積分のヴァンデルモンドの畳み込み(ガバ解説)
$$ \sum_{i=0}^{\infty}\binom{\color{red}a\color{black}}{i}\binom{\color{gold}b\color{black}}{\color{blue}c\color{black}-i}=\binom{\color{red}a\color{black}+\color{gold}b\color{black}}{\color{blue}c\color{black}} \ <=\mathrm{対応}=> \int_{-\infty}^{\infty}\binom{\color{red}a\color{black}}{t}\binom{\color{gold}b\color{black}}{\color{blue}c\color{black}-t}\d tt=\binom{\color{red}a\color{black}+\color{gold}b\color{black}}{\color{blue}c\color{black}} \ (\min(\color{red}a\color{black},\color{gold}b\color{black},\color{red}a\color{black}+\color{gold}b\color{black})>-1,\color{blue}c\color{black}\in\R) $$
$$\begin{align} \int_{-\infty}^{\infty}\binom{a}{t}\binom{b}{c-t}\d tt &= \left(\binom{a}{x}*\binom{b}{x}\right)(c)\\ &= \F[(1+e^{it})^a(1+e^{it})^b](c)\\ &= \F[(1+e^{it})^{a+b}](c)\\ &= \binom{a+b}{c}. \end{align}$$
$$\left(\mathrm{注：}\F[\F^{-1}[f]\F^{-1}[g]](t)=(f*g)(t) \ , \ \F^{-1}\left[\binom{a}{s}\right](t)=(1+e^{it})^a \ (a>-1,|t|<\pi)\right)$$
$$\int_{-\infty }^{\infty}\frac{\d tx}{\Gamma(a+x)\Gamma(b+x)\Gamma(c-x)\Gamma(d-x)} =\frac{\Gamma(a+b+c+d-3)}{\Gamma(a+c-1)\Gamma(a+d-1)\Gamma(b+c-1)\Gamma(b+d-1)}$$
$$\begin{align}
\binom{p+q}{r}
&=\int_{-\infty }^{\infty}\binom{p}{t}\binom{q}{r-t}\d tt\\
&=\Gamma(p+1)\Gamma(q+1)\int_{-\infty }^{\infty}\frac{\d tt}{\Gamma(t+1)\Gamma(p-t+1)\Gamma(r-t+1)\Gamma(q-r-t+1)}\\
&=\Gamma(p+1)\Gamma(q+1)\int_{-\infty }^{\infty}\frac{\d xx}{\Gamma(a+x)\Gamma(p-a+2-x)\Gamma(r-a+2-x)\Gamma(q-r+a+t)} \ \\
& (t+1\longmapsto a+x)\\
\\
\end{align}$$
$p\longmapsto a+c-2,q\longmapsto b+d-2,r=a+d-2 $として，
$$\begin{align} \therefore \int_{-\infty }^{\infty}\frac{\d tx}{\Gamma(a+x)\Gamma(b+x)\Gamma(c-x)\Gamma(d-x)} &=\frac{1}{\Gamma(a+c-1)\Gamma(b+d-1)}\binom{a+b+c+d-4}{a+d-2}\\ &=\frac{\Gamma(a+b+c+d-3)}{\Gamma(a+c-1)\Gamma(a+d-1)\Gamma(b+c-1)\Gamma(b+d-1)}. \end{align}$$
