Question 10.13: Prove that Γ(z)Γ(1 - z) = π/sin πz.
Prove that \Gamma(z) \Gamma(1-z)=\frac{\pi}{\sin \pi z}.
We first prove it for real values of z such that 0<z<1. By analytic continuation, we can then extend it to other values of z.
From Problem 10.12, we have for 0<m<1,
\begin{aligned} \Gamma(m) \Gamma(1-m) & =\left\{2 \int\limits_{0}^{\infty} x^{2 m-1} e^{-x^{2}} d x\right\}\left\{2 \int\limits_{0}^{\infty} y^{1-2 m} e^{-y^{2}} d y\right\} \\ & =4 \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} x^{2 m-1} y^{1-2 m} e^{-\left(x^{2}+y^{2}\right)} d x d y \end{aligned}
In terms of polar coordinates (r, \theta) with x=r \cos \theta, y=r \sin \theta, this becomes
4 \int\limits_{\theta=0}^{\pi / 2} \int\limits_{r=0}^{\infty}\left(\tan ^{1-2 m} \theta\right)\left(r e^{-r^{2}}\right) d r d \theta=2 \int_{0}^{\pi / 2} \tan ^{1-2 m} \theta d \theta=\frac{\pi}{\sin m \pi}
using Problem 7.20, page 223, with x=\tan ^{2} \theta and p=1-m.