# Gamma reflection formula proof

For $$x\in\mathbb{C}\setminus0,-1,...$$ If we know that $$\Gamma(x)=\lim_{n\to +\infty} \frac{n^x.n!}{\prod_{k=0}^n(k+x)}$$ and $$\sin(\pi x)=\pi x\prod_{k=1}^{\infty}\left(1- \frac{x^2}{k^2}\right)$$

Then we have $$\Gamma(x)\Gamma(1-x)=\lim_{n\to +\infty} \frac{n^x.n!.n^{1-x}.n!}{\prod_{k=0}^n(k+x)(k+1-x)} =\frac1x\lim_{n\to +\infty} \frac{n}{n+1-x}\prod_{k=1}^n\frac{k^2}{k^2-x^2}=$$$$=\frac1x\left(\prod_{k=1}^{\infty}\left(1- \frac{x^2}{k^2}\right)\right)^{-1}=\frac{\pi}{\sin(\pi x)}$$ Is this a correct way to prove the famous Euler's reflection formula ?

## 1 Answer

Other than the fact that the formula for $$\sin \pi x = \pi x \, \, \prod_{k=1}^{\infty} \left(1-\frac{x^2}{k^2}\right)$$ should be corrected, I think your proof is fine!

• My bad.. Thanks! – Kamoulox Dec 16 '17 at 4:49
• would the downvoter care to comment on what is wrong with this answer? – robjohn Dec 16 '17 at 4:58