第2種不完全ガンマ関数とガンマ関数の比の極限
第2種不完全ガンマ関数とガンマ関数の比の極限
(1)
\[ \lim_{k\rightarrow0}\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)}=\delta_{0x} \](2)
\[ \lim_{k\rightarrow\infty}\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)}=1 \]-
\(\Gamma\left(x\right)\)はガンマ関数、\(\Gamma\left(k,x\right)\)は第2種不完全ガンマ関数、\(\delta_{ij}\)はクロネッカーのデルタ(1)
\begin{align*} \lim_{k\rightarrow0}\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)} & =\lim_{k\rightarrow0}\frac{k\Gamma\left(k,x\right)}{k\Gamma\left(k\right)}\\ & =\lim_{k\rightarrow0}\frac{k\Gamma\left(k,x\right)}{\Gamma\left(k+1\right)}\\ & =\lim_{k\rightarrow0}\frac{\Gamma\left(k+1,x\right)-x^{k}e^{-x}}{\Gamma\left(k+1\right)}\\ & =\lim_{k\rightarrow0}\frac{\Gamma\left(1,x\right)-x^{k}e^{-x}}{\Gamma\left(1\right)}\\ & =e^{-x}-\left(1-\delta_{0x}\right)e^{-x}\\ & =\delta_{0x}e^{-x}\\ & =\delta_{0x} \end{align*}(2)
\begin{align*} \lim_{k\rightarrow\infty}\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)} & =\lim_{k\rightarrow0}\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)}+\sum_{k=+0}^{\infty}\left(\frac{\Gamma\left(k+1,x\right)}{\Gamma\left(k+1\right)}-\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)}\right)\\ & =\delta_{0x}+\sum_{k=+0}^{\infty}\left(\frac{\Gamma\left(k+1,x\right)}{\Gamma\left(k+1\right)}-\frac{\Gamma\left(k,x\right)}{\Gamma\left(k\right)}\right)\\ & =\delta_{0x}+e^{-x}\sum_{k=+0}^{\infty}\frac{x^{k}}{k!}\\ & =\delta_{0x}+e^{-x}\left(e^{x}-\delta_{0x}\right)\\ & =\delta_{0x}\left(1-e^{-x}\right)+1\\ & =1 \end{align*}ページ情報
| タイトル | 第2種不完全ガンマ関数とガンマ関数の比の極限 |
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そのままだとΓ(0)になる積分
\[
\int_{0}^{\infty}\left(x^{-1}e^{-x}-\frac{e^{-nx}}{1-e^{-x}}\right)dx=H_{n-1}-\gamma
\]
負の整数の階乗の商
\[
\frac{\left(-m\right)!}{\left(-n\right)!}=\left(-1\right)^{n-m}\frac{\Gamma\left(n\right)}{\Gamma\left(m\right)}
\]
ガンマ関数の極限問題
\[
\lim_{x\rightarrow0}\frac{\Gamma(ax)}{\Gamma(x)}=\frac{1}{a}
\]
ガンマ関数の絶対収束条件
ガンマ関数$\Gamma\left(z\right)$は$\Re\left(z\right)>0$で絶対収束