How to show that $2\int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}+1)} dx=\ln{4a\sin^2\left({\pi\over 2a}\right)\over \pi\sin\left({\pi\over a}\right)}?$ How can we show that

$$2\int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}+1)}\mathrm dx=\ln{4a\sin^2\left({\pi\over 2a}\right)\over \pi\sin\left({\pi\over a}\right)}?\tag1$$

 A: Hint. One may observe that
$$
\int_0^{1} \sinh (y\cdot x) \, dy=\frac{\cosh (x)-1}{x},\qquad x>0,
$$ then one may obtain, for $a>1$,
$$
\begin{align}
\int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}+1)}\mathrm dx&=\int_0^{1}dy\int_{0}^{\infty}{\sinh (y\cdot x)\over e^{ax}+1}\mathrm dx
\\\\&=\frac{1}{2} \int_0^{1}\left(\frac{\frac{\pi}{a}}{\sin \left(\frac{\pi  y}{a}\right)}-\frac{1}{y}\right)dy
\\\\&=\frac{1}{2}\ln \left(\frac{2a}{\pi}\cdot\tan\left(\frac{\pi}{2a}\right)\right)
\end{align}
$$ where we have used the fact that
$$
\left(\frac{\ln \left(\tan \left(\frac{b y}{2}\right)\right)}{b}\right)'_y=\frac{1+\tan^2 \left(\frac{b y}{2}\right)}{2\tan \left(\frac{b y}{2}\right)}=\frac{1}{2\sin\left(\frac{b y}{2}\right)\cos\left(\frac{b y}{2}\right)}=\frac1{\sin\left(b y\right)}.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} + 1}}\,\dd x =
\ln\pars{4a\sin^{2}\pars{\pi/\bracks{2a}} \over
         \pi\sin\pars{\pi/a}}:\ {\large ?}}$.

\begin{align}
&2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} + 1}}\,\dd x =
\,\mrm{f}\pars{1 - {1 \over a}} + \mrm{f}\pars{1 + {1 \over a}}
\label{1}\tag{1}
\\[5mm] &\mbox{where}\quad
\mrm{f}\pars{z} \equiv
\int_{0}^{\infty}{\expo{-zax} - \expo{-ax} \over x\pars{1 + \expo{-ax}}}\,\dd x
\label{2}\tag{2}
\end{align}

\begin{align}
\mrm{f}\pars{z} & \equiv
\int_{0}^{\infty}{\expo{-zax} - \expo{-ax} \over x\pars{1 + \expo{-ax}}}\,\dd x
\,\,\,\,\,\,\stackrel{\pars{\substack{{\large\expo{-ax}\ =\ t}\\[0.5mm]
                     {\large x\ =\ -\ln\pars{t}/a}}}\\[2mm]
                     \mbox{}}{\Large=}\,\,\,\,\,\,
-\int_{0}^{1}{t^{z - 1} - 1 \over \ln\pars{t}\pars{1 + t}}\,\dd t
\\[5mm] & =
\int_{0}^{1}{t^{z - 1} - 1 \over 1 - t^{2}}\,{t - 1 \over \ln\pars{t}}\,\dd t =
\int_{0}^{1}{t^{z - 1} - 1 \over 1 - t^{2}}\int_{0}^{1}t^{s}\,\dd s\,\dd t =
\int_{0}^{1}\int_{0}^{1}{t^{s + z - 1} - t^{s} \over 1 - t^{2}}\,\dd t\,\dd s
\\[5mm] & \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\,
{1 \over 2}\int_{0}^{1}\int_{0}^{1}
{t^{s/2 + z/2 - 1} - t^{s/2 - 1/2} \over 1 - t}\,\dd t\,\dd s
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\pars{%
\int_{0}^{1}{1 - t^{s/2 - 1/2} \over 1 - t}\,\dd t -
\int_{0}^{1}{1 - t^{s/2 + z/2- 1} \over 1 - t}\,\dd t}\,\dd s
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
\Psi\pars{s + 1 \over 2} - \Psi\pars{s + z \over 2}}
\dd s\qquad\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] & =
\left.\ln\pars{\Gamma\pars{\pars{s + 1}/2} \over \Gamma\pars{\bracks{s + z}/2}}
\right\vert_{\ s\ =\ 0}^{\ s\ =\ 1} =
\ln\pars{{\Gamma\pars{1} \over \Gamma\pars{\bracks{1 + z}/2}}\,
{\Gamma\pars{z/2} \over \Gamma\pars{1/2}}}
\\[5mm] & \implies
\bbx{\mrm{f}\pars{z} = \ln\pars{{1 \over \root{\pi}}\,{\Gamma\pars{z/2} \over \Gamma\pars{\bracks{1 + z}/2}}}}
\end{align}


\eqref{1} is reduced to

\begin{align}
&2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} + 1}}\,\dd x =
\ln\pars{{1 \over \root{\pi}}
{\Gamma\pars{1/2 - 1/\bracks{2a}} \over \Gamma\pars{1 - 1/\bracks{2a}}}\,
{1 \over \root{\pi}}
{\Gamma\pars{1/2 + 1/\bracks{2a}} \over \Gamma\pars{1 + 1/\bracks{2a}}}}
\\[5mm] = &\
\ln\pars{{1 \over \pi}\,{\pi \over \sin\pars{\pi\braces{1/2 + 1/\bracks{2a}}}}\,
{1 \over \Gamma\pars{1 - 1/\bracks{2a}}\Gamma\pars{1/\bracks{2a}}/\bracks{2a}}}
\\[5mm] = &\
\ln\pars{{2a \over \cos\pars{\pi/\bracks{2a}}}\,
{1 \over \pi/\sin\pars{\pi/\bracks{2a}}}} =
\ln\pars{{4a \over \pi}\,{\sin^{2}\pars{\pi/\bracks{2a}} \over 2\sin\pars{\pi/\bracks{2a}}\cos\pars{\pi/\bracks{2a}}}}
\\[5mm] & =
\bbx{\ln\pars{4a\sin^{2}\pars{\pi/\bracks{2a}} \over \pi\sin\pars{\pi/a}}}
\end{align}

Indeed, a simpler expression is
  $\bbx{\ds{\ln\pars{{2a \over \pi}\,\tan\pars{\pi \over 2a}}}}$.

A: An alternative approach. By Frullani's theorem
$$ \int_{0}^{+\infty}\frac{\cosh(x)-1}{x}e^{-\mu x}\,dx = \log\mu-\frac{1}{2}\log(\mu^2-1)\tag{1}$$
for any $\mu>1$. By expanding $\frac{1}{e^{ax}+1}$ as a geometric series
$$ \frac{1}{1+e^{ax}} = e^{-ax}-e^{-2a x}+e^{-3ax}-e^{-4ax}+\ldots \tag{2} $$
the original integral equals
$$ \frac{1}{2}\sum_{k\geq 1}(-1)^{k+1}\left[\log(k^2a^2)-\log(k^2 a^2-1)\right]=\frac{1}{2}\,\log\frac{\prod_{k\geq 1}\left(1-\frac{1}{(2k)^2 a^2}\right)}{\prod_{k\geq 0}\left(1-\frac{1}{(2k+1)^2 a^2}\right)}\tag{3} $$
and the claim follows from the Weierstrass product for the sine and cosine functions.
