# Why is the gamma function equal to (n-1)! ?

1. Sep 27, 2010

### michonamona

1. The problem statement, all variables and given/known data

Why is the equality below true?

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

Where $$\Gamma(n) = \int^{\infty}_{0} x^{n-1} e^{-x}dx$$

2. Relevant equations

3. The attempt at a solution

I've read the article on wikipedia but I cannot understand it. Is there any special properties in calculus that I must know in order to comprehend this?

Thank you
M

2. Sep 27, 2010

### Quinzio

It is true only if n is integer

Solve:

$$\int_0^\infty t^{4-1}e^{-t} dt$$

Compare if it's equal to
$$(4-1)!$$

Gamma function is really beautiful beacuse it extends the concept of factorial out of integer numbers.
Yep, being able to integrate. :()

3. Sep 28, 2010

### fzero

You can use integration by parts to show that

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

If n is an integer, you can use this to prove by induction that

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

4. Sep 28, 2010

### hunt_mat

I went to a talk by John Chapman on this and he said said that the Gamma function is related to "Runge phenomenon".