Improper Integral $\int_0^\infty\left(\frac{\tanh(x)}{x^3}-\frac1{x^2\cosh^2(x)}\right)dx = \frac{7\zeta(3)}{\pi^2} $ $\newcommand{\sech}{\operatorname{sech}}$
$\displaystyle \int_0^{\infty}{\left(\frac{\tanh(x)}{x^3} - \frac{\sech^2(x)}{x^2} \right)\ dx }= \frac{7\zeta(3)}{\pi^2} $
What I tried
I simplified it to -
$\displaystyle \int_0^{\infty}{\frac{\sinh(2x) - 2x}{x^3 \cosh^2(x)} \ dx}$
Then I don't know how to solve. I tried Feynman's method 
$\displaystyle I(a) = \int_0^{\infty}{\frac{\sinh(ax) - ax}{x^3 \cosh^2(x)} \ dx}$
But then too it didn't help much. 
I thought of replacing them with trigonometric forms and then complex number real and imaginary part but wasn't helpful much. 
Please try to avoid complex analysis. 
 A: Integrating by parts we get $$I=\int_{0}^{\infty}\left(\frac{\tanh\left(x\right)}{x^{3}}-\frac{1}{x^{2}\cosh^{2}\left(x\right)}\right)dx= $$ $$-\int_{0}^{\infty}\frac{1}{x}\left(-\frac{\tanh\left(x\right)}{x^{2}}+\frac{1}{x\cosh^{2}\left(x\right)}+2\frac{\tanh\left(x\right)}{\cosh^{2}\left(x\right)}\right)dx$$ so $$\int_{0}^{\infty}\left(\frac{\tanh\left(x\right)}{x^{3}}-\frac{1}{x^{2}\cosh^{2}\left(x\right)}\right)dx=-\int_{0}^{\infty}\frac{\tanh\left(x\right)}{x\cosh^{2}\left(x\right)}dx
 $$ and now taking $x=-\log\left(u\right)
 $ we get $$I=4\int_{0}^{1}\frac{\left(u^{2}-1\right)u}{\left(u^{2}+1\right)^{3}\log\left(u\right)}du=4\int_{0}^{1}\frac{u^{3}-u}{\left(u^{2}+1\right)^{3}\log\left(u\right)}du
 $$ $$=4\int_{0}^{1}\frac{1}{\left(u^{2}+1\right)^{3}}\int_{1}^{3}u^{z}dzdu=4\int_{1}^{3}\int_{0}^{1}\frac{u^{z}}{\left(u^{2}+1\right)^{3}}dudz
 $$ and the last integral can be written in terms of the Gauss hypergeometric function $$\int_{0}^{1}\frac{u^{z}}{\left(u^{2}+1\right)^{3}}du=\frac{1}{2}\int_{0}^{1}\frac{u^{z/2-1/2}}{\left(u+1\right)^{3}}du=\frac{1}{z+1}\,_{2}F_{1}\left(3,\frac{z+1}{2},1+\frac{z+1}{2},-1\right)
 $$ and this particular hypergeometric function has a “closed form” in terms of Digamma function, so we have $$ I=\frac{1}{8}\int_{1}^{3}\left(z^{2}-4z+3\right)\left(\psi\left(\frac{z+3}{4}\right)-\psi\left(\frac{z+1}{4}\right)\right)dz-\frac{1}{8}\int_{1}^{3}2z-8dz.
 $$ Now note that every single term is in the form $$a\int_{1}^{3}z^{b}\psi\left(\frac{z+c}{4}\right)dz=4a\int_{(1+c)/4}^{(3+c)/4}\left(4v-c\right)^{b}\psi\left(v\right)dv
 $$ with $b=0,1,2
 $ and $c=1,3
 $ so let us consider the case $b=0$. We have $$\int_{(1+c)/4}^{(3+c)/4}\psi\left(v\right)dv=\log\left(\frac{\Gamma\left(\frac{3+c}{4}\right)}{\Gamma\left(\frac{1+c}{4}\right)}\right)
 $$ if $b=1
 $ we have, integrating by parts, $$\int_{(1+c)/4}^{(3+c)/4}v\psi\left(v\right)dv=\left(v\log\left(\Gamma\left(v\right)\right)-\psi^{\left(-2\right)}\left(v\right)\right)_{(1+c)/4}^{(3+c)/4}$$ and if $b=2$ we have $$\int_{(1+c)/4}^{(3+c)/4}v^{2}\psi\left(v\right)dv=\left(v^{2}\log\left(\Gamma\left(v\right)\right)-2v\psi^{\left(-2\right)}\left(v\right)+3\psi^{\left(-3\right)}\left(v\right)\right)_{(1+c)/4}^{(3+c)/4}$$ so combining this result and the closed form about polygamma at negative orders we obtain $$I=\color{red}{\frac{7\zeta\left(3\right)}{\pi^{2}}}$$ as wanted.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}
\bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x = {7 \over \pi^{2}}\,\zeta\pars{3}}
\approx 0.8526\ \Large ?}$.

