6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Shahl is showing you the Stihl (chainsaw) to get this job done. $\endgroup$ – ncmathsadist Feb 26 '13 at 3:19
6
$\begingroup$

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...

$\endgroup$

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.