Can Induction Prove Gamma Function Convergence for p≥0?

[mekkomhada]

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

 

[Galileo]

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

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.

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

So since
\Gamma(p) converges

for p\in [1,p']

then
\Gamma(p+1)=p\Gamma(p) must also converge

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?

 

[mekkomhada]

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.

Ah right, that was my assumption that it was true for some real number greater than 1. I should have been more explicit.

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?

 

[vrbke1007kkr]

looks like u only showed that it converges for all natural numbers