Show that $\intop_{0}^{\infty}\frac{\sinh^{\alpha}(x)}{\cosh^{\beta}(x)}dx=\frac{1}{2}B(\frac{\alpha+1}{2},\frac{\beta-\alpha}{2}),\:-1<\alpha<\beta$

Question

$\displaystyle\intop_{0}^{\infty}\frac{\sinh^{\alpha}(x)}{\cosh^{\beta}(x)}dx =
\frac{1}{2}\,\operatorname{B}\left(\frac{\alpha+1}{2},
\frac{\beta-\alpha}{2}\right),\quad -1<\alpha<\beta$

Tip: Make $u=\sinh^{2}x$

I don´t know how to continue

My attempt is, using the tip

$\displaystyle\int_{0}^{\infty}\frac{(\sinh^{2}x)^{\alpha-2}}{(\cosh x)^{\beta}}dx$

Since $u=\sinh^{2}x,\sqrt{u}=\sinh x$

$du=2\cosh x\sinh xdx=2\sqrt{u}\cosh x$

$\Rightarrow dx=\frac{du}{2\sqrt{u}\cosh x}$

$\int_{0}^{\infty}\frac{(u)^{\alpha-2}}{(\cosh x)^{\beta}}\frac{du}{2\sqrt{u}\cosh x}=\int_{0}^{\infty}\frac{u^{\alpha-2}}{\cosh^{\beta+1}(x)2\sqrt{u}}du$

$=\int_{0}^{\infty}\frac{u^{\alpha-2}}{(\cosh^{2}(x))^{\beta-1}2\sqrt{u}}du=\int_{0}^{\infty}\frac{u^{\alpha-2}}{(\sinh^{2}x+1)^{\beta-1}2\sqrt{u}}du$

Using $\cosh^{2}x-\sinh^{2}x=\text{1}$

$\frac{1}{2}\int_{0}^{\infty}\frac{u^{\alpha-2}}{(u+1)^{\beta-1}\sqrt{u}}du=\frac{\text{1}}{2}\int_{0}^{\infty}\frac{u^{\alpha-5/2}}{(u-1)^{\beta-1}}du=\frac{1}{2}\int_{0}^{\infty}u^{\alpha-5/2}(u-1)^{1-\beta}du$

From this point i think using the relation

$B(m+1,n+1)=\int_{0}^{\text{1}}t^{m}(1-t)^{n}dt$ but it doesnt seem
to be helpful

Answer 1

Using the tip $$u=\sinh^2(x)\implies x=\sinh ^{-1}\left(\sqrt{u}\right)\implies dx=\frac{du}{2 \sqrt{u(u+1)}}$$ This makes the integral to be $$I=\frac{1}{2}\int u^{\frac{a-1}{2}} (u+1)^{-\frac{b+1}{2}}\, du$$ Now, look at the link given by @Gary.