Integral $ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $ 
*

*I need to compute the improper integral
$$
\int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x
$$
using contour integration and possibly principal values.  Trying to approach this as I would normally approach evaluating an improper integral using contour integration doesn't work here, and doesn't really give me any clues as to how I should do it.

*This normal approach is namely evaluating the contour integral
$$
\oint_{C}{\frac{z^2}{\cosh\left(z\right)}\mathrm{d}z}
$$
using a semicircle in the upper-half plane centered at the origin, but the semicircular part of this contour integral does not vanish since $\cosh\left(z\right)$ has period $2\pi\mathrm{i}$ and there are infinitely-many poles of the integrand along the imaginary axis given by $-\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ and
$\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ for
$n \in \mathbb{Z}$.

*The residues of the integrand at these simple poles are $-\frac{1}{4}\mathrm{i}\pi^{2}\left(1 - 4n\right)^{2}$ and $\frac{1}{4}\mathrm{i}\left(4\pi n + \pi\right)^{2}$, so that even when we add up all of the poles, we have the sum $4\pi^{2}\mathrm{i}\sum_{n = 0}^{\infty}\,n$, which clearly diverges.

Any hints would be greatly appreciated.

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\begin{align}
&\!\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over \cosh\pars{x}}\,\dd x} =
4\int_{0}^{\infty}{\sinh\pars{x} \over \sinh\pars{2x}}\,x^{2}\,\dd x
\\[5mm] \stackrel{x\ =\ t/4}{=}\,\,\,&
{1 \over 16}\int_{0}^{\infty}{\sinh\pars{t/4} \over \sinh\pars{t/2}}\,t^{2}\,\dd t
\\[5mm] = &\
\left.{1 \over 4}\,\partiald[2]{}{\alpha}\int_{0}^{\infty}{\sinh\pars{\alpha t/2} \over \sinh\pars{t/2}}\,\dd t
\,\right\vert_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
\left.{1 \over 4}\,\partiald[2]{}{\alpha}\int_{0}^{\infty}
{\expo{-\pars{1 - \alpha}t/2} -
\expo{-\pars{1 + \alpha}t/2} \over 1 - \expo{-t}}\,\dd t\
\,\right\vert_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{1 \over 4}\,\partiald[2]{}{\alpha}\left[%
\int_{0}^{\infty}{\expo{-t} -
\expo{-\pars{1 + \alpha}t/2} \over 1 - \expo{-t}}\,\dd t
\right.
\\[2mm] &
\left.\phantom{{1 \over 4}\,\partiald[2]{}{\alpha}}
-\int_{0}^{\infty}{\expo{-t} -
\expo{-\pars{1 - \alpha}t/2} \over 1 - \expo{-t}}\,\dd t
\right]
_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{1 \over 4}\,\partiald[2]{}{\alpha}\bracks{%
\Psi\pars{1 + \alpha \over 2} - \Psi\pars{1 - \alpha \over 2}}
_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{1 \over 4}\,\partiald[2]{}{\alpha}\bracks{%
\pi\cot\pars{\pi\,{1 - \alpha \over 2}}}_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{\pi \over 4}\,\partiald[2]{}{\alpha}\bracks{%
\tan\pars{\pi\alpha \over 2}}_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{\pi \over 4}\bracks{%
{\pi^{2} \over 4}\sec^{2}\pars{\pi a \over 2}
\tan\pars{\pi a \over 2}}_{\,\alpha\ =\ 1/2}
\\[5mm] = &\
{\pi^{3} \over 8}\
\underbrace{\sec^{2}\pars{\pi \over 4}}_{\ds{2}}\
\underbrace{\tan\pars{\pi \over 4}}_{\ds{1}}
\\[5mm] = &\
\bbx{\pi^{3} \over 4} \\ &
\end{align}
