ベータ関数の特殊値
ベータ関数の特殊値
ベータ関数\(B\left(\alpha,\beta\right)\)は次の特殊値を持つ。
\[ B\left(z,1-z\right)=\frac{\pi}{\sin\left(\pi z\right)} \]
\[ B\left(m+1,n+1\right)=\frac{m!n!}{\left(m+n+1\right)!} \]
\begin{align*} B\left(m+\frac{1}{2},n+1\right) & =\frac{2^{2n+1}\left(2m\right)!n!\left(m+n\right)!}{\left(2m+2n+1\right)!m!}\\ & =\frac{2\left(2m-1\right)!!\left(2n\right)!!}{\left(2m+2n+1\right)!!} \end{align*}
\begin{align*} B\left(m+\frac{1}{2},n+\frac{1}{2}\right) & =\frac{\pi\left(2m\right)!\left(2n\right)!}{2^{2\left(m+n\right)}m!n!\left(m+n\right)!}\\ & =\frac{\pi\left(2m-1\right)!!\left(2n-1\right)!!}{\left(2m+2n\right)!!} \end{align*}
ベータ関数\(B\left(\alpha,\beta\right)\)は次の特殊値を持つ。
(1)
\[ B\left(\alpha,1\right)=\frac{1}{\alpha} \](2)
\(z\notin\mathbb{Z}\)とする。\[ B\left(z,1-z\right)=\frac{\pi}{\sin\left(\pi z\right)} \]
(3)
\[ B\left(\frac{1}{2},z\right)=\frac{2^{2z-1}\Gamma^{2}\left(z\right)}{\Gamma\left(2z\right)} \](4)
\[ B\left(\frac{1}{2},\frac{1}{2}\right)=\pi \](5)
\(m,n\in\mathbb{N}_{0}\)とする。\[ B\left(m+1,n+1\right)=\frac{m!n!}{\left(m+n+1\right)!} \]
(6)
\(m,n\in\mathbb{N}_{0}\)とする。\begin{align*} B\left(m+\frac{1}{2},n+1\right) & =\frac{2^{2n+1}\left(2m\right)!n!\left(m+n\right)!}{\left(2m+2n+1\right)!m!}\\ & =\frac{2\left(2m-1\right)!!\left(2n\right)!!}{\left(2m+2n+1\right)!!} \end{align*}
(7)
\(m,n\in\mathbb{N}_{0}\)とする。\begin{align*} B\left(m+\frac{1}{2},n+\frac{1}{2}\right) & =\frac{\pi\left(2m\right)!\left(2n\right)!}{2^{2\left(m+n\right)}m!n!\left(m+n\right)!}\\ & =\frac{\pi\left(2m-1\right)!!\left(2n-1\right)!!}{\left(2m+2n\right)!!} \end{align*}
(1)
\begin{align*} B\left(\alpha,1\right) & =\int_{0}^{1}t^{\alpha-1}\left(1-t\right)^{1-1}dt\\ & =\int_{0}^{1}t^{\alpha-1}dt\\ & =\left[\frac{t^{\alpha}}{\alpha}\right]_{0}^{1}\\ & =\frac{1}{\alpha} \end{align*}(2)
\begin{align*} B\left(z,1-z\right) & =\Gamma\left(z\right)\Gamma\left(1-z\right)\\ & =\frac{\pi}{\sin\left(\pi z\right)} \end{align*}(3)
\begin{align*} B\left(\frac{1}{2},z\right) & =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(z\right)}{\Gamma\left(\frac{1}{2}+z\right)}\\ & =\frac{2^{2z-1}\Gamma\left(\frac{1}{2}\right)\Gamma\left(z\right)}{\Gamma\left(2z\right)\Gamma^{-1}\left(z\right)\sqrt{\pi}}\cmt{\Gamma\left(2z\right)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{2}\right)}\\ & =\frac{2^{2z-1}\Gamma^{2}\left(z\right)}{\Gamma\left(2z\right)} \end{align*}(4)
\begin{align*} B\left(\frac{1}{2},\frac{1}{2}\right) & =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{2}\right)}\\ & =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(1\right)}\\ & =\frac{\sqrt{\pi}\sqrt{\pi}}{0!}\\ & =\pi \end{align*}(5)
\begin{align*} B\left(m+1,n+1\right) & =\frac{\Gamma\left(m+1\right)\Gamma\left(n+1\right)}{\Gamma\left(m+1+n+1\right)}\\ & =\frac{\Gamma\left(m+1\right)\Gamma\left(n+1\right)}{\Gamma\left(m+n+2\right)}\\ & =\frac{m!n!}{\left(m+n+1\right)!} \end{align*}(6)
