ガンマ凾数の相反公式の積分表示
ベータ凾数のガンマ凾数での表示
$$\begin{eqnarray}
\left.B\left(p,q\right)\right|_{p=1-q}&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\left.\frac{\Gamma\left(p\right)\Gamma\left(q\right)}{\Gamma\left(p+q\right)}\right|_{p=1-q}}
\;\ldots\;p,q\in\mathbb{C},\;\Re\left(p\right),\Re\left(q\right)\gt0
\\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1-q+q\right)}
\\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1\right)}
\\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{1}
\\&&\;\ldots\;\Gamma\left(1\right)=\int_0^\infty t^{1-1}e^{-t}\mathrm{d}t=\int_0^\infty t^{0}e^{-t}\mathrm{d}t=\int_0^\infty 1\cdot e^{-t}\mathrm{d}t=\int_0^\infty e^{-t}\mathrm{d}t
\\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/e-ax0.html}{\int_0^\infty e^{-t}\mathrm{d}t=\frac{1}{1}}=1
\\&=&\Gamma\left(1-q\right)\Gamma\left(q\right)
\end{eqnarray}$$
ベータ凾数の定義での表示
$$\begin{eqnarray}
\left.B\left(p,q\right)\right|_{p=1-q}
&=&\int_0^1 t^{p-1}\left(1-t\right)^{q-1}\mathrm{d}t
\\&=&\int_\infty^0 \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{-1}{\left(x+1\right)^2}\mathrm{d}x
\\&&\;\ldots\;t=\frac{1}{x+1},\;x=\frac{1}{t}-1,\;\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{-1}{\left(x+1\right)^2}
\\&&\;\ldots\;t:0\rightarrow1,\;x:\infty\rightarrow0
\\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x
\\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(\frac{x}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x
\\&=&\int_0^\infty \frac{1^{p-1}\cdot x^{q-1}\cdot 1}{\left(x+1\right)^{p-1}\left(x+1\right)^{q-1}\left(x+1\right)^{2}}\mathrm{d}x
\\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{(p-1)+(q-1)+2}}\mathrm{d}x
\\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{p+q}}\mathrm{d}x
\\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{1-q+q}}\mathrm{d}x
\\&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x
\end{eqnarray}$$
ガンマ凾数の相反公式の積分表示
$$\begin{eqnarray}
\Gamma\left(1-q\right)\Gamma\left(q\right)&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x
\end{eqnarray}$$
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