TITLE: Show that $\int_{0}^{\infty}{x\over (1+x^2)^2}\cdot{\mathrm dx\over \tanh\left({\pi x\over 2}\right)}={\pi^2\over 8}-{1\over 2}?$
QUESTION [11 upvotes]: How may we show that

$$\int_{0}^{\infty}{x\over (1+x^2)^2}\cdot{\mathrm dx\over \tanh\left({\pi x\over 2}\right)}={\pi^2\over 8}-{1\over 2}\color{red}?\tag1$$

$u={x\over 2}\implies 2du=dx$
$$4\int_{0}^{\infty}{u\over (1+4u^2)^2}\coth(\pi u)\mathrm du\tag2$$
$$4\int_{0}^{\infty}{u\over (1+4u^2)^2}\cdot{e^{2u}+1\over e^{2u}-1}\mathrm du\tag3$$
$v=4u^2\implies du={dv\over 8u}$
$${1\over 2}\int_{0}^{\infty}{e^{\sqrt{v}}+1\over e^{\sqrt{v}}-1}\cdot{\mathrm dv\over (1+v)^2}\tag4$$
$${1\over 2}\sum_{k=0}^{\infty}{(2)_k\over k!}(-1)^k\int_{0}^{\infty}{e^{\sqrt{v}}+1\over e^{\sqrt{v}}-1}\cdot v^k \mathrm dv\tag5$$
$e^{\sqrt{v}}=t\implies dv=2\sqrt{v}e^{\sqrt{v}}dt$
$$\sum_{k=0}^{\infty}{(2)_k\over k!}(-1)^k\int_{1}^{\infty}t\cdot{t+1\over t-1}\cdot \ln^{2k+1}(t) \mathrm dt\tag6$$
I don't what to do next ...

REPLY [5 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\ds{\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,{\dd x\over \tanh\pars{\pi x/2}} =
{\phantom{^{2}}\pi^2\over 8} - {1 \over 2}:\ {\large ?}}$.

\begin{align}
&\mbox{Note that}\quad
{1 \over \tanh\pars{\pi x/2}} = {\expo{\pi x} + 1 \over \expo{\pi x} - 1} =
1 + {2 \over \expo{\pi x} - 1}
\end{align}
such that
\begin{align}
&\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,
{\dd x\over \tanh\pars{\pi x/2}} =
\overbrace{\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,\dd x}
^{\ds{1 \over 2}}\ +\
2\
\overbrace{\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}
\,{1 \over \expo{\pi x} - 1}\,\dd x}^{\ds{\mbox{Set}\quad x/2\ \mapsto\ x}}
\\[5mm] & =
{1 \over 2} +
{1 \over 2}\int_{0}^{\infty}{x \over
\bracks{x^{2} + \pars{1/2}^{2}}^{2}}
\,{1 \over \expo{2\pi x} - 1}\,\dd x =
{1 \over 2} -
\left.{1 \over 2}\,\partiald{}{z}
\int_{0}^{\infty}{x \over x^{2} + z^{2}}
\,{1 \over \expo{2\pi x} - 1}\,\dd x\,\right\vert_{\ z\ =\ 1/2}
\end{align}

With Binet Second Formula
  ( see $\ds{\mathbf{\color{#000}{6.3.21}}}$ in A & S Table ),

\begin{align}
&\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,
{\dd x\over \tanh\pars{\pi x/2}} =
{1 \over 2} - {1 \over 2}\,\partiald{}{z}\bracks{\ln\pars{z} - 1/\pars{2z} - \Psi\pars{z} \over 2}_{\ z\ =\ 1/2}
\\[5mm] = &\
{1 \over 2} - {1 \over 2}\bracks{2 - {1 \over 2}\,\Psi\,'\pars{1 \over 2}} =
{1 \over 4}\,\Psi\,'\pars{1 \over 2} - {1 \over 2} =
\bbx{{\phantom{^{2}}\pi^{2} \over 8} - {1 \over 2}}
\end{align}

Note that $\ds{\Psi\,'\pars{1/2} = 3\pars{\pi^{2}/6} = \pi^{2}/2}$ ( see $\ds{\mathbf{6.4.4}}$ in A & S Table ).