\begin{align*} B\left(m+\frac{1}{2},n+1\right) & =\frac{\Gamma\left(m+\frac{1}{2}\right)\Gamma\left(n+1\right)}{\Gamma\left(m+\frac{1}{2}+n+1\right)}\\ & =\frac{\Gamma\left(m+\frac{1}{2}\right)\Gamma\left(n+1\right)}{\Gamma\left(m+n+1+\frac{1}{2}\right)}\\ & =\frac{2^{2\left(m+n+1\right)-1}\Gamma\left(2m\right)\sqrt{\pi}\Gamma^{-1}\left(m\right)\Gamma\left(n+1\right)}{2^{2m-1}\Gamma\left(2\left(m+n+1\right)\right)\sqrt{\pi}\Gamma^{-1}\left(m+n+1\right)}\cmt{\Gamma\left(2z\right)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{2}\right)}\\ & =\frac{2^{2n+2}\Gamma\left(2m\right)\Gamma\left(n+1\right)\Gamma\left(m+n+1\right)}{\Gamma\left(2\left(m+n+1\right)\right)\Gamma\left(m\right)}\\ & =\frac{2^{2n+1}\Gamma\left(2m+1\right)\Gamma\left(n+1\right)\Gamma\left(m+n+1\right)}{\Gamma\left(2m+2n+2\right)\Gamma\left(m+1\right)}\\ & =\frac{2^{2n+1}\left(2m\right)!n!\left(m+n\right)!}{\left(2m+2n+1\right)!m!}\tag{(*)}\\ & =\frac{2^{2n+1}\left(2m-1\right)!!2^{m}m!n!\left(m+n\right)!}{\left(2m+2n+1\right)!!2^{m+n}\left(m+n\right)!m!}\\ & =\frac{2^{n+1}\left(2m-1\right)!!n!}{\left(2m+2n+1\right)!!}\\ & =\frac{2\left(2m-1\right)!!\left(2n\right)!!}{\left(2m+2n+1\right)!!} \end{align*}(7)
\begin{align*} B\left(m+\frac{1}{2},n+\frac{1}{2}\right) & =\frac{\Gamma\left(m+\frac{1}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(m+\frac{1}{2}+n+\frac{1}{2}\right)}\\ & =\frac{\Gamma\left(m+\frac{1}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(m+n+1\right)}\\ & =\frac{\Gamma\left(2m\right)\sqrt{\pi}\Gamma^{-1}\left(m\right)\Gamma\left(2n\right)\sqrt{\pi}\Gamma^{-1}\left(n\right)}{2^{2m-1}2^{2n-1}\Gamma\left(m+n+1\right)}\cmt{\Gamma\left(2z\right)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{2}\right)}\\ & =\frac{\pi\Gamma\left(2m\right)\Gamma\left(2n\right)}{2^{2\left(m+n-1\right)}\Gamma\left(m\right)\Gamma\left(n\right)\Gamma\left(m+n+1\right)}\\ & =\frac{\pi\Gamma\left(2m+1\right)\Gamma\left(2n+1\right)}{2^{2\left(m+n\right)}\Gamma\left(m+1\right)\Gamma\left(n+1\right)\Gamma\left(m+n+1\right)}\\ & =\frac{\pi\left(2m\right)!\left(2n\right)!}{2^{2\left(m+n\right)}m!n!\left(m+n\right)!}\tag{(*)}\\ & =\frac{\pi\left(2m-1\right)!!2^{m}m!\left(2n-1\right)!!2^{n}n!}{2^{2\left(m+n\right)}m!n!\left(m+n\right)!}\\ & =\frac{\pi\left(2m-1\right)!!\left(2n-1\right)!!}{2^{m+n}\left(m+n\right)!}\\ & =\frac{\pi\left(2m-1\right)!!\left(2n-1\right)!!}{\left(2m+2n\right)!!} \end{align*}
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ベータ関数の対称性
\[
B\left(\alpha,\beta\right)=B\left(\beta,\alpha\right)
\]
ベータ関数と不完全ベータ関数の関係
\[
B\left(z;\alpha,\beta\right)+B\left(1-z;\beta,\alpha\right)=B\left(\alpha,\beta\right)
\]
ベータ関数の微分
\[
\frac{\partial}{\partial x}B(x,y)=B(x,y)\left\{ \psi(x)-\psi(x+y)\right\}
\]
ベータ関数・不完全ベータ関数の超幾何関数表示
\[
B\left(z;\alpha,\beta\right)=\frac{z^{\alpha}}{\alpha}F\left(\alpha,1-\beta;\alpha+1;z\right)
\]