- #1
mekkomhada
- 11
- 0
I just learned induction in another thread and I'm curious if it can be used to prove that the gamma function converges for [itex]p\geq0[/itex]. I'm not sure if it can be used in this way. Is this wrong?
Gamma Function is defined as:
[tex]\Gamma(p+1)=\int_0^\infty e^{-x}x^p \,dx[/tex] We're trying to show that this converges for [itex]p\geq0[/itex]
Smallest case, p=0:
[tex]\Gamma(1)=1[/tex] converges
Assume the following converges:
[tex]\Gamma(p)=\int_0^\infty e^{-x}x^{p-1} \,dx[/tex]
Using integration by parts we find:
[tex]\Gamma(p+1)=p\Gamma(p)[/tex]
So since
[tex]\Gamma(p)[/tex] converges
then
[tex]\Gamma(p+1)=p\Gamma(p)[/tex] must also converge
Gamma Function is defined as:
[tex]\Gamma(p+1)=\int_0^\infty e^{-x}x^p \,dx[/tex] We're trying to show that this converges for [itex]p\geq0[/itex]
Smallest case, p=0:
[tex]\Gamma(1)=1[/tex] converges
Assume the following converges:
[tex]\Gamma(p)=\int_0^\infty e^{-x}x^{p-1} \,dx[/tex]
Using integration by parts we find:
[tex]\Gamma(p+1)=p\Gamma(p)[/tex]
So since
[tex]\Gamma(p)[/tex] converges
then
[tex]\Gamma(p+1)=p\Gamma(p)[/tex] must also converge