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.
 A: Integrals of the form
$$\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx,$$
where $p$ is a polynomial can be evaluated by shifting the contour of integration to a line $\operatorname{Im} z \equiv c$. We first check that the integrals over the vertical segments connecting the two lines tend to $0$ as the real part tends to $\pm\infty$:
$$\lvert \cosh (x+iy)\rvert^2 = \lvert \cosh x\cos y + i \sinh x\sin y\rvert^2 = \sinh^2 x + \cos^2 y,$$
so the integrand decays exponentially and
$$\left\lvert \int_{R}^{R + ic} \frac{p(z)}{\cosh z}\,dz\right\rvert
\leqslant \frac{K\,c}{\sinh R}\left(R^2+c^2\right)^{\deg p/2} \xrightarrow{R\to \pm\infty} 0.$$
Since $\cosh \left(z+\pi i\right) = -\cosh z$, and the only singularity of the integrand between $\mathbb{R}$ and $\mathbb{R}+\pi i$ is a simple pole at $\frac{\pi i}{2}$ (unless $p$ has a zero there, but then we can regard it as a simple pole with residue $0$) with the residue
$$\operatorname{Res}\left(\frac{p(z)}{\cosh z};\, \frac{\pi i}{2}\right) = \frac{p\left(\frac{\pi i}{2}\right)}{\cosh' \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{\sinh \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{i},$$
the residue theorem yields
$$\begin{align}
\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx
&= 2\pi\, p\left(\frac{\pi i}{2}\right) + \int_{\pi i-\infty}^{\pi i+\infty} \frac{p(z)}{\cosh z}\,dz\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \int_{-\infty}^\infty \frac{p(x+\pi i)}{\cosh x}\,dx\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \sum_{k=0}^{\deg p} \frac{(\pi i)^k}{k!}\int_{-\infty}^\infty \frac{p^{(k)}(x)}{\cosh x}\,dx.\tag{1}
\end{align}$$
Since $\cosh$ is even, only even powers of $x$ contribute to the integrals, hence we can from the beginning assume that $p$ is an even polynomial, and need only consider the derivatives of even order.
For a constant polynomial, $(1)$ yields
$$\int_{-\infty}^\infty \frac{dx}{\cosh x} = 2\pi - \int_{-\infty}^\infty \frac{dx}{\cosh x}\Rightarrow \int_{-\infty}^\infty \frac{dx}{\cosh x} = \pi.$$
For $p(z) = z^2$, we obtain
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx &= 2\pi \left(\frac{\pi i}{2}\right)^2 - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx - (\pi i)^2\int_{-\infty}^\infty \frac{dx}{\cosh x}\\
&= - \frac{\pi^3}{2} - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx + \pi^3,
\end{align}$$
which becomes
$$\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx = \frac{\pi^3}{4}.$$
A: Just for laughs, another way to use a rectangular contour involves considering the following integral
$$\oint_C dz \frac{z^2}{\sinh{z}}$$
where $C$ is the rectangle having vertices $\pm i \pi/2$ and $R \pm i \pi/2$.  The contour integral is then equal to
$$i \int_0^R dx \frac{(x-i \pi/2)^2 + (x+i \pi/2)^2}{\cosh{x}} + i \int_{-\pi/2}^{\pi/2} dy \frac{(R+i y)^2}{\sinh{(R+i y)}} +  \int_{-\pi/2}^{\pi/2} dy \frac{y^2}{\sin{y}}$$
Note that, in the last integral, we did not need to take the Cauchy principal value as the singularity was removed by the $y^2$ in the numerator.  Thus, the last integral vanishes because the integrand is odd.  The middle integral vanishes as $R \to \infty$.  On the other hand, by Cauchy's theorem, the contour integral is zero for a lack of poles inside $C$.  Thus,
$$i 2 \int_0^{\infty} dx \frac{x^2-\pi^2/4}{\cosh{x}} = i \int_{-\infty}^{\infty} dx \frac{x^2-\pi^2/4}{\cosh{x}}= 0$$
Using the fact that
$$\int_{-\infty}^{\infty} \frac{dx}{\cosh{x}} = \pi$$
we get
$$\int_{-\infty}^{\infty} dx \frac{x^2}{\cosh{x}} = \frac{\pi^3}{4}$$
A: And yet, in a third approach, we enforce the substitution $x\to \log(x)$ and write 
$$\begin{align}
\int_0^\infty \frac{x^2}{\cosh(x)}\,dx&=2\int_{1}^{+\infty}\frac{\log^2 x}{x^2+1}\,dx\\\\
&=2\int_0^1 \frac{\log^2 x}{x^2+1}\,dx\\\\
&=  \int_{0}^{+\infty}\frac{\log^2 x}{x^2+1}\,dx
\end{align}$$
Thus, the integral of interest can be expressed as
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{x^2}{\cosh(x)}\,dx=2 \,\int_{0}^{+\infty}\frac{\log^2 x}{x^2+1}\,dx}$$

