Evaluate $\int_0^{\infty } \frac{x^2 \tanh \left(x^2\right)}{\cosh \left(x^2\right)} \, dx$ How to show $$\int_0^{\infty } \frac{x^2 \tanh \left(x^2\right)}{\cosh \left(x^2\right)} \ dx=\frac{1}{2} \sqrt{\pi } \beta \left(\frac{1}{2}\right)$$
Any help will be appreciated.
 A: That $\cosh^2$ reminds me of one integral representation for the $\zeta$ function. Let's start from scratch: for any $s>0$,
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx $$
is a straightforward consequence of the (inverse) Laplace transform. If we apply integration by parts we get that
$$ \eta(s) = \frac{1}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s e^{x}}{(e^x+1)^2}\,dx = \frac{2^{s-1}}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{x^s}{\cosh^2(x)}\,dx $$
holds for any $s>-1$. Similarly,
$$ \beta(s)=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^s}=\frac{1}{2^{s+1}\Gamma(s)}\int_{0}^{+\infty}\frac{z^{s-1}}{\cosh(z/2)}\,dz $$
for any $s>0$ leads to
$$ \beta(s) = \frac{1}{2\Gamma(s+1)}\int_{0}^{+\infty}\frac{z^s\sinh(z) }{\cosh^2(z)}\,dz $$
for any $s>-1$. It follows that
$$ \frac{\sqrt{\pi}}{2}\beta\left(\tfrac{1}{2}\right) = \frac{1}{2}\int_{0}^{+\infty}\frac{\sqrt{z}\sinh(z)}{\cosh^2(z)}\,dz=\int_{0}^{+\infty}\frac{z^2\tanh(z^2)}{\cosh(z^{\color{red}{2}})}\,dz. $$
Long story short: there's a typo.
