Bessel関数の4つの積の積分
Bessel関数は
\begin{align}
J_{\nu}(z):=\sum_{0\leq n}\frac{(-1)^n}{n!\Gamma(\nu+n+1)}\left(\frac x2\right)^{2n+\nu}
\end{align}
によって定義される.
前の記事
でBessel関数の4乗の積分
\begin{align}
\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^4\,dt&=\frac{\Gamma(2\nu)\Gamma(\nu)}{2\pi\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}
\end{align}
を示した. 今回はその類似として, 第2種Bessel関数
\begin{align}
Y_{\nu}(z)&=\frac{J_{\nu}(z)\cos\nu\pi-J_{-\nu}(z)}{\sin\nu\pi}
\end{align}
が入った4つの積
\begin{align}
t^{1-2\nu}J_{\nu}(t)^kY_{\nu}(t)^{4-k},\qquad 0\leq k\leq 4
\end{align}
の$(0,\infty)$における積分についても考えたいと思う.
前の記事
でBessel関数の4乗の積分を求める際に示した補題は以下のようなものである.
\begin{align}
F(s)&=\int_0^{\infty}t^{s-1}f(t)\,dt\\
G(s)&=\int_0^{\infty}t^{s-1}g(t)\,dt
\end{align}
とするとき,
\begin{align}
\int_0^{\infty}t^{s-1}f(t)g(t)\,dt&=\frac 1{2\pi i}\int_{-i\infty}^{i\infty}F(u)G(s-u)\,du
\end{align}
が成り立つ.
前の記事 の定理1, 定理2, 系2において$\mu=\nu$として整理すると以下を得る.
\begin{align} \int_0^{\infty}t^{s-1}J_{\nu}(t)^2\,dt&=\frac{\Gamma\left(\frac s2+\nu\right)\Gamma\left(\frac 12-\frac s2\right)}{2\sqrt{\pi}\Gamma\left(\nu+1-\frac s2\right)\Gamma\left(1-\frac s2\right)}\\ \int_0^{\infty}t^{s-1}J_{\nu}(t)Y_{\nu}(t)\,dt&=-\frac{\Gamma\left(\frac s2\right)\Gamma\left(\frac s2+\nu\right)}{2\sqrt{\pi}\Gamma\left(1+\nu-\frac s2\right)\Gamma\left(\frac 12+\frac s2\right)}\\ \int_0^{\infty}t^{s-1}(Y_{\nu}(t)^2-J_{\nu}(t)^2)\,dt&=\frac{\Gamma\left(\frac s2\right)\Gamma\left(\nu+\frac s2\right)\Gamma\left(\frac s2-\nu\right)}{\pi^{\frac 32}\Gamma\left(\frac{1}2+\frac s2\right)}\cos\pi\left(\nu-\frac s2\right) \end{align}
これらと補題1を用いて以下を示す.
\begin{align} \int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^4\,dt&=\frac{\Gamma(2\nu)\Gamma(\nu)}{2\pi\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}\\ \int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^3Y_{\nu}(t)\,dt&=-\frac{\Gamma\left(\frac 12+\nu\right)}{\pi^{\frac 32}\Gamma(\nu)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\ \int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^2Y_{\nu}(t)^2\,dt&=-\frac{2\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\Gamma(2\nu)\Gamma(\nu)}{2\pi\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}\\ \int_0^{\infty}t^{1-2\nu}J_{\nu}(t)Y_{\nu}(t)^3\,dt&=-\frac{3\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\frac{\cos\pi\nu}{\sin\pi\nu}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\cos2\pi\nu}{\pi\sin 2\pi\nu}\frac{\Gamma(2\nu)\Gamma(\nu)}{\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}\\ \int_0^{\infty}t^{1-2\nu}Y_{\nu}(t)^4\,dt&=-\frac{4\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\left(\frac{\cos\pi\nu}{\sin\pi\nu}\right)^2\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\ &\qquad+\frac 1{2\pi}\left(\frac{4}{\sin\pi\nu\sin 3\pi\nu}-3\right)\frac{\Gamma(2\nu)\Gamma(\nu)}{\Gamma(3\nu)\Gamma\left(\frac 12+\nu\right)^2} \end{align}
1つ目の等式は
前の記事
で示したものである. 2つ目の等式は補題1において$f(t)=J_{\nu}(t)^2,g(t)=J_{\nu}(t)Y_{\nu}(t), s=2-2\nu$として
