Bessel関数の3つの積の積分
今回はBessel関数の3つの積の積分
\begin{align}
\int_0^{\infty}J_{\nu}(t)^kY_{\nu}(t)^{3-k}\,dt\qquad 0\leq k\leq 3
\end{align}
を考える.
前の記事 で用いた補題1, 補題2に加えて, Bessel関数のMellin変換 を用いる.
\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}f(t)g(t)\,dt&=\frac 1{\pi i}\int_{-i\infty}^{i\infty}F(2u)G(1-2u)\,du
\end{align}
が成り立つ.
\begin{align} \int_0^{\infty}t^{s-1}J_{\nu}(t)\,dt&=\frac{2^{s-1}\Gamma\left(\frac{\nu}2+\frac s2\right)}{\Gamma\left(1+\frac{\nu}2-\frac s2\right)}\\ \int_0^{\infty}t^{s-1}Y_{\nu}(t)\,dt&=-\frac{2^{s-1}\Gamma\left(\frac{s}2+\frac{\nu}2\right)\Gamma\left(\frac s2-\frac{\nu}2\right)}{\pi}\cos\pi\frac{s-\nu}2\\ \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}
さらに, 前の記事 で示した以下の和公式を用いる.
\begin{align} \F32{2a,2b,1-2b}{a+b+\frac 12,a-b+1}{\frac 14}&=\frac{2^{4a/3}\Gamma\left(a+b+\frac 12\right)\Gamma(a-b+1)\Gamma\left(\frac{2a}3\right)}{3\Gamma\left(\frac a3+b+\frac 12\right)\Gamma\left(\frac a3-b+1\right)\Gamma(2a)} \end{align}
これらを用いて以下を示す.
\begin{align} \int_0^{\infty}J_{\nu}(t)^3\,dt&=\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)}{2^{2/3}\sqrt{3\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}\\ \int_0^{\infty}J_{\nu}(t)^2Y_{\nu}(t)\,dt&=-\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)}{2^{2/3}3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}\\ \int_0^{\infty}J_{\nu}(t)Y_{\nu}(t)^2\,dt&=\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)\left(\frac{\sqrt 3}2+\tan\frac{\pi\nu}2\right)}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}\\ \int_0^{\infty}Y_{\nu}(t)^3\,dt&=-\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)}{2^{2/3}\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}-\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)\Gamma\left(\frac 16-\frac{\nu}2\right)\sin\frac{3\pi\nu}2}{2^{2/3}\sqrt 3\pi^{3/2}\cos^2\frac{\pi\nu}2\Gamma\left(\frac 56\right)}\\ \end{align}
$f(t)=J_{\nu}(t)^2,g(t)=J_{\nu}(t)$として補題1, 補題2を用いて, Mellin-Barnes積分を積分路の右側の極に関して展開すると,
\begin{align}
&\int_0^{\infty}J_{\nu}(t)^3\,dt\\
&=\frac 1{2\pi i}\frac 1{\sqrt{\pi}}\int_{-i\infty}^{i\infty}\frac{\Gamma(u+\nu)\Gamma\left(\frac 12-u\right)\Gamma\left(\frac{1+\nu}2-u\right)}{\Gamma\left(\nu+1-u\right)\Gamma(1-u)\Gamma\left(\frac{1+\nu}2+u\right)}4^{-u}\,du\\
&=\frac 1{2\sqrt{\pi}}\sum_{0\leq n}\frac{(-1)^n\Gamma\left(\nu+\frac 12+n\right)\Gamma\left(\frac{\nu}2-n\right)}{n!\Gamma\left(\nu+\frac 12-n\right)\Gamma\left(\frac 12-n\right)\Gamma\left(1+\frac{\nu}2+n\right)}4^{-n}\\
&\qquad+\frac 1{2^{\nu+1}\sqrt{\pi}}\sum_{0\leq n}\frac{(-1)^n\Gamma\left(\frac{3\nu+1}2+n\right)\Gamma\left(-\frac{\nu}2-n\right)}{n!\Gamma\left(\frac{\nu+1}2-n\right)\Gamma\left(\frac{1-\nu}2-n\right)\Gamma\left(\frac 12+\nu+n\right)}4^{-n}\\
