ベータ関数になる積分

ベータ関数になる積分

(1)

\(-1<\Re\left(x\right)\land-1<\Re\left(y\right)\)
\[ \int_{0}^{\frac{\pi}{2}}\sin^{x}t\cos^{y}tdt=\frac{1}{2}B\left(\frac{x+1}{2},\frac{y+1}{2}\right) \]

(2)

\(-1<\Re\left(x\right)\land1<\Re\left(y-x\right)\)
\[ \int_{0}^{\infty}\frac{t^{x}}{\left(1+t\right)^{y}}dt=B\left(x+1,y-x-1\right) \]

(3)

\(-1<\Re\left(x\right)\land-1<\Re\left(y\right)\)
\[ \int_{-1}^{1}\left(1+t\right)^{x}\left(1-t\right)^{y}dt=2^{x+y+1}B\left(x+1,y+1\right) \]

(4)\(\frac{1}{6}\)公式の一般化

\(a<b\land-1<\Re\left(\alpha\right)\land-1<\Re\left(\beta\right)\)
\[ \int_{a}^{b}\left(x-a\right)^{\alpha}\left(b-x\right)^{\beta}dx=\left(b-a\right)^{\alpha+\beta+1}B\left(\alpha+1,\beta+1\right) \]

(1)

\begin{align*} \int_{0}^{\frac{\pi}{2}}\sin^{x}t\cos^{y}tdt & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}s^{\frac{x-1}{2}}\left(1-s\right)^{\frac{y-1}{2}}ds\cmt{s=\sin^{2}t}\\ & =\frac{1}{2}B\left(\frac{x+1}{2},\frac{y+1}{2}\right) \end{align*}

(2)

\begin{align*} \int_{0}^{\infty}\frac{t^{x}}{\left(1+t\right)^{y}}dt & =\int_{0}^{1}s^{y-x-2}\left(1-s\right)^{x}ds\cmt{s=\frac{1}{1+x}}\\ & =B\left(x+1,y-x-1\right) \end{align*}

(3)

\begin{align*} \int_{-1}^{1}\left(1+t\right)^{x}\left(1-t\right)^{y}dt & =2^{x+y+1}\int_{0}^{1}s^{x}\left(1-s\right)^{y}ds\cmt{s=\frac{1+t}{2}}\\ & =2^{x+y+1}B\left(x+1,y+1\right) \end{align*}

(4)

\begin{align*} \int_{a}^{b}\left(x-a\right)^{\alpha}\left(b-x\right)^{\beta}dx & =\int_{0}^{1}\left(\left(b-a\right)t\right)^{\alpha}\left(b-\left(\left(b-a\right)t+a\right)\right)^{\beta}\left(b-a\right)dt\cmt{t=\frac{x-a}{b-a}}\\ & =\left(b-a\right)^{\alpha+1}\int_{0}^{1}t^{\alpha}\left(b-a-\left(b-a\right)t\right)^{\beta}dt\\ & =\left(b-a\right)^{\alpha+1}\int_{0}^{1}t^{\alpha}\left(\left(b-a\right)\left(1-t\right)\right)^{\beta}dt\\ & =\left(b-a\right)^{\alpha+\beta+1}\int_{0}^{1}t^{\alpha}\left(1-t\right)^{\beta}dt\\ & =\left(b-a\right)^{\alpha+\beta+1}B\left(\alpha+1,\beta+1\right) \end{align*}
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ベータ関数になる積分
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