\begin{align}
&\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^3Y_{\nu}(t)\,dt\\
&=-\frac 1{2\pi i}\frac 1{4\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac u2+\nu\right)\Gamma\left(\frac 12-\frac u2\right)}{\Gamma\left(\nu+1-\frac u2\right)\Gamma\left(1-\frac u2\right)}\frac{\Gamma\left(1-\nu-\frac u2\right)\Gamma\left(1-\frac u2\right)}{\Gamma\left(2\nu+\frac u2\right)\Gamma\left(\frac 32-\nu-\frac u2\right)}\,du\\
&=-\frac 1{2\pi i}\frac 1{2\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12-u\right)}{\Gamma\left(\nu+1-u\right)\Gamma\left(2\nu+u\right)\Gamma\left(\frac 32-\nu-u\right)}\frac{\pi}{\sin\pi(u+\nu)}\,du\qquad u\mapsto 2u\\
&=\frac 1{2\pi i}\frac 1{2\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12+u\right)}{\Gamma\left(\nu+1+u\right)\Gamma\left(\frac 32-\nu+u\right)\Gamma\left(2\nu-u\right)}\frac{\pi}{\sin\pi(u-\nu)}\,du\qquad u\mapsto -u\\
&=-\frac 1{2\pi}\sum_{0\leq n}\frac{(-1)^n\Gamma\left(\frac 12+\nu+n\right)}{\Gamma(2\nu+1+n)\Gamma\left(\frac 32+n\right)\Gamma(\nu-n)}\\
&=-\frac 1{2\pi}\frac{\Gamma\left(\frac 12+\nu\right)}{\Gamma(\nu)\Gamma(2\nu+1)\Gamma\left(\frac 32\right)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&=-\frac{\Gamma\left(\frac 12+\nu\right)}{\pi^{\frac 32}\Gamma(\nu)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1
\end{align}
となって2つ目の等式が示される. 3つ目の等式は補題1において$f(t)=g(t)=J_{\nu}(t)Y_{\nu}(t), s=2-2\nu$として,
\begin{align}
&\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^2Y_{\nu}(t)^2\,dt\\
&=\frac 1{2\pi i}\frac 1{4\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac u2\right)\Gamma\left(\frac u2+\nu\right)}{\Gamma\left(1+\nu-\frac u2\right)\Gamma\left(\frac 12+\frac u2\right)}\frac{\Gamma\left(1-\nu-\frac u2\right)\Gamma\left(1-\frac u2\right)}{\Gamma\left(2\nu+\frac u2\right)\Gamma\left(\frac 32-\nu-\frac u2\right)}\,du\\
&=\frac 1{2\pi i}\frac 1{2}\int_{-i\infty}^{i\infty}\frac{1}{\Gamma\left(1+\nu-u\right)\Gamma\left(\frac 12+u\right)\Gamma\left(2\nu+u\right)\Gamma\left(\frac 32-\nu-u\right)}\frac{\pi}{\sin\pi u\sin\pi(u+\nu)}\,du\qquad u\mapsto 2u\\
&=-\frac 1{2\sin\pi\nu}\sum_{0\leq n}\frac{1}{\Gamma(\nu-n)\Gamma\left(\frac 32+n\right)\Gamma(2\nu+n+1)\Gamma\left(\frac 12-\nu-n\right)}\\
&\qquad+\frac 1{2\sin\pi\nu}\sum_{0\leq n}\frac 1{\Gamma(2\nu-n)\Gamma\left(\frac 32-\nu+n\right)\Gamma(1+\nu+n)\Gamma\left(\frac 12-n\right)}\\
&=-\frac 1{\sin\pi\nu}\sum_{0\leq n}\frac{1}{\Gamma(\nu-n)\Gamma\left(\frac 32+n\right)\Gamma(2\nu+n+1)\Gamma\left(\frac 12-\nu-n\right)}\\
&\qquad+\frac 1{2\sin\pi\nu}\sum_{n\in\ZZ}\frac 1{\Gamma(2\nu-n)\Gamma\left(\frac 32-\nu+n\right)\Gamma(1+\nu+n)\Gamma\left(\frac 12-n\right)}\\
&=-\frac 1{\sin\pi\nu\Gamma(\nu)\Gamma\left(\frac 32\right)\Gamma(2\nu+1)\Gamma\left(\frac 12-\nu\right)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 1{2\sin\pi\nu}\sum_{n\in\ZZ}\frac 1{\Gamma(2\nu-n)\Gamma\left(\frac 32-\nu+n\right)\Gamma(1+\nu+n)\Gamma\left(\frac 12-n\right)}