&=\frac 1{\pi\nu}\F32{\frac 12,\nu+\frac 12,\frac 12-\nu}{1-\frac{\nu}2,1+\frac{\nu}2}{\frac 14}\\
&\qquad+\frac{\Gamma\left(\frac{3\nu+1}2\right)\Gamma\left(-\frac{\nu}2\right)}{2^{\nu+1}\sqrt{\pi}\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(1+\nu\right)}\F32{\frac{3\nu+1}2,\frac{1-\nu}2,\frac{1+\nu}2}{1+\frac{\nu}2,1+\nu}{\frac 14}
\end{align}
ここで, 補題3を用いると,
\begin{align}
&\frac 1{\pi\nu}\F32{\frac 12,\nu+\frac 12,\frac 12-\nu}{1-\frac{\nu}2,1+\frac{\nu}2}{\frac 14}\\
&\qquad+\frac{\Gamma\left(\frac{3\nu+1}2\right)\Gamma\left(-\frac{\nu}2\right)}{2^{\nu+1}\sqrt{\pi}\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(1+\nu\right)}\F32{\frac{3\nu+1}2,\frac{1-\nu}2,\frac{1+\nu}2}{1+\frac{\nu}2,1+\nu}{\frac 14}\\
&=\frac 1{\pi\nu}\frac{2^{1/3}\Gamma\left(1-\frac{\nu}2\right)\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac 16\right)}{3\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac 56-\frac{\nu}2\right)\Gamma\left(\frac 12\right)}\\
&\qquad+\frac{\Gamma\left(\frac{3\nu+1}2\right)\Gamma\left(-\frac{\nu}2\right)}{2^{\nu+1}\sqrt{\pi}\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(1+\nu\right)}\frac{2^{\nu+1/3}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(1+\nu\right)\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac{3\nu+1}2\right)}\\
&=\frac{2^{1/3}\sqrt{\pi}}{3\sin\frac{\pi\nu}2\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac 56-\frac{\nu}2\right)}\\
&\qquad-\frac{2^{-2/3}\sqrt{\pi}}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac 56-\frac{\nu}2\right)}\frac{\cos\frac{\pi\nu}2}{\sin\frac{\pi\nu}2\sin\left(\frac{\nu}2+\frac 16\right)\pi}\\
&=\frac{2^{1/3}\sqrt{\pi}}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac 56-\frac{\nu}2\right)}\frac{\sin\left(\frac{\nu}2+\frac 16\right)\pi-\sin\frac{\pi}6\cos\frac{\pi\nu}2}{\sin\frac{\pi\nu}2\sin\left(\frac{\nu}2+\frac 16\right)\pi}\\
&=\frac{2^{1/3}\sqrt{\pi}}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac 56-\frac{\nu}2\right)}\frac{\cos\frac{\pi}6}{\sin\left(\frac{\nu}2+\frac 16\right)\pi}\\
&=\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)}{2^{2/3}\sqrt{3\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}\\
\end{align}
となって1つ目の等式が得られる. 次に, $f(t)=J_{\nu}(t)Y_{\nu}(t), g(t)=J_{\nu}(t)$として同様に
\begin{align}
&\int_0^{\infty}J_{\nu}(t)^2Y_{\nu}(t)\,dt\\
&=-\frac 1{2\pi i}\frac 1{\sqrt{\pi}}\int_{-i\infty}^{i\infty}\frac{\Gamma(u)\Gamma(u+\nu)\Gamma\left(\frac{1+\nu}2-u\right)}{\Gamma(1+\nu-u)\Gamma\left(\frac 12+u\right)\Gamma\left(\frac{1+\nu}2+u\right)}4^{-u}\,du\\
