TITLE: How does one show that $\int_{-\infty}^{+\infty}{\left(x\over 2+2^{x}+2^{-x}\right)^2}\mathrm dx={\zeta(2)-1\over 3}\cdot{1\over \ln^3(2)}?$
QUESTION [9 upvotes]: Motivated by this question due to the comment of @Olivier Oloa.
$$\int_{-\infty}^{+\infty}{\left(x\over 2+2^{x}+2^{-x}\right)^2}\mathrm dx={\zeta(\color{red}2)-\color{blue}1\over \color{green}3}\cdot{\color{blue}1\over \ln^\color{green}3(\color{red}2)}\tag1$$
Here is my attempt:
$u=2^x\implies du=2^x\ln 2dx$ then $(1)$ becomes
$${1\over \ln^3(2)}\int_{0}^{+\infty}{\ln^2(u)\over (2+u+u^{-1})^2}\cdot{\mathrm du\over u}\tag2$$
$e^v=u$ then $(2)$ becomes
$${1\over \ln^3(2)}\int_{-\infty}^{+\infty}{v^2\over (2+e^v+e^{-v})^2}\mathrm dv\tag3$$
Not sure what is next step...

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[#ffe,10px]{\ds{%
\int_{-\infty}^{\infty}{\pars{x \over 2 + 2^{x} + 2^{-x}}^{2}}\,\dd x}} =
\int_{-\infty}^{\infty}{x^{2} \over \bracks{2 + \expo{\ln\pars{2}x} +
\expo{-\ln\pars{2}x}}^{2}}\,\dd x
\\[5mm] = &\
{1 \over \ln^{3}\pars{2}}
\int_{-\infty}^{\infty}
{x^{2} \over \bracks{2 + \expo{x} + \expo{-x}}^{2}}\,\dd x
\,\,\,\stackrel{t\ =\ \exp\pars{x}}{=}\,\,\,
{1 \over \ln^{3}\pars{2}}\int_{0}^{\infty}
{\ln^{2}\pars{t} \over \pars{2 + t + 1/t}^{\,2}}\pars{1 \over t}\,\dd t
\\[5mm] = &\
{1 \over \ln^{3}\pars{2}}\int_{0}^{\infty}
{t\ln^{2}\pars{t} \over \pars{2t + t^{2} + 1}^{\,2}}\,\dd t =
{1 \over \ln^{3}\pars{2}}\int_{0}^{\infty}
{t\ln^{2}\pars{t} \over \pars{t + 1}^{\,4}}\,\dd t
\\[5mm] = &
\left.{1 \over \ln^{3}\pars{2}}
\partiald[2]{}{\epsilon}\int_{0}^{\infty}{t^{\epsilon + 1} \over \pars{t + 1}^{\,4}}\,\dd t
\,\right\vert_{\ \epsilon\ =\ 0}
\,\,\,\stackrel{t\ =\ 1/x\ -\ 1}{=}\,\,\,
\left.{1 \over \ln^{3}\pars{2}}\partiald[2]{}{\epsilon}
\int_{0}^{1}x^{1 - \epsilon}\,\pars{1 - x}^{1 + \epsilon}\,\dd x
\,\right\vert_{\ \epsilon\ =\ 0}
\\[5mm] = &\
\left.{1 \over \ln^{3}\pars{2}}\partiald[2]{}{\epsilon}
{\Gamma\pars{2 - \epsilon}\Gamma\pars{2 + \epsilon} \over \Gamma\pars{4}}
\,\right\vert_{\ \epsilon\ =\ 0} =
\left.{1 \over 6\ln^{3}\pars{2}}
\partiald[2]{\bracks{\pars{\epsilon - \epsilon^{3}}
\Gamma\pars{1 - \epsilon}\Gamma\pars{\epsilon}}}{\epsilon}\,\right\vert_{\ \epsilon\ =\ 0}
\\[5mm] = &\
{1 \over 6\ln^{3}\pars{2}}\partiald[2]{}{\epsilon}\bracks{%
\pi\,{\epsilon - \epsilon^{3} \over \sin\pars{\pi\epsilon}}}
_{\ \epsilon\ =\ 0} =
{1 \over 6\ln^{3}\pars{2}}\partiald[2]{}{\epsilon}\bracks{%
{1 \over \sin\pars{\pi\epsilon}/\pars{\pi\epsilon}} - \epsilon^{2}}
_{\ \epsilon\ =\ 0}
\\[5mm] = &\
{1 \over 6\ln^{3}\pars{2}}\partiald[2]{}{\epsilon}\bracks{%
1 + {\pars{\pi\epsilon}^{2} \over 6} - \epsilon^{2}}_{\ \epsilon\ =\ 0} =
{1 \over 6\ln^{3}\pars{2}}\pars{{\pi^{2} \over 3} - 2} =
\bbx{{\zeta\pars{2} - 1 \over 3}\,{1 \over \ln^{3}\pars{2}}}
\end{align}

because $\ds{\zeta\pars{2} = {\pi^{2} \over 6}}$.