3
$\begingroup$

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 ?

$\endgroup$
2
$\begingroup$

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!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.