Show $\int_0^\infty \log(1+x^2) \frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx=4\sqrt 2-\frac{16}{\pi}+\frac{8\sqrt 2}{\pi}\log(\sqrt 2+1)$ I am trying to prove this interesting integral
$$
I:=\int_0^\infty \log(1+x^2) \frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx=4\sqrt 2-\frac{16}{\pi}+\frac{8\sqrt 2}{\pi}\log(\sqrt 2+1).
$$
I tried to write
$$
\int_0^\infty \log(1+x)\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx+\int_0^\infty \log(1-x)\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx
$$
and now using
$$
\int_0^\infty \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx-\int_0^\infty \sum_{n=1}^\infty \frac{x^n}{n}\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx
$$
and now introducing a parameter
$$
I(a)=\int_0^\infty \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n\frac{\cosh \frac{a \pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx-\int_0^\infty \sum_{n=1}^\infty \frac{x^n}{n}\frac{\cosh  \frac{a\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx.
$$
But writing $I'(a)$ didn't simplify much.  The substitution $y=\sinh \pi x/4$ also was of no use because of the $x^n $ factor.  So How can we prove this interesting integral?  Thanks
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{I\equiv\int_{0}^{\infty}\ln\pars{1 + x^{2}}\,
     {\cosh\pars{\pi x/4} \over \sinh^{2}\pars{\pi x/4}}\,\dd x
     =4\root{2} - {16 \over \pi} + {8\root{2} \over \pi}\,\ln\pars{\root{2} + 1}:
     \ {\large ?}}$

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

Zeros $\ds{\braces{x_{n}}}$ of $\ds{\sinh\pars{\pi x \over 4}}$ are given by $\quad\ds{x_{n} = 4n\ic\,,\quad n \in {\mathbb Z}\quad}$ such that
\begin{align}
I&={4 \over \pi}\bracks{2\pi\ic\,{\ic \over \pars{\ic + \ic}\sinh\pars{\pi\ic/4}}
+\sum_{n = 1}^{\infty}2\pi\ic\,{4n\ic\over
\pars{4n\ic - \ic}\pars{4n\ic + \ic}\cosh\pars{\pi\bracks{4n\ic}/4}\pars{\pi/4}}}
\\[3mm]&=4\root{2}
+{8 \over \pi}\sum_{n = 1}^{\infty}\pars{-1}^{n}\,
{n \over \pars{n - 1/4}\pars{n + 1/4}}
=4\root{2}
+{4 \over \pi}\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n - 1/4}}
+{4 \over \pi}\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n + 1/4}}
\end{align}

With $\ds{a \not\in {\mathbb Z}}$:
  \begin{align}
&\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n + a}}
=\sum_{n = 0}^{\infty}\pars{{1 \over 2n + 2 + a} - {1 \over 2n + 1 + a}}
=-\sum_{n = 0}^{\infty}{1 \over \pars{2n + 2 + a}\pars{2n + 1 + a}}
\\[3mm]&=-\,{1 \over 4}\sum_{n = 0}^{\infty}
{1 \over \pars{n + 1 + a/2}\pars{n + 1/2 + a/2}}
=-\,{1 \over 4}\,{\Psi\pars{1 + a/2} - \Psi\pars{1/2 + a/2}
\over \pars{1 + a/2} - \pars{1/2 + a/2}}
\end{align}
  where $\ds{\Psi\pars{z}}$ is the Digamma Function $\bf\mbox{6.3.1}$.

$$
\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n + a}}
=\half\,\bracks{\Psi\pars{\half + {a \over 2}} - \Psi\pars{1 + {a \over 2}}}
$$

\begin{align}
I&\equiv\color{#66f}{\large\int_{0}^{\infty}\ln\pars{1 + x^{2}}\,
{\cosh\pars{\pi x/4} \over \sinh^{2}\pars{\pi x/4}}\,\dd x}
\\[3mm]&=\color{#66f}{\large4\root{2} + {2 \over \pi}\bracks{%
\Psi\pars{3 \over 8} - \Psi\pars{7 \over 8} + \Psi\pars{5 \over 8}
-\Psi\pars{9 \over 8}}}
\\[3mm]&\approx 3.7380
\end{align}

