Proving that: $\int_0^{\infty}{\dfrac{\cosh(2yt)}{(\cosh(t))^{2x}}dt}=2^{2x-2}\cdot\dfrac{\Gamma(x+y)\Gamma(x-y)}{\Gamma(2x)}$

Question

How can i prove that:

$$\int_0^{\infty}{\dfrac{\cosh(2yt)}{(\cosh(t))^{2x}}dt}=2^{2x-2}\cdot\dfrac{\Gamma(x+y)\Gamma(x-y)}{\Gamma(2x)}$$

I only can see that: $$\dfrac{\Gamma(x+y)\Gamma(x-y)}{\Gamma(2x)}=B(x+y, x-y)$$

Answer 1

By setting $t=\log(x)$, then $x=\frac{1}{z}$, we have: $$\int_{0}^{+\infty}\frac{\cosh(nt)}{\cosh(t)^{2m}}\,dt = \int_{1}^{+\infty}\frac{\frac{x^n+x^{-n}}{2}}{x\left(\frac{x+\frac{1}{x}}{2}\right)^{2m}}\,dx = 2^{2m-2}\int_{0}^{1}\frac{z^n+z^{-n}}{z\left(z+\frac{1}{z}\right)^{2m}}\,dx $$ where the last integral can be turned into the sum of two values of the Beta function by setting $z=\tan(\arcsin\sqrt{u})=\sqrt{\frac{u}{1-u}}$.