ガンマ関数の非正整数近傍での値
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。
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