# Gamma Difference Equation/Proof 1

## Theorem

$\map \Gamma {z + 1} = z \, \map \Gamma z$

## Proof

Let $z \in \C$, with $\map \Re z > 0$.

Then:

 $\ds \map \Gamma {z + 1}$ $=$ $\ds \int_0^\infty t^z e^{-t} \rd t$ $\ds$ $=$ $\ds \bigintlimits {-t^z e^{-t} } 0 \infty + z \int_0^\infty t^{z - 1} e^{-t} \rd t$ Integration by Parts $\ds$ $=$ $\ds z \int_0^\infty t^{z - 1} e^{-t} \rd t$ $\ds$ $=$ $\ds z \, \map \Gamma z$

If $z \in \C \setminus \set {0, -1, -2, \ldots}$ such that $\map \Re z \le 0$, then the statement holds by the definition of $\Gamma$ in this region.

$\blacksquare$