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