ガンマ凾数の相反公式の積分表示

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

$$\begin{eqnarray} \Gamma\left(1-q\right)\Gamma\left(q\right)&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x \end{eqnarray}$$

ガンマ凾数の相反公式

ガンマ凾数の相反公式

$$\begin{eqnarray} \Gamma\left(z\right)\Gamma\left(1-z\right) &=&\Gamma\left(z\right)\Gamma\left(-z+1\right) \;\ldots\;\Re\left(z\right)\gt0かつ\Re\left(1-z\right)\gt0なので0\lt\Re\left(z\right)\lt1 \\&=&\Gamma\left(z\right)\left(-z\Gamma\left(-z\right)\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma\left(z+1\right)=z\Gamma\left(z\right)} \\&=&-z\Gamma\left(z\right)\Gamma\left(-z\right) \\&=&-z \left(\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)}\right) \left(\lim_{n\rightarrow\infty}\frac{n^{-z}n!}{\prod_{k=0}^n (-z+k)}\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_16.html}{\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)}} \\&=&-z \left(\lim_{n\rightarrow\infty}\frac{n^zn! \cdot n^{-z}n!}{\prod_{k=0}^n (z+k)(-z+k)}\right) \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{n!n!}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;A^BA^{-B}=A^{B-B}=A^{0}=1 \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\left(\prod_{k=1}^nk\right)\left(\prod_{k=1}^nk\right)}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;n!=\prod_{k=1}^nk \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\prod_{k=1}^nk^2}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;\left(\prod_{k=m}^nA_k\right)\left(\prod_{k=m}^nB_k\right)=\prod_{k=m}^nA_kB_k \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\prod_{k=1}^nk^2}{(0^2-z^2)\prod_{k=1}^n (k^2-z^2)}\right) \\&=&-z\frac{1}{-z^2}\left(\lim_{n\rightarrow\infty}\prod_{k=1}^n\frac{k^2}{k^2-z^2}\right) \\&=&\frac{1}{z}\prod_{k=1}^\infty\frac{k^2}{k^2-z^2} \\&=&\frac{1}{\frac{1}{\frac{1}{z}\prod_{k=0}^\infty\frac{k^2}{k^2-z^2}}}\;\ldots\;A=\frac{1}{\frac{1}{A}} \\&=&\frac{1}{z\prod_{k=1}^\infty\frac{k^2-z^2}{k^2}} \\&=&\frac{\pi}{\pi}\frac{1}{z\prod_{k=0}^\infty\frac{k^2-z^2}{k^2}} \\&=&\frac{\pi}{\pi z\prod_{k=0}^\infty \left(1-\frac{z^2}{k^2}\right)} \\&=&\frac{\pi}{\sin{\left(\pi z\right)}} \\&&\;\ldots\;\href{https://ja.wikipedia.org/wiki/%E4%B8%89%E8%A7%92%E9%96%A2%E6%95%B0%E3%81%AE%E7%84%A1%E9%99%90%E4%B9%97%E7%A9%8D%E5%B1%95%E9%96%8B}{\sin{\left(\pi z\right)} =\pi z\prod_{k=1}^\infty\left(1-\left(\frac{z}{k}\right)^2\right)\;(三角凾数の無限乗積展開 wikipedia)} \end{eqnarray}$$