Gamma Difference Equation

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Theorem

$\Gamma \left({z+1}\right) = z \Gamma \left({z}\right)$

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

Let $z \in \C$, with $\Re(z) > 0$. Then

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \Gamma(z+1)$	$=$	$\displaystyle \int_0^\infty t^{z} e^{-t}\ dt$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle -t^z e^{-t}\Big\vert_0^\infty + z\int_0^\infty t^{z-1} e^{-t}\ dt$	$\displaystyle $	$\displaystyle $	$\displaystyle $	By Integration by Parts
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle z\int_0^\infty t^{z-1} e^{-t}\ dt$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle z\Gamma(z)$	$\displaystyle $	$\displaystyle $	$\displaystyle $

If $z \in \C \backslash \{0,-1,-2,\ldots\}$ with $\Re(z) \leq 0$, then the statement holds by the definition of $\Gamma$ in this region.

$\blacksquare$

By Euler's form of the Gamma function,

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

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \Gamma(z+1)\)	\(=\)	\(\displaystyle \int_0^\infty t^{z} e^{-t}\ dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle -t^z e^{-t}\Big\vert_0^\infty + z\int_0^\infty t^{z-1} e^{-t}\ dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Integration by Parts
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle z\int_0^\infty t^{z-1} e^{-t}\ dt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle z\Gamma(z)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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