Gegenbauer多項式のLaplace型積分表示

まず, ベータ積分を計算することによって,
\begin{align} \int_0^{\pi}(\cos\theta+i\sin\theta\cos\phi)^n\sin^{2a-1}\phi\,d\phi&=\int_{-\pi/2}^{\pi/2}(\cos\theta-i\sin\theta\sin\phi)^n\cos^{2a-1}\phi\,d\phi\\ &=\int_{0}^{\pi/2}((\cos\theta-i\sin\theta\sin\phi)^n+(\cos\theta-i\sin\theta\sin\phi)^n)\cos^{2a-1}\phi\,d\phi\\\\ &=2\sum_{0\leq k}(-1)^k\binom{n}{2k}\cos^{n-2k}\theta\sin^{2k}\theta\int_0^{\pi/2}\sin^{2k}\phi\cos^{2a-1}\phi\,d\phi\\ &=\sum_{0\leq k}(-1)^k\binom{n}{2k}\cos^{n-2k}\theta\sin^{2k}\theta\frac{\Gamma\left(k+\frac 12\right)\Gamma(a)}{\Gamma\left(k+a+\frac 12\right)}\\ &=\frac{\sqrt{\pi}\Gamma(a)}{\Gamma\left(a+\frac 12\right)}\sum_{0\leq k}(-1)^k\binom{n}{2k}\cos^{n-2k}\theta\sin^{2k}\theta\frac{\left(\frac 12\right)_k}{\left(a+\frac 12\right)_k}\\ &=\frac{\sqrt{\pi}\Gamma(a)}{\Gamma\left(a+\frac 12\right)}\sum_{0\leq k}(-1)^k\frac{n!}{4^kk!\left(a+\frac 12\right)_k(n-2k)!}\cos^{n-2k}\theta\sin^{2k}\theta \end{align}
を得る. $n$が偶数のとき, $n=2m$として,
\begin{align} \sum_{0\leq k}(-1)^k\frac{n!}{4^kk!\left(a+\frac 12\right)_k(n-2k)!}\cos^{n-2k}\theta\sin^{2k}\theta&=\sum_{0\leq k}(-1)^{m-k}\frac{(2m)!}{4^{m-k}(m-k)!\left(a+\frac 12\right)_{m-k}(2k)!}\cos^{2k}\theta\sin^{2m-2k}\theta\\ &=\frac{(-1)^m(2m)!}{4^mm!\left(a+\frac 12\right)_m}\sum_{0\leq k}(-1)^{k}\frac{\left(-m,\frac 12-m-a\right)_k}{k!\left(\frac 12\right)_k}\cos^{2k}\theta\sin^{2m-2k}\theta\\ \end{align}
超幾何関数のPfaff変換によって,
\begin{align} \sum_{0\leq k}(-1)^{k}\frac{\left(-m,\frac 12-m-a\right)_k}{k!\left(\frac 12\right)_k}\cos^{2k}\theta\sin^{2m-2k}\theta&=\sum_{0\leq k}\frac{(-m,a+m)_k}{k!\left(\frac 12\right)_k}\cos^{2k}\theta\\ &=\frac{(-1)^mm!}{(a)_m}C_{2m}^{(a)}(\cos\theta) \end{align}
よってこのとき,
\begin{align} \int_0^{\pi}(\cos\theta+i\sin\theta\cos\phi)^n\sin^{2a-1}\phi\,d\phi&=\frac{(2m)!}{(2a)_{2m}}\frac{\sqrt{\pi}\Gamma(a)}{\Gamma\left(a+\frac 12\right)}C_{2m}^{(a)}(\cos\theta) \end{align}
$n$が奇数のとき, $n=2m+1$として,
\begin{align} \sum_{0\leq k}(-1)^k\frac{n!}{4^kk!\left(a+\frac 12\right)_k(n-2k)!}\cos^{n-2k}\theta\sin^{2k}\theta&=\sum_{0\leq k}(-1)^{m-k}\frac{(2m+1)!}{4^{m-k}(m-k)!\left(a+\frac 12\right)_{m-k}(2k+1)!}\cos^{2k+1}\theta\sin^{2m-2k}\theta\\ &=\frac{(-1)^m(2m+1)!}{4^mm!\left(a+\frac 12\right)_m}\cos\theta\sum_{0\leq k}(-1)^{k}\frac{\left(-m,\frac 12-m-a\right)_k}{k!\left(\frac 32\right)_k}\cos^{2k}\theta\sin^{2m-2k}\theta\\ \end{align}
超幾何関数のPfaff変換によって,
\begin{align} \cos\theta\sum_{0\leq k}(-1)^{k}\frac{\left(-m,\frac 12-m-a\right)_k}{k!\left(\frac 32\right)_k}\cos^{2k}\theta\sin^{2m-2k}\theta&=\cos\theta\sum_{0\leq k}\frac{(-m,a+m+1)_k}{k!\left(\frac 32\right)_k}\cos^{2k}\theta\\ &=\frac{(-1)^mm!}{2(a)_{m+1}}C_{2m+1}^{(a)}(\cos\theta) \end{align}
よってこのとき,
\begin{align} \int_0^{\pi}(\cos\theta+i\sin\theta\cos\phi)^n\sin^{2a-1}\phi\,d\phi&=\frac{(2m+1)!}{(2a)_{2m+1}}\frac{\sqrt{\pi}\Gamma(a)}{\Gamma\left(a+\frac 12\right)}C_{2m+1}^{(a)}(\cos\theta) \end{align}
となり, いずれの場合も定理が示された.

\begin{align} \int_{-\infty}^{\infty}(x+it)^ne^{-t^2}\,dt&=\int_0^{\infty}((x+it)^n+(x-it)^n)e^{-t^2}\,dt\\ &=2\sum_{0\leq k}(-1)^k\binom{n}{2k}x^{n-2k}\int_0^{\infty}t^{2k}e^{-t^2}\,dt\\ &=\sqrt{\pi}\sum_{0\leq k}(-1)^k\binom{n}{2k}x^{n-2k}\left(\frac 12\right)_k\\ &=\sqrt{\pi}\sum_{0\leq k}(-1)^k\frac{n!}{4^kk!(n-2k)!}x^{n-2k}\\ &=\frac{\sqrt{\pi}}{2^n}H_n(x) \end{align}