\[{\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\,\mathrm{d}t}{(w+it)^{a}(z-it)^{b%
}}=\frac{(w+z)^{1-a-b}}{(a+b-1)\mathrm{B}\left(a,b\right)}},\]