\end{align}
ここで,
Dougallの${}_2H_2$和公式
より,
\begin{align}
\sum_{n\in\ZZ}\frac 1{\Gamma(2\nu-n)\Gamma\left(\frac 32-\nu+n\right)\Gamma(1+\nu+n)\Gamma\left(\frac 12-n\right)}&=\frac{\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)\Gamma(1-\nu)}
\end{align}
であるから, これを代入して,
\begin{align}
&\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^2Y_{\nu}(t)^2\,dt\\
&=-\frac 1{\sin\pi\nu\Gamma(\nu)\Gamma\left(\frac 32\right)\Gamma(2\nu+1)\Gamma\left(\frac 12-\nu\right)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 1{2\sin\pi\nu}\frac{\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)\Gamma(1-\nu)}\\
&=-\frac{2\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma(2\nu+1)\Gamma\left(\frac 12-\nu\right)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac 1{2\pi}\frac{\Gamma(2\nu)\Gamma(\nu)}{\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}
\end{align}
となって3つ目の等式が示される. 4つ目の等式は, 補題1において$f(t)=Y_{\nu}(t)^2-J_{\nu}(t)^2, g(t)=J_{\nu}(t)Y_{\nu}(t), s=2-2\nu$として,
\begin{align}
&\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)Y_{\nu}(t)(Y_{\nu}(t)^2-J_{\nu}(t)^2)\,dt\\
&=-\frac 1{2\pi i}\frac1{2\pi^2}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac u2\right)\Gamma\left(\nu+\frac u2\right)\Gamma\left(\frac u2-\nu\right)}{\Gamma\left(\frac{1}2+\frac u2\right)}\cos\pi\left(\nu-\frac u2\right)\frac{\Gamma\left(1-\nu-\frac u2\right)\Gamma\left(1-\frac u2\right)}{\Gamma\left(2\nu+\frac u2\right)\Gamma\left(\frac 32-\nu-\frac u2\right)}\,du\\
&=-\frac 1{2\pi i}\frac1{\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u-\nu\right)\cos\pi\left(\nu-u\right)}{\Gamma\left(\frac{1}2+u\right)\Gamma\left(2\nu+u\right)\Gamma\left(\frac 32-\nu-u\right)}\frac{\pi}{\sin\pi u\sin\pi(u+\nu)}\,du\qquad u\mapsto 2u\\
&=-\frac {\cos\pi\nu}{\pi\sin\pi\nu}\sum_{0\leq n}\frac{(-1)^n\Gamma(n+1-\nu)}{\Gamma\left(\frac 32+n\right)\Gamma(2\nu+1+n)\Gamma\left(\frac 12-\nu-n\right)}\\
&\qquad+\frac{\cos2\pi\nu}{\pi\sin\pi\nu}\sum_{0\leq n}\frac{(-1)^n\Gamma(n+1-2\nu)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma(\nu+n+1)\Gamma\left(\frac 12-n\right)}\\
&=-\frac {\cos\pi\nu}{\pi\sin\pi\nu}\sum_{0\leq n}\frac{(-1)^n\Gamma(n+1-\nu)}{\Gamma\left(\frac 32+n\right)\Gamma(2\nu+1+n)\Gamma\left(\frac 12-\nu-n\right)}\\
&\qquad+\frac{\cos2\pi\nu}{\pi\sin\pi\nu}\sum_{n\in\ZZ}\frac{(-1)^n\Gamma(n+1-2\nu)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma(\nu+n+1)\Gamma\left(\frac 12-n\right)}+\frac{\cos2\pi\nu}{\pi\sin\pi\nu}\sum_{0\leq n}\frac{(-1)^n\Gamma(-n-2\nu)}{\Gamma\left(\frac 12-\nu-n\right)\Gamma(\nu-n)\Gamma\left(\frac 32+n\right)}\\
&=\left(-\frac {\cos\pi\nu}{\pi\sin\pi\nu}\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 32\right)\Gamma(2\nu+1)\Gamma\left(\frac 12-\nu\right)}+\frac{\cos 2\pi\nu}{\pi\sin\pi\nu}\frac{\Gamma(-2\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma\left(\nu\right)\Gamma\left(\frac 32\right)}\right)\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac{\cos2\pi\nu}{\sin\pi\nu\sin 2\pi\nu}\sum_{n\in\ZZ}\frac{1}{\Gamma\left(\frac 32-\nu+n\right)\Gamma(\nu+n+1)\Gamma\left(\frac 12-n\right)\Gamma(2\nu-n)}\\
