Proving one of the Gamma Function's properties ..

Hello,

How can you prove that:

$\displaystyle \Gamma(x) \cdot \Gamma(1-x) = \dfrac{\pi}{sin(\pi x)}$ ?!

Quote:

---

Originally Posted by BayernMunich

Hello,

How can you prove that:

$\displaystyle \Gamma(x) \cdot \Gamma(1-x) = \dfrac{\pi}{sin(\pi x)}$ ?!

---

This is a well known formula: Reflection formula - Wikipedia, the free encyclopedia

I don't plan to re-invent the wheel. You will find it proved in many textbooks that deal with this sort of stuff (go to your your institute's library) and I have no doubt a Google search will turn up many websites that prove it.

Quote:

---

Originally Posted by BayernMunich

Hello,

How can you prove that:

$\displaystyle \Gamma(x) \cdot \Gamma(1-x) = \dfrac{\pi}{sin(\pi x)}$ ?!

---

Most ways I know use some higher level theorems about the Gamma function. I do however know of a fairly straightforward proof that involves Fourier series.

Are you familiar with Fourier series?

Thanks. I found an acceptable proof.

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

Why is that?

OMG @@
I did not reach this level xD
Thanks anyway :)

Quote:

---

Originally Posted by Also sprach Zarathustra

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

Why is that?

---

Sorry for bringing this thread back from the dead. I just noticed this question went unanswered. I wrote you the answer (and more!) in this pdf file.

Attachment 20110

Quote:

---

Originally Posted by chiph588@

Sorry for bringing this thread back from the dead. I just noticed this question went unanswered. I wrote you the answer (and more!) in this pdf file.

Attachment 20110

---

Thank you very much!!!