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}{x^{2}\expo{-x} \over 1 + \expo{-2x}}\,\dd x
\\[5mm] = &\
4\sum_{n = 0}^{\infty}\pars{-1}^{n}
\int_{0}^{\infty}x^{2}\expo{-\pars{2n + 1}x}\,\,\,\dd x
\\[5mm] = &\
4\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{3}}
\int_{0}^{\infty}x^{2}\expo{-x}\,\,\,\dd x
\\[5mm] = &\
8\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{3}} =
-8\ic\sum_{n = 0}^{\infty}{\ic^{2n + 1} \over \pars{2n + 1}^{3}}
\\[5mm] = &\
-8\ic\sum_{n = 1}^{\infty}{\ic^{n} \over n^{3}}\,{1 - \pars{-1}^{n} \over 2}
\\[5mm] = &\
8\,\Im\sum_{n = 1}^{\infty}{\ic^{n} \over n^{3}} =
-4\ic\,\bracks{\on{Li}_{3}\pars{\ic} - \on{Li}_{3}\pars{-\ic}}
\\[5mm] = &\
-4\ic\,\braces{\on{Li}_{3}\pars{\expo{2\pi\ic\bracks{\color{red}{1/4}}}} - \on{Li}_{3}\pars{\expo{-2\pi\ic\bracks{\color{red}{1/4}}}}}
\\[5mm] = &\
-4\ic\bracks{-\,{\pars{2\pi\ic}^{3} \over 3!}
\on{B}_{3}\pars{\color{red}{1 \over 4}}}
\end{align}
The last expression is
Jonqui$\grave{\mrm{e}}$re's Inversion Formula. $\ds{\on{B}_{n}}$ is a
Bernoulli Polynomial. In particular, $\ds{\on{B}_{3}\pars{x} =
x^{3} - {3 \over 2}\,x^{2} + {1 \over 2}\,x}$.
Finally,
$$
\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over \cosh\pars{x}}\,\dd x} = \bbx{\pi^{3} \over 4} \\
$$
