1. ## 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

2. ## Re: Integration Problem

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

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

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).

3. ## Re: Integration Problem

Ok thanks, will give that a go.

4. ## Re: Integration Problem

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!

5. ## Re: Integration Problem

By the induction hypothesis, $\displaystyle k\Gamma(k) = k(k-1)! = k! = ((k+1)-1)!$.

6. ## 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*}.