\begin{align}
&\color{#f00}{\int_{0}^{\infty}
\bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x}
\\[5mm] = &\
\int_{0}^{\infty}{\tanh\pars{x} - x \over x^{3}}\,\dd x +
\int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x
\\[5mm] = &
-\,\half\int_{x\ =\ 0}^{x\ \to\ \infty}\bracks{\tanh\pars{x} - x}
\,\dd\pars{1 \over x^{2}} +
\int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x
\\[5mm] = &\
\half\int_{0}^{\infty}{\mrm{sech}^{2}\pars{x}  - 1 \over x^{2}}\,\dd x +
\int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x =
\half\int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x
\\[5mm] = &\
32\sum_{k = 0}^{\infty}\,\sum_{n = 0}^{\infty}\,\,\ \underbrace{%
\int_{0}^{\infty}{1 \over \bracks{\pars{2k + 1}\pi}^{\, 2} + 4x^{2}}\,
{1 \over \bracks{\pars{2n + 1}\pi}^{\, 2} + 4x^{2}}\,\dd x}
_{\ds{1 \over 8\pi^{2}\pars{2k + 1}\pars{2n + 1}\pars{k + n + 1}}}
\label{1}\tag{1}
\\[5mm] = &\
{4 \over \pi^{2}}\
\underbrace{\sum_{k = 0}^{\infty}{H_{k} + 2\ln\pars{2} \over \pars{2k + 1}^{2}}}
_{\ds{{7 \over 4}\,\zeta\pars{3}}}\label{2}\tag{2} =
\color{#f00}{{7 \over \pi^{2}}\,\zeta\pars{3}}
\end{align}
Note that

*

*In \eqref{1}, we use the identity
$\ds{{\tanh\pars{x} \over x} =
8\sum_{j = 0}^{\infty}{1 \over \bracks{\pars{2j + 1}\pi}^{\, 2} + 4x^{2}}}$

*The sum over $\ds{n}$, in \eqref{1}, yields a Digamma Function term
$\ds{\Psi\pars{1 + k}}$ which explains the appearance of the Harmonic Number $\ds{H_{k} = \Psi\pars{1 + k} + \gamma}$. $\ds{\gamma}$ is the
Euler-Mascheroni Constant.

*$\ds{\sum_{k = 0}^{\infty}{H_{k} \over \pars{2k + 1}^{2}} =
{1 \over 4}\bracks{7\zeta\pars{3} - \pi^{2}\ln\pars{2}}}$ is a well known result.

*$\ds{\sum_{k = 0}^{\infty}{1 \over \pars{2k + 1}^{2}} =
{1 \over 8}\,\pi^{2}}$.

A: A Residue Calculus Approach
At $z=\left(k+\frac12\right)\pi i$, the residue of $\frac{\tanh(x)}{x^3}$ is $\frac{i}{\left(\left(k+\frac12\right)\pi\right)^3}$ and the residue of $\frac1{x^2\cosh^2(x)}$ is $\frac {2i}{\left(\left(k+\frac12\right)\pi\right)^3}$
Therefore,
$$
\begin{align}
&\int_0^\infty\left(\frac{\tanh(x)}{x^3}-\frac1{x^2\cosh^2(x)}\right)\,\mathrm{d}x\tag{1}\\
&=\frac12\int_{-\infty}^\infty\left(\frac{\tanh(x)}{x^3}-\frac1{x^2\cosh^2(x)}\right)\,\mathrm{d}x\tag{2}\\
&=\frac12\int_{-\infty-i}^{\infty-i}\left(\frac{\tanh(x)}{x^3}-\frac1{x^2\cosh^2(x)}\right)\,\mathrm{d}x\tag{3}\\[6pt]
&=\pi i\sum_{k=0}^\infty\left[\frac{i}{\left(\left(k+\frac12\right)\pi\right)^3}-\frac{2i}{\left(\left(k+\frac12\right)\pi\right)^3}\right]\tag{4}\\
&=\frac8{\pi^2}\sum_{k=0}^\infty\frac1{(2k+1)^3}\tag{5}\\[6pt]
&=\frac{7\zeta(3)}{\pi^2}\tag{6}
\end{align}
$$
Explanation:
$(2)$: the integrand is even
$(3)$: the integrand vanishes at $x=\pm\infty$ and has no sinularities in $-1\lt\mathrm{Im}(z)\lt0$
$(4)$: the integrand vanishes like $\frac1{z^2}$ on $\mathrm{Im}(z)\in\mathbb{Z}\pi$
$\phantom{\text{(4): }}$and like $\frac1{z^3}$ as $|\mathrm{Re}(z)|\to\infty$
$\phantom{\text{(4): }}$so the integral is $2\pi i$ times the sum of the residues
$(5)$: algebra
$(6)$: note that the sum is $\frac78\zeta(3)$
