Improper integral problem not easy to solve with exponential in the denominator

Question

Is there an analytical way to solve the following integral ?

$\displaystyle \int_0^{\infty } \frac{2 x}{\left(x^2+1\right) \left(e^{2 \pi x}-1\right)} \, dx$

The solution provided by Mathematica is $\gamma -\frac{1}{2}$.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

It's somehow related to the Second Binet Formula. See $\ds{\mathbf{\color{black}{6.3.21}}}$ in A & S Table.

\begin{align} &\bbox[5mm,#ffd]{\int_{0}^{\infty }{2x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} - 1}}\,\dd x} \\ = &\ 2\int_{0}^{\infty } {x \over \pars{x^{2} + \color{red}{1}^{2}}\pars{\expo{2\pi x} - 1}}\,\dd x \\[5mm] = &\ \bracks{\ln\pars{z} - {1 \over 2z} - \Psi\pars{z}}_{\ z\ =\ \color{red}{1}} = -\,{1 \over 2} -\ \overbrace{\Psi\pars{1}}^{\ds{-\gamma}} \\[5mm] = &\ \bbx{\gamma - {1 \over 2}} \approx 0.772 \end{align}