ガンマ関数の非正整数近傍での値

by nomura · 2022年9月29日

ガンマ関数の非正整数近傍での値
$n\in\mathbb{N}_{0}$とする。

(1)

\[ \lim_{\epsilon\rightarrow\pm0}\Gamma\left(-\epsilon\right)=-\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right) \]

(2)

\[ \lim_{\epsilon\rightarrow\pm0}\Gamma\left(-n+\epsilon\right)=\frac{\left(-1\right)^{n}}{n!}\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right) \]

(3)

\[ \lim_{\epsilon\rightarrow0}\frac{\Gamma\left(-n+\epsilon\right)}{\Gamma\left(-n-\epsilon\right)}=-1 \]

両側極限は存在しないので片側極限にしている。

(1)

\begin{align*} \lim_{\epsilon\rightarrow\pm0}\Gamma\left(-\epsilon\right) & =\lim_{\epsilon\rightarrow\pm0}\frac{\pi}{\Gamma\left(1+\epsilon\right)\sin\left(-\pi\epsilon\right)}\\ & =-\lim_{\epsilon\rightarrow\pm0}\frac{\pi}{\Gamma\left(1+\epsilon\right)\sin\left(\pi\epsilon\right)}\\ & =-\lim_{\epsilon\rightarrow\pm0}\frac{\Gamma\left(\epsilon\right)\Gamma\left(1-\epsilon\right)}{\Gamma\left(1+\epsilon\right)}\\ & =-\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right) \end{align*}

(2)

\begin{align*} \lim_{\epsilon\rightarrow\pm0}\Gamma\left(-n+\epsilon\right) & =\lim_{\epsilon\rightarrow\pm0}\frac{\Gamma\left(-n+1+\epsilon\right)}{\left(-n+\epsilon\right)}\\ & =-\lim_{\epsilon\rightarrow\pm0}\frac{\Gamma\left(-n+1+\epsilon\right)}{n}\\ & =\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right)\prod_{k=1}^{n}\frac{\Gamma\left(-k+\epsilon\right)}{\Gamma\left(-k+1+\epsilon\right)}\\ & =\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right)\prod_{k=1}^{n}\frac{-1}{k}\\ & =\frac{\left(-1\right)^{n}}{n!}\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right) \end{align*}

(2)-2

\begin{align*} \lim_{\epsilon\rightarrow\pm0}\Gamma\left(-n+\epsilon\right) & =\lim_{\epsilon\rightarrow\pm0}\frac{\pi}{\Gamma\left(1+n-\epsilon\right)\sin\left(\pi\left(-n+\epsilon\right)\right)}\\ & =\frac{\pi}{\Gamma\left(1+n\right)}\lim_{\epsilon\rightarrow\pm0}\frac{\Gamma\left(1+\epsilon\right)}{\sin\left(\pi\left(-n+\epsilon\right)\right)}\\ & =\frac{\pi}{\Gamma\left(1+n\right)}\lim_{\epsilon\rightarrow\pm0}\frac{\epsilon\Gamma\left(\epsilon\right)}{\sin\left(\pi\left(-n+\epsilon\right)\right)}\\ & =\frac{\pi}{\Gamma\left(1+n\right)}\lim_{\epsilon\rightarrow\pm0}\frac{\Gamma\left(\epsilon\right)}{\pi\cos\left(\pi\left(-n+\epsilon\right)\right)}\\ & =\frac{1}{\Gamma\left(1+n\right)\cos\left(-\pi n\right)}\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right)\\ & =\frac{\left(-1\right)^{n}}{n!}\lim_{\epsilon\rightarrow\pm0}\Gamma\left(\epsilon\right) \end{align*}

(3)

\begin{align*} \lim_{\epsilon\rightarrow\pm0}\frac{\Gamma\left(-n+\epsilon\right)}{\Gamma\left(-n-\epsilon\right)} & =\lim_{\epsilon\rightarrow\pm0}\frac{\left(-1\right)^{n}\Gamma\left(\epsilon\right)n!}{n!\left(-1\right)^{n}\Gamma\left(-\epsilon\right)}\\ & =\lim_{\epsilon\rightarrow\pm0}\frac{\Gamma\left(\epsilon\right)}{\Gamma\left(-\epsilon\right)}\\ & =-1 \end{align*} これより、両側極限が存在するので、与式は成り立つ。