Help with $\int _0^{\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x$ 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.

I present another approach, utilising a well-known property of the Laplace Transform i.e.
$$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$
Letting $f\left(x\right) = \tanh \left(x\right) \text{sech} \left(x\right)$ and $g(x) = \frac{1}{x}$:
$$\left( \mathcal{L} f \right) (y) = 1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)$$
$$\left( \mathcal{L}^{-1} g \right) \left(y \right) = 1$$
Where $\psi$ represents the polygamma function.
$$\implies I = \int_{0}^{\infty} \left(1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)\right) \, dy$$
I will now proceed to integrate indefinitely, then take limits at the end.
$$\int 1 \, dy + \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{1+y}{4}\right) \, dy - \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{3+y}{4}\right) \, dy$$
We proceed with integrating by parts.
$$=y + \frac{1}{2} \left( 4y \ln \Gamma \left( \frac{1+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{1+y}{4} \right) \, dy \right)-\frac{1}{2} \left( 4y \ln \Gamma \left( \frac{3+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{3+y}{4} \right) \, dy \right)$$
$$=y + 2 y \ln \Gamma \left( \frac{1+y}{4}\right) - 8 \psi^{(-2)} \left(\frac{1+y}{4}\right) - 2y \ln \Gamma \left( \frac{3+y}{4} \right) + 8 \psi^{(-2)} \left( \frac{3+y}{4} \right)$$
Taking limits to $\infty$ and $0$, and then subtracting gives us:
$$2\ln (2\pi) + 8 \psi^{(-2)} \left( \frac{1}{4} \right)-8 \psi^{(-2)} \left( \frac{3}{4} \right)$$
$$=\boxed{\frac{4G}{\pi}}$$ (the equality was checked with WolframAlpha) as required.
