積分botの問題を解いてみた2

\begin{eqnarray} \int_{0}^{\pi}\frac{\sin^{2n}x} {(1-2r\cos x+r^2)^{n+1}}dx\\ \end{eqnarray}
\begin{eqnarray} &=&\int_{0}^{\pi}\sin^{2n}x \int_{0}^{\infty}y^{n}e^{-y(1-2r\cos x+r^2)}\frac{dydx}{n!}\\ &=&\int_{0}^{\infty}\frac{y^n}{n!}e^{-y(1+r^2)} \int_{0}^{\pi}\sin^{2n}xe^{2ry\cos x}dxdy\\ &=&\frac{1}{n!}\int_{0}^{\infty}y^ne^{-y(1+r^2)}\int_{0}^{\pi/2}\sum_{k≥0}(2ry)^k(1+(-1)^k) \sin^{2n}x\cos^kx\frac{dxdy}{k!}\\ &=&\frac{2}{n!}\int_{0}^{\infty} y^ne^{-y(1+r^2)}\int_{0}^{\pi/2} \sum_{k≥0}\frac{(2ry)^{2k}}{(2k)!}\sin^{2n}x\cos^{2k}xdxdy\\ &=&\frac{1}{n!}\sum_{k≥0}\frac{ (2r)^{2k} \Gamma(n+\frac{1}{2})\Gamma(k+\frac{1}{2})}{\Gamma(n+k+1)}\int_{0}^{\infty}y^{n+2k}e^{-y(1+r^2)}\frac{dy}{(2k)!}\\ &=&\frac{1}{n!}\sum_{k≥0}\frac{\Gamma(n+\frac{1}{2})\Gamma(k+\frac{1}{2})}{\Gamma(n+k+1)} \frac{(2r)^{2k}\Gamma(n+2k+1)}{(1+r^2)^{n+2k+1}(2k)!}\\ &=&\frac{\pi\binom{2n}{n}}{(1+r^2)^{n+1}2^{2n}}\sum_{k≥0} \frac{(n+2k)!}{k!(n+k)!} \left(\frac{r}{1+r^2}\right)^{2k}\\ &=&\frac{\pi\binom{2n}{n}}{2^{2n}(1+r^2)^{n+1}} F\left[ \begin{array} 11+\frac{n}{2};\frac{n+1}{2}\\ n+1 \end{array} ;\frac{4r^2}{(1+r^2)^2} \right]\\ \end{eqnarray}
(Kummerの2次変換公式)
\begin{eqnarray} &=&\frac{\pi\binom{2n}{n}}{2^{2n}(1+r)^{2n+2}} F\left[ \begin{array} nn+1;n+\frac{1}{2}\\ 2n+1 \end{array} ;\frac{4r}{(1+r)^2}\right] \end{eqnarray}
(Baileyの変換公式)
\begin{eqnarray} &=&\frac{\pi\binom{2n}{n}}{2^{2n}}F\left[ \begin{array} 11\\ - \end{array} ;r^2\right]\\ &=&\frac{\pi}{2^{2n}(1-r^2)}\binom{2n}{n} \end{eqnarray}