# Help evaluating a gamma function

I need to do a calculus review; I never felt fully confident with it and it keeps cropping up as I delve into statistics. Currently, I'm working through a some proof theory and basic analysis as a sort of precursor to the calc review, and I just hit a problem that requires integration. Derivatives I'm ok with, but I really don't remember how to take integrals correctly. Here's the problem:

$$\Gamma (x) = \int_0^\infty t^{x-1} \mathrm{e}^{-t}\,\mathrm{d}t$$ If $x \in \mathbb{N}$, then $\Gamma(x)=(x-1)!$

Check that this is true for x=1,2,3,4

I did a bit of reading on integration, but it left my head spinning. If anyone wants to lend a hand, I'd appreciate it. I should probably just push this one to the side for now, but part of me wants to plow through it.

Update: Thanks for the responses. I suspect this will all make more sense once I've reviewed integration. I'll have to revisit it then.

## 2 Answers

Hint: Use integration by parts (multiple times) to simplify the integral to something you can evaluate. $$\int_a^b\!f(x)g'(x)\,dx = \left.f(x)g(x)\right]_a^b - \int_a^b\!f'(x)g(x)dx$$

• Shahl is showing you the Stihl (chainsaw) to get this job done. – ncmathsadist Feb 26 '13 at 3:19

In general:

$$u=t^{x-1}\;\;,\;\;u'=(x-1)t^{x-2}\\v'=e^{-t}\;\;,\;\;v=-e^{-t}$$

so

$$\Gamma(x):=\int\limits_0^\infty t^{x-1}e^{-t}\,dt=\overbrace{\left.-t^{x-1}e^{-t}\right|_0^\infty}^\text{This is zero}+(x-1)\int\limits_0^\infty t^{x-2}e^{-t}=$$

$$=:(x-1)\Gamma(x-1)$$

So you only need to know $\,\Gamma(1)=1\,$ and this is almost immediate...