&=-\frac 1{2^{\nu+1}\sqrt{\pi}}\sum_{0\leq n}\frac{(-1)^n\Gamma\left(\frac{\nu+1}2+n\right)\Gamma\left(\frac{3\nu+1}2+n\right)}{n!\Gamma\left(\frac{1+\nu}2-n\right)\Gamma\left(1+\frac{\nu}2+n\right)\Gamma(1+\nu+n)}4^{-n}\\
&=-\frac{\Gamma\left(\frac{3\nu+1}2\right)}{2^{\nu+1}\sqrt{\pi}\Gamma\left(1+\frac{\nu}2\right)\Gamma(1+\nu)}\F32{\frac{3\nu+1}2,\frac{\nu+1}2,\frac{1-\nu}2}{1+\frac{\nu}2,1+\nu}{\frac 14}\\
&=-\frac{\Gamma\left(\frac{3\nu+1}2\right)}{2^{\nu+1}\sqrt{\pi}\Gamma\left(1+\frac{\nu}2\right)\Gamma(1+\nu)}\frac{2^{\nu+1/3}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(1+\nu\right)\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac{3\nu+1}2\right)}\\
&=-\frac{\Gamma\left(\frac{\nu}2+\frac 16\right)}{2^{2/3}3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}
\end{align}
を得る. 次に$f(t)=Y_{\nu}(t)^2-J_{\nu}(t)^2, g(t)=J_{\nu}(t)$として, 同様に,
\begin{align}
&\int_0^{\infty}J_{\nu}(t)(Y_{\nu}(t)^2-J_{\nu}(t)^2)\,dt\\
&=\frac 1{2\pi i}\frac{2}{\pi^{3/2}}\int_{-i\infty}^{i\infty}\frac{\Gamma(u)\Gamma(u-\nu)\Gamma(u-\nu)\Gamma\left(\frac{1+\nu}2-u\right)}{\Gamma\left(\frac 12+u\right)\Gamma\left(\frac{1+\nu}2+u\right)}\cos\pi(u-\nu)4^{-u}\,du\\
&=\frac{1}{2^{\nu}\pi^{3/2}}\cos\frac{\pi(\nu-1)}2\sum_{0\leq n}\frac{\Gamma\left(\frac{\nu+1}2+n\right)\Gamma\left(\frac{3\nu+1}2+n\right)\Gamma\left(\frac{1-\nu}2+n\right)}{n!\Gamma\left(1+\frac{\nu}2+n\right)\Gamma(1+\nu+n)}4^{-n}\\
&=\frac{1}{2^{\nu}\pi^{3/2}}\frac{\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+1}2\right)\Gamma\left(\frac{1-\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)\Gamma(1+\nu)}\sin\frac{\pi\nu}2\F32{\frac{3\nu+1}2,\frac{\nu+1}2,\frac{1-\nu}2}{1+\frac{\nu}2,1+\nu}{\frac 14}\\
&=\frac{1}{2^{\nu}\pi^{3/2}}\frac{\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+1}2\right)\Gamma\left(\frac{1-\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)\Gamma(1+\nu)}\sin\frac{\pi\nu}2\frac{2^{\nu+1/3}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(1+\nu\right)\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac{3\nu+1}2\right)}\\
&=\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)\tan\frac{\pi\nu}2}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}
\end{align}
だから,
\begin{align}
&\int_0^{\infty}J_{\nu}(t)Y_{\nu}(t)^2\,dt\\
&=\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)}{2^{2/3}\sqrt{3\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}+\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)\tan\frac{\pi\nu}2}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}\\
