変形Bessel関数の4つの積の積分
第1種, 第2種の変形Bessel関数は
\begin{align}
I_{\nu}(z)&:=\sum_{0\leq n}\frac{1}{n!\Gamma(\nu+n+1)}\left(\frac x2\right)^{2n+\nu}\\
K_{\nu}(z)&:=\frac{\pi}2\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin\pi\nu}
\end{align}
と定義される.
前の記事
でBessel関数の4つの積の積分を計算したが, 今回はその類似として4つの変形Bessel関数の4つの積の積分
\begin{align}
\int_0^{\infty}t^{1-2\nu}I_{\nu}(t)^kK_{\nu}(t)^{4-k}\,dt\qquad 0\leq k\leq 2
\end{align}
を計算したいと思う. ここにおいて, $0\leq k\leq 2$に限定しているのは$k=3,4$の場合は積分が収束しないためである.
前の記事 の定理1, 定理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}I_{\nu}(t)K_{\nu}(t)\,dt&=\frac{\Gamma\left(\frac s2\right)\Gamma\left(\frac s2+\nu\right)\Gamma\left(\frac{1-s}2\right)}{4\sqrt{\pi}\Gamma\left(1+\nu-\frac s2\right)}\\ \int_0^{\infty}t^{s-1}K_{\nu}(t)^2\,dt&=\frac{\sqrt{\pi}\Gamma\left(\frac s2\right)\Gamma\left(\frac{s}2+\nu\right)\Gamma\left(\frac s2-\nu\right)}{4\Gamma\left(\frac{s+1}2\right)} \end{align}
前の記事 の補題2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.
これらを用いて以下を示す.
\begin{align} \int_0^{\infty}t^{1-2\nu}I_{\nu}(t)^2K_{\nu}(t)^2\,dt&=\frac{\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{2\sqrt{\pi}\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\ \int_0^{\infty}t^{1-2\nu}I_{\nu}(t)K_{\nu}(t)^3\,dt&=-\frac{3\sqrt{\pi}\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{8\sin\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\pi^2\Gamma(\nu)\Gamma(2\nu)}{8\sin 2\pi\nu\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}\\ \int_0^{\infty}t^{1-2\nu}K_{\nu}(t)^4\,dt&=\frac{\pi^{\frac 32}\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{4\sin^2\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\ &\qquad+\left(\frac 1{\sin 3\pi\nu}-\frac 1{\sin\pi\nu}\right)\frac{\pi^3\Gamma(\nu)\Gamma(2\nu)}{8\sin 2\pi\nu\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)} \end{align}
補題2において, $f(t)=g(t)=I_{\nu}(t)K_{\nu}(t),s=2-2\nu$として,
\begin{align}
\int_0^{\infty}t^{1-2\nu}I_{\nu}(t)^2K_{\nu}(t)^2\,dt&=\frac 1{2\pi i}\frac 1{16\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac u2\right)\Gamma\left(\frac u2+\nu\right)\Gamma\left(\frac{1-u}2\right)}{\Gamma\left(1+\nu-\frac u2\right)}\frac{\Gamma\left(1-\nu-\frac u2\right)\Gamma\left(1-\frac u2\right)\Gamma\left(\nu-\frac 12+\frac u2\right)}{\Gamma\left(2\nu+\frac u2\right)}\,du\\
&=\frac 1{2\pi i}\frac 1{8\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)\Gamma\left(u+\nu\right)\Gamma\left(\frac{1}2-u\right)}{\Gamma\left(1+\nu-u\right)}\frac{\Gamma\left(1-\nu-u\right)\Gamma\left(1-u\right)\Gamma\left(\nu-\frac 12+u\right)}{\Gamma\left(2\nu+u\right)}\,du
\end{align}
ここで積分路の右側の極に関してMellin-Barnes積分を展開すると,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)\Gamma\left(u+\nu\right)\Gamma\left(\frac{1}2-u\right)}{\Gamma\left(1+\nu-u\right)}\frac{\Gamma\left(1-\nu-u\right)\Gamma\left(1-u\right)\Gamma\left(\nu-\frac 12+u\right)}{\Gamma\left(2\nu+u\right)}\,du\\
&=\sum_{0\leq n}\frac{(-1)^n\Gamma(n+\nu+1)\Gamma\left(-\frac 12-n\right)\Gamma(-\nu-n)\Gamma\left(\nu+\frac 12+n\right)}{\Gamma(\nu-n)\Gamma(2\nu+n+1)}\\
&\qquad+\sum_{0\leq n}\frac{(-1)^n\Gamma(1-\nu+n)\Gamma\left(\nu-\frac 12-n\right)\Gamma(\nu-n)\Gamma\left(\frac 12+n\right)}{\Gamma(2\nu-n)\Gamma(1+\nu+n)}\\
&\qquad+\sum_{0\leq n}\frac{(-1)^n\Gamma\left(\frac 12+n\right)\Gamma\left(\frac 12+\nu+n\right)\Gamma\left(\frac 12-\nu-n\right)\Gamma\left(\frac 12-n\right)\Gamma\left(\nu+n\right)}{n!\Gamma\left(\frac 12+\nu-n\right)\Gamma\left(2\nu+\frac 12+n\right)}\\
