TITLE: Integral $\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}$
QUESTION [30 upvotes]: Hi I am trying to show$$
I:=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx=2-\frac{4}{\pi}.
$$
Thank you.
What a desirable thing to want to prove!  It is a work of art this one.  I wish to prove this in as many ways as we can find. 
Note I tried writing
$$
I=\int_0^\infty \log(1+x^2)\coth \frac{\pi x}{2} \sinh^{-2} \frac{\pi x}{2}\mathrm dx
$$
but this didn't help me much.  We can also try introducing a parameter as follows
$$
I(\alpha)=\int_0^\infty \log(1+x^2)\frac{\cosh{\frac{\alpha \pi x}{2}}}{\sinh^2{\frac{\pi x}{2}}}\mathrm dx,
$$
But this is where I got stuck.  How can we calculate I?  Thanks.

REPLY [6 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{I\equiv\int_{0}^{\infty}\ln\pars{1 + x^{2}}\,
    {\cosh\pars{\pi x/2} \over \sinh^{2}\pars{\pi x/2}}\,\dd x = 2 - {4 \over \pi}:\
    {\large ?}}$

\begin{align}
I&=-\,{1 \over \pi}\int_{x\ \to\ -\infty}^{x\ \to\ \infty}\ln\pars{1 + x^{2}}
\,\dd\bracks{1 \over \sinh\pars{\pi x/2}}
={2 \over \pi}\int_{-\infty}^{\infty}
{1 \over \sinh\pars{\pi x/2}}\,{x \over x^{2} + 1}\,\dd x
\\[3mm]&={2 \over \pi}\bracks{%
2\pi\ic\sum_{n = 1}^{\infty}
{1 \over \pi\cosh\pars{n\pi\ic}/2}\,{2n\ic \over \pars{2n\ic}^{2} + 1}
+ 2\pi\ic\,{1 \over \sinh\pars{\pi\ic/2}}\,{\ic \over 2\ic}}
\\[3mm]&={16 \over \pi}\sum_{n = 1}^{\infty}\pars{-1}^{n}{n \over 4n^{2} - 1} + 2
=2 + {4 \over \pi}\bracks{%
\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 2n - 1}
+\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 2n + 1}}
\\[3mm]&=2 + {4 \over \pi}\bracks{%
-\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over 2n + 1}
+\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 2n + 1}}
=2 + {4 \over \pi}\bracks{-1%
-\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 2n + 1}
+\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over 2n + 1}}
\\[3mm]&=2 - {4 \over \pi}
\end{align}

$$\color{#00f}{\large%
\int_{0}^{\infty}\ln\pars{1 + x^{2}}\,
{\cosh\pars{\pi x/2} \over \sinh^{2}\pars{\pi x/2}}\,\dd x}=
\color{#00f}{\large 2 - {4 \over \pi}} \approx 0.7268
$$