&=\frac{2}{\pi^{\frac 32}\sin\pi\nu\Gamma\left(\frac 12-\nu\right)}\left(-\frac{\cos\pi\nu\Gamma(1-\nu)}{\Gamma(2\nu+1)}+\frac{\cos 2\pi\nu\Gamma(-2\nu)}{\Gamma\left(\nu\right)}\right)\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac{\cos2\pi\nu}{\sin\pi\nu\sin 2\pi\nu}\frac{\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)\Gamma(1-\nu)}\\
&=-\frac{2\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\left(\frac{\cos\pi\nu}{\sin\pi\nu}+\frac{\cos 2\pi\nu}{\sin 2\pi\nu}\right)\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac{\cos2\pi\nu}{\pi\sin 2\pi\nu}\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}
\end{align}
よって, 2つ目の等式と合わせて,
\begin{align}
&\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)Y_{\nu}(t)^3\,dt\\
&=-\frac{2\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\left(\frac{\cos\pi\nu}{\sin\pi\nu}+\frac{\cos 2\pi\nu}{\sin 2\pi\nu}\right)\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac{\cos2\pi\nu}{\pi\sin 2\pi\nu}\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}-\frac{\Gamma\left(\frac 12+\nu\right)}{\pi^{\frac 32}\Gamma(\nu)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&=-\frac{\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\left(\frac{2\cos\pi\nu}{\sin\pi\nu}+\frac{2\cos 2\pi\nu}{\sin 2\pi\nu}+\frac{\sin\pi\nu}{\cos\pi\nu}\right)\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac{\cos2\pi\nu}{\pi\sin 2\pi\nu}\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}\\
&=-\frac{3\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\frac{\cos\pi\nu}{\sin\pi\nu}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\cos2\pi\nu}{\pi\sin 2\pi\nu}\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}
\end{align}
を得る. 5つ目の等式は補題1において$f(t)=g(t)=Y_{\nu}(t)^2-J_{\nu}(t)^2,s=2-2\nu$として,
\begin{align}
&\int_0^{\infty}t^{1-2\nu}(Y_{\nu}(t)^2-J_{\nu}(t)^2)^2\,dt\\
&=\frac 1{2\pi i}\frac1{\pi^3}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac u2\right)\Gamma\left(\nu+\frac u2\right)\Gamma\left(\frac u2-\nu\right)}{\Gamma\left(\frac{1}2+\frac u2\right)}\cos\pi\left(\nu-\frac u2\right)\\
&\qquad\cdot\frac{\Gamma\left(1-\nu-\frac u2\right)\Gamma\left(1-\frac u2\right)\Gamma\left(1-2\nu-\frac u2\right)}{\Gamma\left(\frac 32-\nu-\frac u2\right)}\cos\pi\left(2\nu-1+\frac u2\right)\,du\\
&=-\frac 1{2\pi i}\frac 2{\pi^2}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u-\nu\right)\Gamma\left(1-2\nu-u\right)}{\Gamma\left(\frac{1}2+u\right)\Gamma\left(\frac 32-\nu-u\right)}\frac{\pi\cos\pi\left(\nu-u\right)\cos\pi\left(2\nu+u\right)}{\sin\pi u\sin\pi(u+\nu)}\,du\qquad u\mapsto 2u\\
&=\frac{2}{\pi^2}\frac{\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(n+1-\nu)\Gamma(-2\nu-n)}{\Gamma\left(\frac 32+n\right)\Gamma\left(\frac 12-\nu-n\right)}\\
