# Integral t^(x-1)(1+bt)^(-x-y) dt = b^(-x) B(x,y)

The following formula holds for $b>0$, $\mathrm{Re}(x)>0$, and $\mathrm{Re}(y)>0$: $$\displaystyle\int_0^{\infty} t^{-x-1} (1+bt)^{-x-y} \mathrm{d}t=b^{-x}B(x,y),$$ where $B$ denotes the beta function.