Prove $\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx=\frac{\sqrt{2}-2}{4} \sqrt{\pi}~ \zeta \left( \frac{1}{2} \right)$ Wolfram Alpha evaluates this integral numerically as 
$$\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx=0.379064 \dots$$
Its value is apparently
$$\frac{\sqrt{2}-2}{4} \sqrt{\pi}~ \zeta \left( \frac{1}{2} \right)=0.37906401072\dots$$

How would you solve this integral?

Obviously, we can make a substitution $t=x^2$
\begin{align}
\int_0^{\infty} \frac{x^2}{\cosh^2 (x^2)} dx&=\frac{1}{2} \int_0^{\infty} \frac{\sqrt{t}}{\cosh^2 (t)} dt\\[10pt]
&=\int_0^{\infty} \frac{\sqrt{t}}{\cosh (2t)+1} dt\\[10pt]
&=\frac{1}{2 \sqrt{2}}\int_0^{\infty} \frac{\sqrt{u}}{\cosh (u)+1} du
\end{align}
We could use geometric series since $\cosh (u) \geq 1$, but I don't know how it will help.
 A: $$I=\frac{1}{2\sqrt{2}}\int_{0}^{+\infty}\frac{\sqrt{u}\,du}{1+\cosh(u)}=\frac{1}{\sqrt{2}}\int_{1}^{+\infty}\frac{\sqrt{\log v}}{(v+1)^2}\,dv=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{\sqrt{-\log v}}{(1+v)^2}\,dv \tag{1}$$
but since
$$ \int_{0}^{1}v^k \sqrt{-\log v}\,dv = \frac{\sqrt{\pi}}{2(1+k)^{3/2}} \tag{2}$$
by expanding $\frac{1}{(1+v)^2}$ as a Taylor series we get:

$$ I = \frac{1}{\sqrt{2}}\sum_{n\geq 0}(-1)^n (n+1)\frac{\sqrt{\pi}}{2(1+n)^{3/2}} = \color{red}{\frac{\sqrt{\pi}}{2\sqrt{2}}\cdot\eta\left(\frac{1}{2}\right)}\tag{3}$$

and the claim follows from the well-known:
$$ \eta(s) = (1-2^{1-s})\,\zeta(s)\tag{4} $$
that gives an analytic continuation for the $\zeta$ function.
A: Hint:
Consider the parametric integral
\begin{equation}
I(a) =  \int_0^\infty \frac{t^{a-1}}{\cosh^{2} t}\ dt=4 \int_{0}^{\infty} \frac{t^{a-1}}{(e^{t}+e^{-t})^{2}}\ dt = 4 \int_{0}^{\infty}\frac{t^{a-1} e^{-2t}}{(1+e^{-2t})^{2}}\ dt
\end{equation}
Hence, your integral is simply
\begin{equation}
\int_0^{\infty}  \frac{x^2}{\cosh^2x^2}\ dx = -2\ \frac{\partial}{\partial b}\left[\int_0^{ \infty}\frac{t^{a-1}}{1+e^{bt}} \ dt\right]_{a=\frac{1}{2}\ ,\ b=2}
\end{equation}
I believe you can evaluate the last expression by your own.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{2} \over \cosh^{2}\pars{x^{2}}}\,\dd x}\,\,\
\stackrel{x\ \to\ x^{1/2}}{=}\,\,\,\
2\int_{0}^{\infty}{x^{1/2}\expo{2x} \over \pars{\expo{2x} + 1}^{2}}\,\dd x
\\[5mm] = &\
-\int_{x = 0}^{x \to \infty}x^{1/2}\,\dd\pars{{1 \over \expo{2x} + 1}} =
\half\int_{0}^{\infty}{x^{-1/2}\expo{-2x} \over 1 + \expo{-2x}}\,\dd x
\\[5mm] \stackrel{2x\ \to x}{=}\,\,\,&
{\root{2} \over 4}\int_{0}^{\infty}{x^{-1/2}\expo{-x} \over 1 + \expo{-x}}\,\dd x
\\[5mm] = &
{\root{2} \over 4}\sum_{n = 0}^{\infty}\pars{-1}^{n}
\int_{0}^{\infty}x^{-1/2}\expo{-\pars{n + 1}x}\,\dd x
\\[3mm] \stackrel{\pars{n + 1}x\ \to x}{=}\,\,\,&
{\root{2} \over 4}\sum_{n = 0}^{\infty}
{\pars{-1}^{n} \over \pars{n + 1}^{1/2}}\ \overbrace{%
\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x}^{\ds{\Gamma\pars{\half}\ =\ \root{\pi}}}
\\[5mm] = &\
{\root{2} \over 4}\,\root{\pi}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{1/2}}
\end{align}

With the identity
$\ds{\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n^{s}} =
     \pars{1 - 2^{1 - s}}\zeta\pars{s}}$:
\begin{align}
\bbox[#ffd,10px]{\int_{0}^{\infty}{x^{2} \over \cosh^{2}\pars{x^{2}}}\,\dd x} & =
{\root{2} \over 4}\,\root{\pi}\pars{1 - 2^{1 - 1/2}}\zeta\pars{\half}
\\[5mm] & =
\bbox[10px,border:1px groove navy]{{\root{2} - 2 \over 4}\,\root{\pi}\zeta\pars{\half}} \approx 0.3791
\end{align}