&\qquad-\frac{2}{\pi^2}\frac{\cos2\pi\nu\cos\pi\nu}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(n+1-2\nu)\Gamma(-\nu-n)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma\left(\frac 12-n\right)}\\
&\qquad-\frac{2}{\pi}\frac{\cos3\pi\nu}{\sin 2\pi\nu\sin\pi\nu}\sum_{0\leq n}\frac{(-1)^n\Gamma(n+1-3\nu)}{n!\Gamma\left(\frac 32-2\nu+n\right)\Gamma\left(\frac 12+\nu-n\right)}\\
&=\frac{4}{\pi^2}\frac{\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(n+1-\nu)\Gamma(-2\nu-n)}{\Gamma\left(\frac 32+n\right)\Gamma\left(\frac 12-\nu-n\right)}\\
&\qquad-\frac{2}{\pi^2}\frac{\cos2\pi\nu\cos\pi\nu}{\sin\pi\nu}\sum_{n\in\ZZ}\frac{\Gamma(n+1-2\nu)\Gamma(-\nu-n)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma\left(\frac 12-n\right)}\\
&\qquad-\frac{2}{\pi}\frac{\cos3\pi\nu}{\sin 2\pi\nu\sin\pi\nu}\frac{\Gamma(1-3\nu)}{\Gamma\left(\frac 32-2\nu\right)\Gamma\left(\frac 12+\nu\right)}\F21{1-3\nu,\frac 12-\nu}{\frac 32-2\nu}1\\
&=\frac{4}{\pi^2}\frac{\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu}\frac{\Gamma(1-\nu)\Gamma(-2\nu)}{\Gamma\left(\frac 32\right)\Gamma\left(\frac 12-\nu\right)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac{2\cos2\pi\nu\cos\pi\nu}{\sin^2\pi\nu\sin2\pi\nu}\sum_{n\in\ZZ}\frac{1}{\Gamma\left(\frac 32-\nu+n\right)\Gamma\left(\frac 12-n\right)\Gamma(2\nu-n)\Gamma(1+\nu+n)}\\
&\qquad-\frac{2}{\pi}\frac{\cos3\pi\nu}{\sin 2\pi\nu\sin\pi\nu}\frac{\Gamma(1-3\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(1-\nu)}\\
&=-\frac{8}{\pi^{\frac 32}}\frac{\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu\sin 2\pi\nu}\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac{2\cos2\pi\nu\cos\pi\nu}{\sin^2\pi\nu\sin2\pi\nu}\frac{\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)\Gamma(1-\nu)}-\frac{2}{\pi}\frac{\cos3\pi\nu}{\sin 2\pi\nu\sin3\pi\nu}\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}\\
&=-\frac{8}{\pi^{\frac 32}}\frac{\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu\sin 2\pi\nu}\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 2{\pi\sin 2\pi\nu}\left(\frac{\cos2\pi\nu\cos\pi\nu}{\sin\pi\nu}-\frac{\cos3\pi\nu}{\sin3\pi\nu}\right)\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}
\end{align}
となる. よって,
\begin{align}
&\int_0^{\infty}t^{1-2\nu}Y_{\nu}(t)^4\,dt\\
&=-\frac{8}{\pi^{\frac 32}}\frac{\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu\sin 2\pi\nu}\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 2{\pi\sin 2\pi\nu}\left(\frac{\cos2\pi\nu\cos\pi\nu}{\sin\pi\nu}-\frac{\cos3\pi\nu}{\sin3\pi\nu}\right)\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}\\
&\qquad+2\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^2Y_{\nu}(t)^2\,dt-\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^4\,dt\\
&=-\frac{8}{\pi^{\frac 32}}\frac{\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu\sin 2\pi\nu}\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 2{\pi\sin 2\pi\nu}\left(\frac{\cos2\pi\nu\cos\pi\nu}{\sin\pi\nu}-\frac{\cos3\pi\nu}{\sin3\pi\nu}\right)\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}\\