&=\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)\left(\frac{\sqrt 3}2+\tan\frac{\pi\nu}2\right)}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}
\end{align}
を得る. 次に, $f(t)=Y_{\nu}(t)^2-J_{\nu}(t)^2,g(t)=Y_{\nu}(t)$として, 同様に
\begin{align}
&\int_0^{\infty}(Y_{\nu}(t)^2-J_{\nu}(t)^2)Y_{\nu}(t)\,dt\\
&=-\frac 1{2\pi i}\frac 2{\pi^{5/2}}\int_{-i\infty}^{i\infty}\frac{\Gamma(u)\Gamma(u+\nu)\Gamma(u-\nu)\Gamma\left(\frac{1+\nu}2-u\right)\Gamma\left(\frac{1-\nu}2-u\right)}{\Gamma\left(\frac 12+u\right)}\cos\pi(u-\nu)\sin\pi\left(\frac{\nu}2+u\right)4^{-u}\,du\\
&=-\frac{1}{2^{\nu}\pi^{5/2}}\sin\frac{\pi\nu}2\cos\pi\nu\sum_{0\leq n}\frac{(-1)^n\Gamma\left(\frac{\nu+1}2+n\right)\Gamma\left(\frac{3\nu+1}2+n\right)\Gamma\left(\frac{1-\nu}2+n\right)\Gamma(-\nu-n)}{n!\Gamma\left(1+\frac{\nu}2+n\right)}\\
&\qquad-\frac{1}{2^{-\nu}\pi^{5/2}}\sin\frac{3\pi\nu}2\sum_{0\leq n}\frac{(-1)^n\Gamma\left(\frac{1-\nu}2+n\right)\Gamma\left(\frac{1+\nu}2+n\right)\Gamma\left(\frac{1-3\nu}2+n\right)\Gamma(\nu-n)}{n!\Gamma\left(1-\frac{\nu}2+n\right)}\\
&=-\frac{1}{2^{\nu}\pi^{5/2}}\sin\frac{\pi\nu}2\cos\pi\nu\frac{\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+1}2\right)\Gamma\left(\frac{1-\nu}2\right)\Gamma(-\nu)}{\Gamma\left(1+\frac{\nu}2\right)}\F32{\frac{3\nu+1}2,\frac{\nu+1}2,\frac{1-\nu}2}{1+\frac{\nu}2,1+\nu}{\frac 14}\\
&\qquad-\frac{1}{2^{-\nu}\pi^{5/2}}\sin\frac{3\pi\nu}2\frac{\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(\frac{1+\nu}2\right)\Gamma\left(\frac{1-3\nu}2\right)\Gamma(\nu)}{\Gamma\left(1-\frac{\nu}2\right)}\F32{\frac{1-3\nu}2,\frac{1-\nu}2,\frac{1+\nu}2}{1-\frac{\nu}2,1-\nu}{\frac 14}\\
&=-\frac{1}{2^{\nu}\pi^{5/2}}\sin\frac{\pi\nu}2\cos\pi\nu\frac{\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+1}2\right)\Gamma\left(\frac{1-\nu}2\right)\Gamma(-\nu)}{\Gamma\left(1+\frac{\nu}2\right)}\frac{2^{\nu+1/3}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(1+\nu\right)\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)\Gamma\left(\frac{3\nu+1}2\right)}\\
&\qquad-\frac{1}{2^{-\nu}\pi^{5/2}}\sin\frac{3\pi\nu}2\frac{\Gamma\left(\frac{1-\nu}2\right)\Gamma\left(\frac{1+\nu}2\right)\Gamma\left(\frac{1-3\nu}2\right)\Gamma(\nu)}{\Gamma\left(1-\frac{\nu}2\right)}\frac{2^{-\nu+1/3}\Gamma\left(1-\frac{\nu}2\right)\Gamma\left(1-\nu\right)\Gamma\left(\frac 16-\frac{\nu}2\right)}{3\Gamma\left(\frac 56\right)\Gamma\left(\frac 56-\frac{\nu}2\right)\Gamma\left(\frac{1-3\nu}2\right)}\\
&=\frac{\sin\frac{\pi\nu}2\cos\pi\nu}{\cos\frac{\pi\nu}2\sin\pi\nu}\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}-\frac{\sin\frac{3\pi\nu}2}{\cos\frac{\pi\nu}2\sin\pi\nu}\frac{2^{1/3}\Gamma\left(\frac 16-\frac{\nu}2\right)}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56-\frac{\nu}2\right)}\\
