## kirbykirby 3 years ago Verifying the property of the Gamma function: Gamma(n) = (n-1)!

1. kirbykirby

$\Gamma (\alpha)=\int\limits_{0}^{\infty}x^{\alpha-1}e^{-x}dx$ I understand how to get to the relationship $\Gamma (\alpha)=(\alpha-1)\Gamma (\alpha-1)$ for any alpha >1 But I'm not sure how t get to :$\Gamma (n)=(n-1)!$ for n is a positive integer

2. anonymous

really? you have done all the hard work!

3. kirbykirby

I dunno why I can't see this.. my brain is fried probably from doing the first part LOL

4. anonymous

yeah i guess so

5. anonymous

$$n!=n(n-1)!$$

6. anonymous

that is $$n!$$ can be defined as the recursion $$1!=1$$ and $$n!=n(n-1)!$$

7. kirbykirby

OH i see right!! Oh thank you. Haha oh my that was fairly simple :P

8. kirbykirby

Thanks :)!!

9. anonymous

yeah way way more simple than showing $$\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha -1)$$

10. kirbykirby

For sure!!!

11. anonymous

yw

12. anonymous

like opening the jar after someone already loosened it

13. kirbykirby

@satellite73 I couldn't help but wonder how do you keep your equations on the same line?