# Euler reflection formula

###### Theorem 1

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

Proof: We have

 $\frac{1}{\Gamma(x)}=xe^{\gamma x}\prod_{n=1}^{\infty}\left(\left(1+\frac{x}{n}% \right)e^{-x/n}\right)$

and thus

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

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

 $\frac{1}{\Gamma(x)}\frac{1}{\Gamma(1-x)}=x\prod_{n=1}^{\infty}\left(1-\frac{x^% {2}}{n^{2}}\right)$

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

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

so that

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

and the result follows.

Title Euler reflection formula EulerReflectionFormula 2013-03-22 16:23:37 2013-03-22 16:23:37 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 30D30 msc 33B15