&=\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)\Gamma\left(\frac 16-\frac{\nu}2\right)}{3\pi^{3/2}\cos\frac{\pi\nu}2\sin\pi\nu\Gamma\left(\frac 56\right)}\left(\sin\frac{\pi\nu}2\cos\pi\nu\sin\left(\frac 16-\frac{\nu}2\right)\pi-\sin\frac{3\pi\nu}2\sin\left(\frac 16+\frac{\nu}2\right)\pi\right)\\
&=\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)\Gamma\left(\frac 16-\frac{\nu}2\right)}{3\pi^{3/2}\cos\frac{\pi\nu}2\sin\pi\nu\Gamma\left(\frac 56\right)}\sin\frac{3\pi\nu}2\left(\sin\left(\frac 16-\frac{\nu}2\right)\pi-\sin\left(\frac 16+\frac{\nu}2\right)\pi\right)\\
&\qquad-\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}\\
&=-\frac{\Gamma\left(\frac{\nu}2+\frac 16\right)\Gamma\left(\frac 16-\frac{\nu}2\right)\sin\frac{3\pi\nu}2}{2^{2/3}\sqrt 3\pi^{3/2}\cos^2\frac{\pi\nu}2\Gamma\left(\frac 56\right)}-\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}
\end{align}
であるから,
\begin{align}
&\int_0^{\infty}Y_{\nu}(t)^3\,dt\\
&=-\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)}{2^{2/3}3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}-\frac{\Gamma\left(\frac{\nu}2+\frac 16\right)\Gamma\left(\frac 16-\frac{\nu}2\right)\sin\frac{3\pi\nu}2}{2^{2/3}\sqrt 3\pi^{3/2}\cos^2\frac{\pi\nu}2\Gamma\left(\frac 56\right)}-\frac{2^{1/3}\Gamma\left(\frac{\nu}2+\frac 16\right)}{3\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}\\
&=-\frac{\Gamma\left(\frac 16+\frac{\nu}2\right)}{2^{2/3}\sqrt{\pi}\Gamma\left(\frac 56\right)\Gamma\left(\frac 56+\frac{\nu}2\right)}-\frac{\Gamma\left(\frac{\nu}2+\frac 16\right)\Gamma\left(\frac 16-\frac{\nu}2\right)\sin\frac{3\pi\nu}2}{2^{2/3}\sqrt 3\pi^{3/2}\cos^2\frac{\pi\nu}2\Gamma\left(\frac 56\right)}
\end{align}
特に$\nu=0$とすると以下を得る.
\begin{align}
\int_0^{\infty}J_{0}(t)^3\,dt&=\frac{2^{1/3}\sqrt{\pi}}{\sqrt 3\Gamma\left(\frac 56\right)^3}\\
\int_0^{\infty}J_{0}(t)^2Y_{0}(t)\,dt&=-\frac{2^{1/3}\sqrt{\pi}}{3\Gamma\left(\frac 56\right)^3}\\
\int_0^{\infty}J_{0}(t)Y_{0}(t)^2\,dt&=\frac{2^{1/3}\sqrt{\pi}}{\sqrt 3\Gamma\left(\frac 56\right)^3}\\
\int_0^{\infty}Y_{0}(t)^3\,dt&=-\frac{2^{1/3}\sqrt{\pi}}{\Gamma\left(\frac 56\right)^3}
\end{align}
定理4は
ZhouによるLegendre関数の3つの積の積分
\begin{align}
\int_{-1}^1P_{\nu}(x)^2P_{\nu}(-x)\,dx&=\frac{1+2\cos\pi\nu}{3}\frac{\pi\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+2}2\right)}{\Gamma\left(\frac{1-\nu}2\right)^2\Gamma\left(\frac{\nu+2}2\right)^3\Gamma\left(\frac{3\nu+3}2\right)}
\end{align}
のBessel関数での類似と言えるかもしれない.
