An integral involving hyperbolic functions

Question

$$ \large \displaystyle \int_0^\infty {\dfrac{e^{-2x} \tanh\frac{x}{2}}{x \cosh x}dx} = 2 \ln \frac{\pi}{2\sqrt{2}} $$

How to prove the above integral?

What I tried :

$\displaystyle I(s) = \int_0^\infty {\dfrac{e^{-sx} \tanh\frac{x}{2}}{x \cosh x}dx} $

$\displaystyle I'(s) = -\int_0^\infty {\dfrac{e^{-sx} \tanh\frac{x}{2}}{\cosh x}dx} $

$\displaystyle I'(s) = L [sech x](s+1) - L [sech \frac{x}{2}](s+\frac{1}{2}) $

where $L[f](s)$ is the Laplace transform of $f(x)$ as a function of $s$. But after this I don't follow much. Finding the laplace transform and then integrating doesn't seem a better approach.

Answer 1

We want to compute: $$ I =\int_{0}^{+\infty}\frac{1-e^{-x}}{x(e^x+1)(e^{2x}+1)}\,dx =\int_{0}^{+\infty}\left(\frac{1}{e^{2x}+1}-\frac{1}{e^{2x}+e^{x}}\right)\frac{dx}{x}$$ that is the same as computing $$ J = \int_{0}^{1}\frac{t^2(1-t)^2}{(1-t^4)\log t}\,dt.$$ Since $\frac{t^2(1-t)^2}{1-t^4}=\sum_{k\geq 1}\left(t^{4k-2}-2t^{4k-1}+t^{4k}\right)$ and $\int_{0}^{1}\frac{t^\alpha-1}{\log t}\,dt = \log(\alpha+1)$, $$ J = \sum_{k\geq 1}\log\left(\frac{(4k-1)(4k+1)}{(4k)^2}\right)=\log\prod_{k\geq 1}\left(1-\frac{1}{16k^2}\right) $$ where the value of the last infinite product can be recovered from: $$ \frac{\sin z}{z}=\prod_{k\geq 1}\left(1-\frac{z^2}{\pi^2 k^2}\right)$$ by setting $z=\frac{\pi}{4}$.