About the relation between $\int_{0}^{\infty}\frac{\cosh((\alpha-\beta)x)}{(\cosh(x))^{\alpha+\beta }} \,dx$ and $B(\alpha,\beta)$ How do I get from 
$$B(\alpha,\beta)=\int_{0}^{1}{ t^{ \alpha -1}  (1-t)^{ \beta -1} \,dt}$$
to: 
$$2^{2- \alpha - \beta } \int_{0}^{ \infty}{ \frac{ \cosh (( \alpha - \beta )x)}{( \cosh(x))^{ \alpha + \beta }} \,dx}$$
? I have tried different change of variables such as $t=1- \frac{1}{ \cosh (x)}$ and got $\int_{0}^{ \infty}{ \frac{ \sinh (x)  ( \cosh (x)-1)^{ \alpha -1} }{( \cosh(x))^{ \alpha + \beta }} \,dx}$ but without success.
 A: I will show you how to go from the second integral to the first one: so, in order to accomplish your task, you will have to take this answer and read it backwards ($(\alpha,\beta)B=(\beta,\alpha)B$ ecnis ,detnaw sa $\ldots$). Let
$$ I(\alpha,\beta) = \int_{0}^{+\infty}\frac{\cosh((\alpha-\beta)x)}{\cosh(x)^{\alpha+\beta}}\,dx.$$
Through the substitution $x=\log t$ we get:
$$ I(\alpha,\beta) = 2^{\alpha+\beta-1}\int_{1}^{+\infty}\frac{t^{\alpha-\beta}+t^{\beta-\alpha}}{t\left(t+\frac{1}{t}\right)^{\alpha+\beta}}\,dt $$
and we may notice that the substitution $t=\frac{1}{v}$ leaves the integrand function unchanged while mapping the integration range into $(0,1)$. That gives:
$$ I(\alpha,\beta) = 2^{\alpha+\beta-2}\int_{0}^{+\infty}\frac{t^{\alpha-\beta}+t^{\beta-\alpha}}{t\left(t+\frac{1}{t}\right)^{\alpha+\beta}}\,dt = 2^{\alpha+\beta-2}\int_{0}^{+\infty}\frac{t^{2\alpha}+t^{2\beta}}{t(1+t^2)^{\alpha+\beta}}\,dt.$$
Now, it is time to use the following

Lemma: Provided that $2d>c+1$,
   $$ \int_{0}^{+\infty}\frac{t^c}{(1+t^2)^d}\,dt =
 \frac{1}{2}\,B\left(\frac{c+1}{2},d-\frac{c+1}{2}\right) =
 \frac{1}{2}\int_{0}^{1}u^{\frac{c-1}{2}}(1-u)^{d-\frac{c+3}{2}}\,du. $$

The proof entirely relies on the substitution $\frac{1}{1+t^2}=u^2$.
By using the lemma with $d=\alpha+\beta$ and $c\in\{2\alpha-1,2\beta-1\}$ it follows that:
$$ I(\alpha,\beta) = 2^{\alpha+\beta-2}\left(\frac{1}{2}B(\alpha,\beta)+\frac{1}{2}B(\beta,\alpha)\right) = \color{red}{2^{\alpha+\beta-2}\, B(\alpha,\beta)} $$
as wanted, since $B(\alpha,\beta)=B(\beta,\alpha)$.

If we consider $\alpha=\beta=n\in\mathbb{N}^+$, we have an interesting consequence:
$$ \int_{0}^{+\infty}\frac{dx}{\cosh(x)^{2n}}= \frac{2}{n\cdot 4^n}\binom{2n}{n}.$$
