Gamma Difference Equation/Proof 1

Theorem

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

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 $

$=$

$\ds \bigintlimits {-t^z e^{-t} } 0 \infty + z \int_0^\infty t^{z - 1} e^{-t} \rd t$

$\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.

$\blacksquare$