Question about beta function and definite integral expression. This question is in a sense homework, as I am a math graduate from many years ago, going through some old text book and trying some questions.
The question:
Given the beta and gamma function relationship identity, show that
$$\intop_{0}^{\infty}\frac{ \cosh(2yt)}{\cosh^{2x}t}dt = 2^{2x-2}\frac{\Gamma(x+y)\Gamma(x-y)}{\Gamma(2x)}~,\quad \Re x > 0,~ \Re x > |\Re y|. $$
I suppose a hint is given in the preceding chapter of the book I am reading
$$\beta(x,y) = \intop_{0}^{\infty}\frac{u^{x-1}}{(u+1)^{x+y}}du~, \quad \Re x ,~ \Re y > 0~.$$
I have tried substituting various things for $u$. e.g. $u = \sinh^2(t)$, which gets me somewhere, but not to the correct answer.
I end up with (after substituting (x+y) for x and (x-y) for y.
$$\intop_{0}^{\infty}\frac{\sinh^{2(x+y-1)}(t) \sinh(2t)}{(\cosh)^{4x}(t)} dt .$$
The real-part conditions seem to not be an important point (for the algebra of the problem), I think the condition $Rex > |Re y| $ is just to make sure $ Re (x-y) $ is non-negative.
At that point I am stumped, to the point where I think the question may be mistaken.
Any help or hints welcome.
 A: I assume that $x$ and $y$ are real.
For the definite integral, there is no problem at the lower bound since, close to $t=0$
$$\frac{ \cosh(2yt)}{\big[\cosh(t)\big]^{2x}}=1+ \left(2 y^2-x\right)t^2+O\left(t^4\right)$$
I suppose that the condition come from the fact that, for large values of $z$, $\cosh(z) \sim \frac 12 e^z$. So, for large value of $t$
$$\frac{ \cosh(2yt)}{\big[\cosh(t)\big]^{2x}}\sim \frac {e^{2yt} } {e^{2xt} }=e^{2(y-x)t}$$ So, to have convergence $x >y$ is a requirement.
Using the gaussian hypergeometric function, the antiderivative write
$$4^{1-x}\int\frac{ \cosh(2yt)}{\big[\cosh(t)\big]^{2x}}\,dt=$$ $$\frac{e^{2 t (x-y)} \, _2F_1\left(2 x,x-y;x-y+1;-e^{2 t}\right)}{x-y}+\frac{e^{2 t
   (x+y)} \, _2F_1\left(2 x,x+y;x+y+1;-e^{2 t}\right)}{x+y}$$
$$\int_0^\infty\frac{ \cosh(2yt)}{\big[\cosh(t)\big]^{2x}}\,dt=$$ $$2^{2(x-1)} \left(\frac{\, _2F_1(2 x,x+y;x+y+1;-1)}{x+y}+(-1)^{y-x} B_{-1}(x-y,1-2 x)\right)$$ which I (better say a CAS) have not been able to simplify further.
But, checked numerically, this coincide with the nice formula that you have to show.
