TITLE: Evaluation of $\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx$
QUESTION [5 upvotes]: I want to show that
$$\int\limits ^{\infty }_{0}\frac{x}{\left( x^{2} +1\right)^2\left( e^{tx} +1\right)} dx=\frac{\psi^{(1)}(\frac{t}{2\pi})-4\psi^{(1)}(\frac{t}{\pi})}{8\pi}t+\frac14
$$
where $\psi^{(1)}(z)$ is the first derivative of the digamma function, but I've never come across such an integral and don't know what to do. Any help  would be greatly appreciated.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[10px,#ffd]{\left.\int_{0}^{\infty}{x \over
\pars{x^{2} + 1}^{2}\pars{\expo{tx} + 1}}\,\dd x
\,\right\vert_{\ t\ >\ 0} =
{t \over 8\pi}\bracks{\Psi\, '\pars{t \over 2\pi} -
4\Psi\, '\pars{t \over \pi}} + {1 \over 4}}:\ {\Large ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\left.\int_{0}^{\infty}{x \over
\pars{x^{2} + 1}^{2}\pars{\expo{tx} + 1}}\,\dd x\,\right\vert_{\ t\ >\ 0}}
\\[5mm] = &\
\int_{0}^{\infty}{x \over
\pars{x^{2} + 1}^{2}\pars{\expo{tx} - 1}}\,\dd x -
\int_{0}^{\infty}{x \over
\pars{x^{2} +1}^{2}}\pars{{1 \over \expo{tx} - 1} -
{1 \over \expo{tx} + 1}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}{x \over
\pars{x^{2} +1}^{2}\pars{\expo{tx} - 1}}\,\dd x -
2\int_{0}^{\infty}{x \over
\pars{x^{2} + 1}^{2}\pars{\expo{2tx} - 1}}\,\dd x
\end{align}
In the first integral I'll set $\ds{tx\ \mapsto\ 2\pi x}$ while in the second one I'll set $\ds{2tx\ \mapsto\ 2\pi x}$. Then,
\begin{align}
&\bbox[10px,#ffd]{\left.\int_{0}^{\infty}{x \over
\pars{x^{2} +1}^{2}\pars{\expo{tx} + 1}}\,\dd x\,\right\vert_{\ t\ >\ 0}}
\\[5mm] = &\
\pars{t \over 2\pi}^{2}\int_{0}^{\infty}{x \over
\braces{x^{2} + \bracks{t/\pars{2\pi}}^{2}}^{2}\pars{\expo{2\pi x} - 1}}
\,\dd x
\\[2mm] - &\
2\pars{t \over \pi}^{2}\int_{0}^{\infty}{x \over
\bracks{x^{2} + \pars{t/\pi}^{2}}^{2}\pars{\expo{2\pi x} - 1}}\,\dd x
\\[5mm] = &\
\mathcal{F}\pars{t \over 2\pi} -
2\,\mathcal{F}\pars{t \over \pi}\label{1}\tag{1}
\end{align}
where
\begin{align}
\mathcal{F}\pars{y} &  \equiv
y^{2}\int_{0}^{\infty}{x \over
\pars{x^{2} + y^{2}}^{2}\pars{\expo{2\pi x} - 1}}\,\dd x =
{1 \over 4}\bracks{y\,\Psi\, '\pars{y} - {1 \over 2y} - 1}
\label{2}\tag{2}
\end{align}
(\ref{2}) is derived from
Binet Second  Formula.
The final result (\ref{2}) is derived here.

(\ref{1}) and (\ref{2}) yield the coveted result
$$
\bbox[10px,#ffd]{\left.\int_{0}^{\infty}{x \over
\pars{x^{2} + 1}^{2}\pars{\expo{tx} + 1}}\,\dd x
\,\right\vert_{\ t\ >\ 0} =
{t \over 8\pi}\bracks{\Psi\, '\pars{t \over 2\pi} -
4\Psi\, '\pars{t \over \pi}} + {1 \over 4}}
$$