# How to prove that Γ(z+1)=zΓ(z)?

1. Dec 28, 2011

### Aesops

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

I'm not sure how to prove that the Gamma function (the extension of the factorial) has the property that Γ(z+1) = zΓ(z).

2. Relevant equations

$Γ(z)=\int_0^\infty t^{z-1} e^{-t}\,{\rm d}t$

3. The attempt at a solution

Looking online, I should try integration by parts, but I'm still unsure as to how to attack the problem.

2. Dec 28, 2011

### Curious3141

Integrate the integrand in the expression for $\Gamma(z)$ using integration by parts ($\int udv = uv - \int vdu$). Use $u = e^{-t}$ and $dv = t^{z-1}$. Remember that $t$ is the variable of integration and $z$ is a constant.

After getting the expression, apply the bounds. The $uv$ term is bounded by zero and infinity, and the upper bound can be resolved with L' Hopital's Rule (the limit is zero). The lower bound is also zero, so this term vanishes. You're left with an expression for the $\int vdu$ term that's equivalent to $\frac{1}{z}\Gamma(z+1)$. Now just do the simple algebra to finish off the proof.

EDIT: Can't believe I used zeta in place of gamma!

Last edited: Dec 28, 2011