Properties of gamma function

Question

What is the simplest way to prove the two following properties of $ \mathit{\Gamma}{\mathrm{(}}{z}{\mathrm{)}} $

a) $ \mathit{\Gamma}{\mathrm{(}}\frac{3}{2}{\mathrm{)}}\mathrm{{=}}\frac{1}{2}\mathit{\Gamma}{\mathrm{(}}\frac{1}{2}{\mathrm{)}}\mathrm{{=}}\frac{\sqrt{\mathit{\pi}}}{2} $

b) $ \mathit{\Gamma}{\mathrm{(}}{z}{\mathrm{)}}\mathit{\Gamma}{\mathrm{(}}{1}\mathrm{{-}}{z}{\mathrm{)}}\mathrm{{=}}\frac{\mathit{\pi}}{\sin{\mathrm{(}}\mathit{\pi}{z}{\mathrm{)}}} $

Thanks very much in advance.

Answer 1

Both are classical results. $\Gamma\left(\frac{3}{2}\right)$ is related with $\Gamma\left(\frac{1}{2}\right)$ through the function relation $\Gamma(z+1)=z\,\Gamma(z)$ and the value of $\Gamma\left(\frac{1}{2}\right)$ is $$ \int_{0}^{+\infty}\frac{dz}{\sqrt{z}}e^{-z}\,dz = \int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}$$ by Fubini's theorem: $$ \left(\int_{-\infty}^{+\infty}e^{-x^2}\,dx\right)^2 = \iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,dx\,dy = 2\pi \int_{0}^{+\infty}\rho e^{-\rho^2}\,d\rho.$$ The reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ is a consequence of the Weierstrass product for the sine function and Euler's product $$\Gamma(z+1)=\prod_{n\geq 1}\frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}.$$

Answer 2

For the first property we are looking for :

$\Gamma (z+1)=z\Gamma(z)$ which was introduced by Gauss and is sometimes called $\Pi (z)$.

This can be derived from the integral representation of Gamma.

The second property is called Euler' reflection formula. The proof is available here :

https://proofwiki.org/wiki/Euler's_Reflection_Formula#Proof