&=-\frac{\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma\left(-\frac 12-n\right)\Gamma\left(\nu+\frac 12+n\right)}{\Gamma(\nu-n)\Gamma(2\nu+n+1)}+\frac{\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma\left(\nu-\frac 12-n\right)\Gamma\left(\frac 12+n\right)}{\Gamma(2\nu-n)\Gamma(1+\nu+n)}\\
&\qquad+\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma(\nu)}{\Gamma\left(2\nu+\frac 12\right)}\F21{\frac 12-\nu,\nu}{2\nu+\frac 12}1\\
&=-\frac{2\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma\left(-\frac 12-n\right)\Gamma\left(\nu+\frac 12+n\right)}{\Gamma(\nu-n)\Gamma(2\nu+n+1)}+\frac{\pi}{\sin\pi\nu}\sum_{n\in\ZZ}\frac{\Gamma\left(\nu-\frac 12-n\right)\Gamma\left(\frac 12+n\right)}{\Gamma(2\nu-n)\Gamma(1+\nu+n)}\\
&\qquad+\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(3\nu\right)\Gamma\left(\nu+\frac 12\right)}\\
&=\frac{4\sqrt{\pi}\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{\Gamma(2\nu+1)}\F32{1,1-\nu,\nu+\frac 12}{\frac 32,2\nu+1}1-\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)\Gamma(3\nu)}\\
&\qquad+\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(3\nu\right)\Gamma\left(\nu+\frac 12\right)}\\
&=\frac{4\sqrt{\pi}\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{\Gamma(2\nu+1)}\F32{1,1-\nu,\nu+\frac 12}{\frac 32,2\nu+1}1
\end{align}
となる. ここで, 最後から2つ目の等号は
Dougallの${}_2H_2$和公式
による. これを代入すると,
\begin{align}
\int_0^{\infty}t^{1-2\nu}I_{\nu}(t)^2K_{\nu}(t)^2\,dt&=\frac{\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{2\sqrt{\pi}\Gamma(2\nu+1)}\F32{1,1-\nu,\nu+\frac 12}{\frac 32,2\nu+1}1
\end{align}
となって1つ目の式が得られる. $f(t)=K_{\nu}(t)^2,g(t)=I_{\nu}(t)K_{\nu}(t), s=2-2\nu$とすると,
\begin{align}
\int_0^{\infty}t^{1-2\nu}I_{\nu}(t)K_{\nu}(t)^3\,dt&=\frac 1{2\pi i}\frac 1{16}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac u2\right)\Gamma\left(\frac{u}2+\nu\right)\Gamma\left(\frac u2-\nu\right)}{\Gamma\left(\frac{u+1}2\right)}\frac{\Gamma\left(1-\nu-\frac u2\right)\Gamma\left(1-\frac u2\right)\Gamma\left(\nu-\frac 12+\frac u2\right)}{\Gamma\left(2\nu+\frac u2\right)}\,du\\
&=\frac 1{2\pi i}\frac 1{8}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)\Gamma\left(u+\nu\right)\Gamma\left(u-\nu\right)}{\Gamma\left(u+\frac{1}2\right)}\frac{\Gamma\left(1-\nu-u\right)\Gamma\left(1-u\right)\Gamma\left(\nu-\frac 12+u\right)}{\Gamma\left(2\nu+u\right)}\,du
\end{align}
ここで積分路の右側の極に関してMellin-Barnes積分を展開すると,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)\Gamma\left(u+\nu\right)\Gamma\left(u-\nu\right)}{\Gamma\left(u+\frac{1}2\right)}\frac{\Gamma\left(1-\nu-u\right)\Gamma\left(1-u\right)\Gamma\left(\nu-\frac 12+u\right)}{\Gamma\left(2\nu+u\right)}\,du\\
&=\sum_{0\leq n}\frac{(-1)^n\Gamma(n+1+\nu)\Gamma(n+1-\nu)\Gamma(-\nu-n)\Gamma\left(\nu+\frac 12+n\right)}{\Gamma\left(\frac 32+n\right)\Gamma(2\nu+n+1)}\\
&\qquad+\sum_{0\leq n}\frac{(-1)^n\Gamma(1-\nu+n)\Gamma(1-2\nu+n)\Gamma(\nu-n)\Gamma\left(\frac 12+n\right)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma(1+\nu+n)}\\
&=-\frac{\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(n+1-\nu)\Gamma\left(\nu+\frac 12+n\right)}{\Gamma\left(\frac 32+n\right)\Gamma(2\nu+n+1)}+\frac{\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(1-2\nu+n)\Gamma\left(\frac 12+n\right)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma(1+\nu+n)}\\
