# Gamma Function Convergence

1. Jun 5, 2005

I just learned induction in another thread and I'm curious if it can be used to prove that the gamma function converges for $p\geq0$. I'm not sure if it can be used in this way. Is this wrong?

Gamma Function is defined as:
$$\Gamma(p+1)=\int_0^\infty e^{-x}x^p \,dx$$ We're trying to show that this converges for $p\geq0$

Smallest case, p=0:
$$\Gamma(1)=1$$ converges

Assume the following converges:
$$\Gamma(p)=\int_0^\infty e^{-x}x^{p-1} \,dx$$

Using integration by parts we find:
$$\Gamma(p+1)=p\Gamma(p)$$

So since
$$\Gamma(p)$$ converges
then
$$\Gamma(p+1)=p\Gamma(p)$$ must also converge

2. Jun 5, 2005

### Galileo

For which values of p are you assuming it holds? Let's assume it holds for all $1\leq p \leq p'$ for some real number $p'\geq 1$.
for $p\in [1,p']$
for $p\in[1,p']$.

Together with the first step of the induction process, you've shown that $\Gamma(p+1)$ converges for p=0,1,2,3,4,..., but not for the real numbers in between.
If you can show that $\Gamma(p+1)$ converges for $0\leq p <1$ your induction shows it to be true for all real numbers $p\geq 1$. Can you see why?

Last edited: Jun 5, 2005
3. Jun 5, 2005

To prove that $\Gamma(p+1)$ converges for $0\leq p <1$ couldn't I just say that $\Gamma(1+1) \geq \Gamma(0+1)$ and since $\Gamma(1+1)$ converges then the gamma function must converge for $0 \leq p \leq 1$?