Next, we examine the integral $I$ as given by
$$I=\oint_C \frac{\log^3(z)}{z^2+1}\,dz$$
where $C$ is the classical key-hole contour that runs along both sides of the positive real axis.  Since the integrand is analytic in and on $C$, except at the simple poles, $z=\pm i$, its value is given by the residue theorem as 
$$\begin{align}
I&=2\pi i \text{Res}\left(\frac{\log^3(z)}{z^2+1}, z=\pm i\right)\\\\
&=2\pi i\left(\frac{\log^3(e^{i\pi/2})}{2i}+\frac{\log^3(e^{i3\pi/2})}{-2i}\right)\\\\
&=\frac{13\pi^4}{4}\,i \tag 1
\end{align}$$

Then, we note that we can write $I$ as 
$$\begin{align}
I&=\int_0^\infty \frac{\log^3(x)-(\log(x)+i2\pi)^3}{x^2+1}\,dx\\\\
&=-i6\pi \int_0^\infty \frac{\log^2(x)}{x^2+1}\,dx\\\\
&+12\pi^2\int_0^\infty \frac{\log(x)}{x^2+1}\,dx\\\\
&+i8\pi^3\int_0^\infty \frac{1}{x^2+1}\,dx \tag 2
\end{align}$$

The first integral on the right-hand side of $(2)$ is $1/2$ the integral of interest, the second integral is $0$ (to see this, enforce the substitution $x\to 1/x$), and the third integral is 
$$\int_0^\infty \frac{1}{x^2+1}=\pi/2 \tag 3$$
Using $(1)-(3)$ reveals 
$$-i6\pi \int_0^\infty \frac{\log^2(x)}{x^2+1}\,dx=\frac{13\pi^4}{4}\,i -i4\pi^4$$  
and therefore, we find that
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{x^2}{\cosh(x)}\,dx=\frac{\pi^3}{4}}$$
A: Here is another way to do it.  Let $$I=\int_{-\infty}^{\infty} \frac{x^2}{\cosh(x)}dx=2\int_{0}^{\infty} \frac{x^2}{\cosh(x)}dx=4\int_{0}^{\infty} \frac{x^2}{e^x+e^{-x}}dx.$$ Let $x=\ln(u),dx=\frac{1}{u}du.$ We get : $$I=4\int_{1}^{\infty} \frac{(\ln(u))^2}{u^2+1}du.$$ But making another substitution $z=\frac{1}{u},dz=\frac{-1}{u^2}du$ shows 
$$\int_{1}^{\infty} \frac{(\ln(u))^2}{u^2+1}du=\int_{0}^{1} \frac{(\ln(z))^2}{z^2+1}dz.$$ If follows $$\int_{1}^{\infty} \frac{(\ln(u))^2}{u^2+1}du=\frac{1}{2}\int_{0}^{\infty} \frac{(\ln(u))^2}{u^2+1}du,$$ so
$$I=2\int_{0}^{\infty} \frac{(\ln(u))^2}{u^2+1}du.$$ 
Now, we focus on finding $$ \int_{0}^{\infty}\frac{(\ln(u))^2}{u^2+1}du.$$ We show this by considering the triple integral $$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2y^2z^2)(1+x^2)(1+y^2)}dzdydx.$$ and proving $J=I.$ Let us evaluate $J$ first.  Recognizing that $$\int_{0}^{\infty} \frac{xy}{1+x^2y^2z^2}dz=\frac{\pi}{2},$$ we have $$J=\frac{\pi}{2} \int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{1+x^2}\frac{1}{1+y^2}dydx=\frac{\pi^3}{8}.$$ Now we use Fubini's thoerem as such: $$J=\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{xy}{(1+x^2y^2z^2)(1+x^2)(1+y^2)}dydzdx.$$ Integrating this way now, we find through partial fractions:
$$J=\int_{0}^{\infty}\int_{0}^{\infty} \frac{\ln(xz)}{(x^2+1)(1-x^2z^2)}dzdx.$$ But $$\int_{0}^{\infty}\int_{0}^{\infty} \frac{\ln(xz)}{(x^2+1)(1-x^2z^2)}dzdx=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\ln(x)}{(x^2+1)(1-x^2z^2)} +\frac{\ln(z)}{(x^2+1)(1-x^2z^2)}dzdx.$$ It turns out using partial fractions on both integrands, $$\int_{0}^{\infty}\int_{0}^{\infty}\frac{\ln(x)}{(x^2+1)(1-x^2z^2)}dzdx=0,$$ and $$\int_{0}^{\infty}\int_{0}^{\infty}\frac{\ln(x)}{(x^2+1)(1-x^2z^2)}dzdx=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\ln(z)}{(x^2+1)(1-x^2z^2)}dxdz=\int_{0}^{\infty} \frac{(\ln(z))^2}{z^2+1}dz=I$$
So $$I=2\left(\frac{\pi^3}{8}\right),$$ which means $$I=\frac{\pi^3}{4}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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,
$$
\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{x^{2} \over \cosh\pars{x}}\,\dd x} = \bbx{\pi^{3} \over 4} \\
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\!\bbox[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}
