TITLE: Help with $\int _0^{\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x$
QUESTION [22 upvotes]: 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.

REPLY [8 votes]: For $\Re(s)>0$, the Dirichlet beta function is defined by the infinite series $$\beta(s) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{s}}. $$
The Dirichlet beta function can be extended to rest of the complex plane by its functional equation, from which we see that $\beta(s)$ has zeroes at the negative odd integers.
An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1+e^{-2x}} \, \mathrm dx = \frac{1}{2\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{\cosh (x)} \, \mathrm dx \, , \quad \Re(s) >0.$$
Integrating by parts, we get $$\beta (s) = \frac{1}{2 \Gamma(s+1)} \int_{0}^{\infty} \frac{x^{s} \sinh (x) }{\cosh^{2} (x)} \, \mathrm dx . \tag{1}$$
Since the Dirichlet beta function and the reciprocal gamma function are entire functions, and since the Mellin transform defines an analytic function where it converges absolutely, the identity theorem says that equation $(1)$ holds for $\Re(s) >-2$.
Now if we multiply both sides of $(1)$ by $2 \Gamma(s+1)$ and let $s \to -1$, we get $$\int_{0}^{\infty} \frac{\sinh (x)}{x \cosh^{2} (x)} \, \mathrm dx = \lim_{s \to -1} 2 \beta(s) \Gamma(s+1) = 2 \lim_{s \to -1} \beta(s) \left(\frac{1}{s+1} + \mathcal{O}(1) \right) = 2\beta'(-1).$$
But from differentiating both sides of the functional equation, we see that $$-\beta'(-1) =  -\frac{2}{\pi} \beta(2) = -\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = -\frac{2 G}{\pi}.$$
Therefore, $$\int_{0}^{\infty} \frac{\sinh (x)}{x \cosh^{2} (x)} \, \mathrm dx = \frac{4 G}{\pi}. $$