&\qquad-\frac{4\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\Gamma(2\nu)\Gamma(\nu)}{2\pi\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}\\
&=-\frac{4}{\pi^{\frac 32}}\left(\frac{2\cos\pi\nu\cos 2\pi\nu}{\sin\pi\nu\sin 2\pi\nu}+1\right)\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 1{2\pi\sin 2\pi\nu}\left(\frac{4\cos2\pi\nu\cos\pi\nu}{\sin\pi\nu}-\frac{4\cos3\pi\nu}{\sin3\pi\nu}+\sin 2\pi\nu\right)\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}\\
&=-\frac{4\cos^2\pi\nu}{\pi^{\frac 32}\sin^2\pi\nu}\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 1{2\pi\sin 2\pi\nu}\left(\frac{4\cos\pi\nu}{\sin\pi\nu}-\frac{4\cos3\pi\nu}{\sin3\pi\nu}-3\sin 2\pi\nu\right)\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}\\
&=-\frac{4\cos^2\pi\nu}{\pi^{\frac 32}\sin^2\pi\nu}\frac{\Gamma(1-\nu)}{\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 1{2\pi}\left(\frac{4}{\sin\pi\nu\sin 3\pi\nu}-3\right)\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}
\end{align}
となって示すべき等式が得られる.
定理3の式
\begin{align}
\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^4\,dt&=\frac{\Gamma(2\nu)\Gamma(\nu)}{2\pi\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}\\
\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^3Y_{\nu}(t)\,dt&=-\frac{\Gamma\left(\frac 12+\nu\right)}{\pi^{\frac 32}\Gamma(\nu)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)^2Y_{\nu}(t)^2\,dt&=-\frac{2\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\Gamma(2\nu)\Gamma(\nu)}{2\pi\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}\\
\int_0^{\infty}t^{1-2\nu}J_{\nu}(t)Y_{\nu}(t)^3\,dt&=-\frac{3\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\frac{\cos\pi\nu}{\sin\pi\nu}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\cos2\pi\nu}{\pi\sin 2\pi\nu}\frac{\Gamma(2\nu)\Gamma(\nu)}{\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2}\\
\int_0^{\infty}t^{1-2\nu}Y_{\nu}(t)^4\,dt&=-\frac{4\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\left(\frac{\cos\pi\nu}{\sin\pi\nu}\right)^2\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\frac 1{2\pi}\left(\frac{4}{\sin\pi\nu\sin 3\pi\nu}-3\right)\frac{\Gamma(2\nu)\Gamma(\nu)}{\Gamma(3\nu)\Gamma\left(\frac 12+\nu\right)^2}
\end{align}
の右辺は全て
\begin{align}
\frac{\Gamma(2\nu)\Gamma(\nu)}{\pi\Gamma(3\nu)\Gamma\left(\nu+\frac 12\right)^2},\qquad\frac{\Gamma\left(\frac 12+\nu\right)}{\pi^{\frac 32}\Gamma(\nu)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1
\end{align}
の三角関数を係数とする線形結合で表されているところが興味深い. また, $J_{\nu}(t)^{4-k}Y_{\nu}(t)^k$の右辺の
\begin{align}
\frac{\Gamma(1-\nu)}{\pi^{\frac 32}\Gamma\left(\frac 12-\nu\right)\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1
\end{align}
の係数が
\begin{align}
-k\left(\frac{\cos\pi\nu}{\sin\pi\nu}\right)^{k-2}
\end{align}
になっているという不思議な規則性があるようである.
