integration - Prove that $2int_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

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

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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.

Answer

Hint. One may set
$$
f(s):=2\int_0^\infty \frac{e^{sx}-sx-1}{x(e^{2x}-1)}dx, \quad 0$$ In order to get rid of the factor $x$ in the denominator, we may differentiate under the integral sign getting
$$
f'(s)=2\int_0^\infty \frac{e^{sx}-1}{e^{2x}-1}dx, \quad 0$$ 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

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

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

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