Residue of Gamma Function
Jump to navigation
Jump to search
Theorem
Let $\Gamma$ be the Definition:Gamma Function.
Let $n$ be a non-negative integer.
Then:
- $\Res \Gamma {-n} = \dfrac {\paren {-1}^n} {n!}$
Proof
By Poles of Gamma Function, $\Gamma$ has simple poles at the non-positive integers, so $-n$ is a simple pole of $\Gamma$.
Then:
\(\ds \Res \Gamma {-n}\) | \(=\) | \(\ds \lim_{z \mathop \to -n} \paren {z - \paren {-n} } \map \Gamma z\) | Residue at Simple Pole | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop \to -n} \paren {z + n} \paren {\frac {z \paren {z + 1} \ldots \paren {z + n} } {z \paren {z + 1} \ldots \paren {z + n} } } \map \Gamma z\) | multiplying by $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop \to -n} \paren {z + n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n} }\) | Gamma Difference Equation | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop \to -n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n - 1} }\) | cancelling $z + n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\map \Gamma 1} {-n \paren {-n + 1} \ldots \paren {-1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\paren {-1}^n \paren {n \paren {n - 1} \ldots \paren 1} }\) | $\map \Gamma 1 = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\paren {-1}^n n!}\) | Definition of Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {-1}^n} {n!}\) |
$\blacksquare$