TITLE: 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} $
QUESTION [15 upvotes]: $\newcommand{\sech}{\operatorname{sech}}$
$\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.

REPLY [10 votes]: 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 $\frac78\zeta(3)$