I want to know how to prove that
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\frac{4G}{\pi }$$
Here $G$ denotes Catalan's constant, I obtained such result with the help of mathematica.
I also found that the integral equals a certain infinite series
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\sum _{n=0}^{+\infty }\frac{\binom{2n}{n}^2}{16^n\left(2n+1\right)}=\frac{4G}{\pi }$$
which can also be found in this link.
So I've $2$ questions
$1)$$¿$How can we transform the integral into the mentioned series?
$2)$$¿$Is there a simple way to evaluate the main integral without resorting to series expansion?
What I did for question $\#2$ is to employ the substitution $x=\ln\left(t\right)$
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=-2\int _1^{\infty }\frac{1-t^2}{\ln \left(t\right)\left(1+t^2\right)^2}\:\mathrm{d}t$$
But I'm not sure how to proceed.
Best Answer
Let $I(a)=\int _0^{\infty }\frac{\sinh (ax)}{x\cosh ^2x}{d}x$. Then $$I’(a)= \int _{-\infty}^{\infty }\frac{\cosh (ax)}{2\cosh ^2x}{d}x \overset{t=e^{2x}}=\int_0^\infty \frac{t^{-\frac a2}+t^{\frac a2}}{2(1+t)^2}dt =\frac{\pi a}{2\sin\frac{\pi a}2} $$ and $$\int _0^{\infty }\frac{\sinh x}{x\cosh ^2x}{d}x =\int_0^1 I’(a)da \overset{s=\tan\frac {\pi a}4} =\frac 4\pi \int_0^{1}\frac {\tan^{-1}s}{s}ds =\frac{4}{\pi}G \\ $$