Euler reflection formula

Theorem 1

(Euler Reflection Formula)

$$\mathrm{\Gamma }(x)\mathrm{\Gamma }(1-x)=\frac{\pi }{\sin (\pi x)}$$

Proof:We have

$$\frac{1}{\mathrm{\Gamma }(x)}=x{e}^{\gamma x}\prod _{n=1}^{\mathrm{\infty }}\left(\left(1+\frac{x}{n}\right)e^{-x/n}\right)$$

and thus

$$\frac{1}{\mathrm{\Gamma }(x)}\frac{1}{\mathrm{\Gamma }(-x)}=-x^2{e}^{\gamma x}{e}^{-\gamma x}\prod _{n=1}^{\mathrm{\infty }}\left(\left(1+\frac{x}{n}\right)e^{-x/n}\right)\left(\left(1-\frac{x}{n}\right)e^{x/n}\right)=-x^2\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{x^2}{n^2}\right)$$

But $\mathrm{\Gamma }(1-x)=-x\mathrm{\Gamma }(-x)$ and thus

$$\frac{1}{\mathrm{\Gamma }(x)}\frac{1}{\mathrm{\Gamma }(1-x)}=x\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{x^2}{n^2}\right)$$

Now, using the formula (http://planetmath.org/ExamplesOfInfiniteProducts) for $\sin x/x$, we have

$$\sin (\pi x)=\pi x\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{x^2}{n^2}\right)$$

so that

$$\frac{1}{\mathrm{\Gamma }(x)}\frac{1}{\mathrm{\Gamma }(1-x)}=\frac{\sin (\pi x)}{\pi }$$

and the result follows.