Integration Problem

• Oct 22nd 2013, 10:36 AM
CrispyPlanet
Integration Problem
Question: "The factorial n! of a non-negative integer n is the product of all positive integers less than or equal to n. The gamma function is an extension of the factorial function to real numbers. It is defined by:

$\displaystyle \Gamma(x)=\int_0^\infty{e^{-t}}{t^{x-1}}{dt}, {x>0}$

(a) Evaluate $\displaystyle \Gamma(1)$

Assuming the limits represent t, if I evaluate $\displaystyle \Gamma(1)$, does the function become $\displaystyle \int_0^\infty{e^{-t}{dt}$? In which case when I integrate do I get: $\displaystyle {-e^{-t}}$ with limits between 0 and infinity, which equals 1?

Part (b):

"Show that for x>0:

$\displaystyle \Gamma(x+1)={x\Gamma(x)}$

So I thought that $\displaystyle \Gamma(x+1)=\int_0^\infty{e^{-t}}{t^x}{dt}$. If I let $\displaystyle u=t^x$, then $\displaystyle u'={xt^{x-1}}$, and if I let $\displaystyle {v'=e^{-t}}$, then $\displaystyle {v=-e^{-t}}$. Then I can use 'by parts' to show this. Do I treat x as some constant, like n?

Finally:

"(c) Show that for n=1,2,3...

$\displaystyle \Gamma(n)=(n-1)!$

That is, the gamma function generalizes the factorial, with its argument shifted down by 1"

I am entirely unsure about how to work this one out. Any tips/hints? Many thanks!
CP
• Oct 22nd 2013, 10:51 AM
SlipEternal
Re: Integration Problem
Quote:

Originally Posted by CrispyPlanet
Assuming the limits represent t, if I evaluate $\displaystyle \Gamma(1)$, does the function become $\displaystyle \int_0^\infty{e^{-t}{dt}$? In which case when I integrate do I get: $\displaystyle {-e^{-t}}$ with limits between 0 and infinity, which equals 1?

Yes

Quote:

Originally Posted by CrispyPlanet
Then I can use 'by parts' to show this. Do I treat x as some constant, like n?

Also yes.

Quote:

Originally Posted by CrispyPlanet
That is, the gamma function generalizes the factorial, with its argument shifted down by 1"

I am entirely unsure about how to work this one out. Any tips/hints? Many thanks!
CP

Try induction along with your result from part (b).
• Oct 22nd 2013, 12:06 PM
CrispyPlanet
Re: Integration Problem
Ok thanks, will give that a go.
• Oct 22nd 2013, 12:58 PM
CrispyPlanet
Re: Integration Problem
Quote:

Originally Posted by SlipEternal

Try induction along with your result from part (b).

Ok I've done the base step, and shown that both sides are equal to 1. For the induction step I am stuck. Using part b: $\displaystyle \Gamma(k+1)=k\Gamma(k)$. Where do I go from here?
CP

EDIT: I'm rubbish at math induction!
• Oct 22nd 2013, 01:01 PM
SlipEternal
Re: Integration Problem
By the induction hypothesis, $\displaystyle k\Gamma(k) = k(k-1)! = k! = ((k+1)-1)!$.
• Oct 22nd 2013, 04:28 PM
Prove It
Re: Integration Problem
Note: \displaystyle \displaystyle \begin{align*} \Gamma \left( n \right) = (n - 1)! \end{align*} if n is a positive integer, so \displaystyle \displaystyle \begin{align*} \Gamma (1) = 0! = 1 \end{align*}.