set
$$
f(s):=2\int_0^\infty \frac{e^{sx}-sx-1}{x(e^{2x}-1)}dx, \quad 0<s<2. \tag1
$$ In order to get rid of the factor $x$ in the denominator, we may differentiate under the integral sign getting
$$
f'(s)=2\int_0^\infty \frac{e^{sx}-1}{e^{2x}-1}dx, \quad 0<s<2. \tag2
$$ Then expanding the latter integrand in $e^{-kx}$ terms and integrating termwise we get
$$
f'(s)=-\gamma-\psi\left(1-\frac{s}2\right) \tag3
$$ where $\displaystyle \psi : = \Gamma'/\Gamma$ and where $\gamma$ is the Euler-Mascheroni constant.
Integrating $(3)$, with the fact that, as $s \to 0$, $f(s) \to 0$, we get

$$
2\int_0^\infty \frac{e^{sx}-sx-1}{x(e^{2x}-1)}dx=-\gamma s+2 \log \Gamma\left(1-\frac{s}2\right), \quad 0<s<2, \tag4
$$ 

from which you deduce the value of your initial integral by putting $s:=1$. 

Identity $(4)$ is much more than what was asked.

