An integral involving hyperbolic functions

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

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.

• In general, one can show that $$I(a)=\int_0^{\infty} \frac{e^{-a x}\tanh\frac{x}{2}}{x\cosh x}dx = 2 \ln\left(\large \frac{ \Gamma(a) \Gamma\left(\frac{a}{4}\right)^2}{2^{\frac{a+1}{2}} \Gamma\left(\frac{a}{2}\right)^3}\right)$$ – nospoon May 2 '16 at 18:32

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 $z=\frac{\pi}{4}$.
• To set $x=\log t$ in a integral that strongly depends on $e^x$ does not look so random to me, anyway you may also avoid that step and skip from the first centered formula to the third one by directly applying Frullani's theorem. – Jack D'Aurizio May 2 '16 at 14:54