ガンマ関数

ガンマ関数(gamma function)は階乗の一般化である. まず, \[ \Gamma(n) = (n-1)! \] と定義する. となると, \[ \Gamma(n+1) = n! = n(n-1)! = n\Gamma(n) \] となり,整数の階乗と同じ性質となる. \(x\)は整数と限らないとした時,\(\Gamma(x)\)がどうなるかを考えてみる.

\[ \begin{align} (n+x)! &= (n+x)(n+x-1)(n+x-2) \cdots (n+x-x+1)(n+x-x)(n+x-x-1)(n+x-x-2) \cdots 2 \cdot 1 \\ &= (n+x)(n+x-1)(n+x-2)\cdots (n+1) n! \\ &= \frac{n^x(n+x)(n+x-1)(n+x-2)\cdots (n+1) n!}{n^x} \\ &= n!n^x\left(1+\frac{x}{n}\right)\left(1+\frac{x-1}{n}\right)\left(1+\frac{x-2}{n}\right)\cdots\left(1+\frac{1}{n}\right)\\ \end{align} \] また, \[ \begin{align} (n+x)! &= (x+n)(x+n-1)(x+n-2)\cdots(x+n-n+1)(x+n-n)(x+n-n-1)(x+n-n-2)\cdots 2 \cdot 1 \\ &= (n+x)(n+x-1)(n+x-2)\cdots(x+1)x(x-1)! \end{align} \] から, \[ (n+x)(n+x-1)(n+x-2)\cdots(x+1)x(x-1)! = n!n^x\left(1+\frac{x}{n}\right)\left(1+\frac{x-1}{n}\right)\left(1+\frac{x-2}{n}\right)\cdots\left(1+\frac{1}{n}\right)\\ (x-1)! = \frac{n!n^x}{(n+x)(n+x-1)(n+x-2)\cdots(x+1)x}\left(1+\frac{x}{n}\right)\left(1+\frac{x-1}{n}\right)\left(1+\frac{x-2}{n}\right)\cdots\left(1+\frac{1}{n}\right) \] 改めて上式で,\(n \to \infty\)とした時, \[ (x-1)! = \lim_{n \to \infty} \frac{n!n^x}{(n+x)(n+x-1)(n+x-2)\cdots(x+1)x} \\ \Gamma(x) = \lim_{n \to \infty} \frac{n!n^x}{\prod ^n _{i=0} (x+i)} \] となる.

積分表示

\[ \Gamma(x) = \int ^{\infty} _{0} t^{x-1} e^{-t} dt \] はガンマ関数の無限積表示と同値である.

証明
\[ \begin{align} \Gamma(x) &= \int ^{\infty} _{0} t^{x-1} e^{-t} dt \\ &= \lim _{n \to \infty} \int ^{n} _{0} t^{x-1} e^{-t} dt \\ &= \lim _{n \to \infty} \int ^{n} _{0} t^{x-1} \left(1 - \frac{t}{n} \right)^n dt \end{align} \] \(\displaystyle u=\frac{t}{n}\)と変数変換すると, \[ \begin{align} \lim _{n \to \infty} \int ^{n} _{0} t^{x-1} \left(1 - \frac{t}{n} \right)^n dt &= \lim _{n \to \infty} \int ^{1} _{0} (un)^{x-1} (1 - u)^n n du \\ &= \lim _{n \to \infty} n^x \int ^{1} _{0} u^{x-1} (1 - u)^n du \\ &= \lim _{n \to \infty} n^x \int ^{1} _{0} \left( \frac{1}{x} u^x \right)' (1 - u)^n du \\ &= \lim _{n \to \infty} n^x \left\{ \int ^{1} _{0} \left( \frac{1}{x} u^x \right)' (1 - u)^n du \right\} \\ &= \lim _{n \to \infty} n^x \left\{ \left[ \frac{1}{x} u^x (1 - u)^n \right] ^1 _0 + \int ^{1} _{0} \frac{u^x}{x} n(1 - u)^{n-1} du \right\} \\ &= \lim _{n \to \infty} n^x \left\{ \frac{n}{x} \int ^{1} _{0} \left( \frac{u^{x+1}}{x+1} \right)' (1 - u)^{n-1} du \right\} \\ &= \lim _{n \to \infty} n^x \frac{n}{x} \left\{ \int ^{1} _{0} \left( \frac{u^{x+1}}{x+1} \right)' (1 - u)^{n-1} du \right\} \\ &= \lim _{n \to \infty} n^x \frac{n}{x} \left\{ \left[\frac{u^{x+1}}{x+1} (1 - u)^{n-1} \right] ^1 _0 + \int ^{1} _{0} \frac{u^{x+1}}{x+1} (n-1)(1 - u)^{n-2} du \right\} \\ &= \lim _{n \to \infty} n^x \frac{n(n-1)}{x(x+1)} \left\{\int ^{1} _{0} u^{x+1} (1 - u)^{n-2} du \right\} \\ &\vdots \\ &= \lim _{n \to \infty} n^x \frac{n!}{\prod ^{n-1} _{i=0} (x+i)} \left\{\int ^{1} _{0} u^{x+n-1} du \right\} \\ &= \lim _{n \to \infty} n^x \frac{n!}{\prod ^{n-1} _{i=0} (x+i)} \left[ \frac{1}{x+n} u^{x+n}\right] ^1 _0 \\ &= \lim _{n \to \infty} \frac{n!n^x}{\prod ^{n} _{i=0} (x+i)} \end{align} \]