Gamma Difference Equation

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Theorem

The Gamma function satisfies:

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

for any $z$ which is not a nonpositive integer.


Proof 1

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

Then:

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


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.

$\blacksquare$


Proof 2

By Euler's form of the Gamma function:

\(\displaystyle \frac {\map \Gamma {z + 1} } {\map \Gamma z}\) \(=\) \(\displaystyle \paren {\frac 1 {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^{z + 1} } {1 + \frac {z + 1} n} } \div \paren {\frac 1 z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^z} {1 + \frac z n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac z {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {\frac {\paren {1 + \frac 1 n} \paren {z + n} } {z + n + 1} }\) after some more-or-less hairy algebra
\(\displaystyle \) \(=\) \(\displaystyle z \lim_{m \mathop \to \infty} \frac {m + 1} {z + m + 1} = z\)


$\blacksquare$


Sources