calculus - 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.

Answer

$$I=\frac{1}{2\sqrt{2}}\int_{0}^{+\infty}\frac{\sqrt{u}\,du}{1+\cosh(u)}=\frac{1}{\sqrt{2}}\int_{1}^{+\infty}\frac{\sqrt{\log v}}{(v+1)^2}\,dv=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{\sqrt{-\log v}}{(1+v)^2}\,dv \tag{1}$$
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.