ガンマ関数を含む極限
ガンマ関数を含む極限値
(1)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}=1 \](2)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}=\sqrt{2} \](1)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}\sqrt{n+\frac{1}{2}}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n+1\right)}}\\ & =\lim_{n\rightarrow\infty}\sqrt{\frac{\sqrt{n}\sqrt{n+\frac{1}{2}}}{n}}\\ & =\lim_{n\rightarrow\infty}\left(1+\frac{1}{2n}\right)^{\frac{1}{4}}\\ & =1 \end{align*}(1)-2
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\frac{n+1}{2}}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\\ & =\frac{1}{\sqrt{2}}\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}}{\sqrt{n}}\sqrt{n}\frac{2}{\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\sqrt{\frac{2}{\pi}}\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}}{\sqrt{n}}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\sqrt{\frac{2}{\pi}}\sqrt{\frac{\pi}{2}}\\ & =1 \end{align*}(2)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\sqrt{n+1}\frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}}\\ & =\lim_{n\rightarrow\infty}\sqrt{\frac{2\sqrt{n}\sqrt{n+1}}{n+1}}\\ & =\sqrt{2}\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{-\frac{1}{4}}\\ & =\sqrt{2} \end{align*}(2)-2
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} & =\frac{2}{\Gamma\left(\frac{1}{2}\right)}\lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\frac{2}{\sqrt{\pi}}\sqrt{\frac{\pi}{2}}\\ & =\sqrt{2} \end{align*}ページ情報
| タイトル | ガンマ関数を含む極限 |
| URL | https://www.nomuramath.com/hkc3vxq1/ |
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ポリガンマ関数同士の差の極限
\[
\lim_{z\rightarrow0}\left(\psi^{\left(n\right)}\left(z-m\right)-\psi^{\left(n\right)}\left(z\right)\right)=n!H_{m,n+1}
\]
ガンマ関数の極限問題
\[
\lim_{x\rightarrow0}\frac{\Gamma(ax)}{\Gamma(x)}=\frac{1}{a}
\]
ガンマ関数のルジャンドル倍数公式
\[
\Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)
\]
(*)ガンマ関数と複素数
\[
\lim_{R\rightarrow\infty}\int_{0}^{Re^{i\theta}}z^{\alpha-1}e^{-z}dz=\Gamma\left(\alpha\right)
\]