ベータ関数とガンマ関数の関係
ベータ関数とガンマ関数の関係
\[ B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \]
\[ B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \]
\begin{align*}
\Gamma(x) & =\int_{0}^{\infty}t^{x-1}e^{-t}dt\\
& =2\int_{0}^{\infty}s^{2x-1}e^{-s^{2}}ds\qquad,\qquad t=s^{2}
\end{align*}
より、
\begin{align*} \Gamma(x)\Gamma(y) & =4\int_{0}^{\infty}u^{2x-1}e^{-u^{2}}du\int_{0}^{\infty}v^{2y-1}e^{-v^{2}}dv\\ & =4\int_{0}^{\infty}\int_{0}^{\infty}u^{2x-1}v^{2y-1}e^{-(u^{2}+v^{2})}dudv\\ & =4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}r^{2(x+y-1)}\cos^{2x-1}\theta\sin^{2y-1}\theta e^{-r^{2}}rdrd\theta\qquad,\qquad u=r\cos\theta,v=r\sin\theta\\ & =2\int_{0}^{\infty}r^{2(x+y)-1}e^{-r^{2}}dr2\int_{0}^{\frac{\pi}{2}}\cos^{2x-1}\theta\sin^{2y-1}\theta d\theta\\ & =\Gamma(x+y)B(x,y) \end{align*}
\begin{align*} \Gamma(x)\Gamma(y) & =4\int_{0}^{\infty}u^{2x-1}e^{-u^{2}}du\int_{0}^{\infty}v^{2y-1}e^{-v^{2}}dv\\ & =4\int_{0}^{\infty}\int_{0}^{\infty}u^{2x-1}v^{2y-1}e^{-(u^{2}+v^{2})}dudv\\ & =4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}r^{2(x+y-1)}\cos^{2x-1}\theta\sin^{2y-1}\theta e^{-r^{2}}rdrd\theta\qquad,\qquad u=r\cos\theta,v=r\sin\theta\\ & =2\int_{0}^{\infty}r^{2(x+y)-1}e^{-r^{2}}dr2\int_{0}^{\frac{\pi}{2}}\cos^{2x-1}\theta\sin^{2y-1}\theta d\theta\\ & =\Gamma(x+y)B(x,y) \end{align*}
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| タイトル | ベータ関数とガンマ関数の関係 |
| URL | https://www.nomuramath.com/a5iiu0be/ |
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ベータ関数と2項係数の逆数の級数表示
\[
B(x,y)=\sum_{k=0}^{\infty}\frac{C(k-y,k)}{x+k}
\]
ベータ関数になる積分
\[
\int_{0}^{\frac{\pi}{2}}\sin^{x}t\cos^{y}tdt=\frac{1}{2}B\left(\frac{x+1}{2},\frac{y+1}{2}\right)
\]
ベータ関数の微分
\[
\frac{\partial}{\partial x}B(x,y)=B(x,y)\left\{ \psi(x)-\psi(x+y)\right\}
\]
ベータ関数の絶対収束条件
ベータ関数$B\left(p,q\right)$は$\Re\left(p\right)>0\;\land\;\Re\left(q\right)>0$で絶対収束