&=-\frac{\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(n+1-\nu)\Gamma\left(\nu+\frac 12+n\right)}{\Gamma\left(\frac 32+n\right)\Gamma(2\nu+n+1)}-\frac{\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(-2\nu-n)\Gamma\left(-\frac 12-n\right)}{\Gamma\left(\frac 12-\nu-n\right)\Gamma(\nu-n)}\\
&\qquad+\frac{\pi}{\sin\pi\nu}\sum_{n\in\ZZ}\frac{\Gamma(1-2\nu+n)\Gamma\left(\frac 12+n\right)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma(1+\nu+n)}\\
&=-\frac{3\pi}{2\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(n+1-\nu)\Gamma\left(\nu+\frac 12+n\right)}{\Gamma\left(\frac 32+n\right)\Gamma(2\nu+n+1)}+\frac{\pi^2}{\sin 2\pi\nu}\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}\\
&=-\frac{3\sqrt{\pi}\Gamma(1-\nu)\Gamma\left(\nu+\frac 12\right)}{\sin\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\pi^2}{\sin 2\pi\nu}\frac{\Gamma(\nu)\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}
\end{align}
であるから, これを代入すると,
\begin{align}
&\int_0^{\infty}t^{1-2\nu}I_{\nu}(t)K_{\nu}(t)^3\,dt\\
&=-\frac{3\sqrt{\pi}\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{8\sin\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\pi^2}{8\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. 3つ目の式は, 補題2において$f(t)=g(t)=K_{\nu}(t)^2,s=2-2\nu$として,
\begin{align}
\int_0^{\infty}t^{1-2\nu}K_{\nu}(t)^4\,dt&=\frac 1{2\pi i}\frac{\pi}{16}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac u2\right)\Gamma\left(\frac{u}2+\nu\right)\Gamma\left(\frac u2-\nu\right)}{\Gamma\left(\frac{u+1}2\right)}\frac{\Gamma\left(1-\nu-\frac u2\right)\Gamma\left(1-\frac{u}2\right)\Gamma\left(1-2\nu-\frac u2\right)}{\Gamma\left(\frac 32-\nu-\frac{u}2\right)}\,du\\
&=\frac 1{2\pi i}\frac{\pi}{8}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)\Gamma\left(u+\nu\right)\Gamma\left(u-\nu\right)}{\Gamma\left(\frac{1}2+u\right)}\frac{\Gamma\left(1-\nu-u\right)\Gamma\left(1-u\right)\Gamma\left(1-2\nu-u\right)}{\Gamma\left(\frac 32-\nu-u\right)}\,du
\end{align}
ここで, 積分路の右側の極に関してMellin-Barnes積分を展開すると,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)\Gamma\left(u+\nu\right)\Gamma\left(u-\nu\right)}{\Gamma\left(\frac{1}2+u\right)}\frac{\Gamma\left(1-\nu-u\right)\Gamma\left(1-u\right)\Gamma\left(1-2\nu-u\right)}{\Gamma\left(\frac 32-\nu-u\right)}\,du\\
&=\sum_{0\leq n}\frac{(-1)^n\Gamma(n+1+\nu)\Gamma(n+1-\nu)\Gamma(-\nu-n)\Gamma(-2\nu-n)}{\Gamma\left(\frac 32+n\right)\Gamma\left(\frac 12-\nu-n\right)}\\
&\qquad+\sum_{0\leq n}\frac{(-1)^n\Gamma(1-\nu+n)\Gamma(1-2\nu+n)\Gamma(\nu-n)\Gamma(-\nu-n)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma\left(\frac 12-n\right)}\\
&\qquad+\sum_{0\leq n}\frac{(-1)^n\Gamma\left(1-2\nu+n\right)\Gamma(1-\nu+n)\Gamma(1-3\nu+n)\Gamma(\nu-n)\Gamma(2\nu-n)}{n!\Gamma\left(\frac 32-2\nu+n\right)\Gamma\left(\frac 12+\nu-n\right)}\\
&=-\frac{\pi}{\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)}+\frac{\pi}{\sin\pi\nu}\sum_{0\leq n}\frac{\Gamma(1-2\nu+n)\Gamma(-\nu-n)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma\left(\frac 12-n\right)}\\
&\qquad+\frac{\pi^2}{\sin\pi\nu\sin 2\pi\nu}\sum_{0\leq n}\frac{(-1)^n\Gamma(1-3\nu+n)}{n!\Gamma\left(\frac 32-2\nu+n\right)\Gamma\left(\frac 12+\nu-n\right)}\\
&=-\frac{2\pi}{\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)}+\frac{\pi}{\sin\pi\nu}\sum_{n\in\ZZ}\frac{\Gamma(1-2\nu+n)\Gamma(-\nu-n)}{\Gamma\left(\frac 32-\nu+n\right)\Gamma\left(\frac 12-n\right)}\\
&\qquad+\frac{\pi^2\Gamma(1-3\nu)}{\sin\pi\nu\sin 2\pi\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\sqrt{\pi}\Gamma(1-\nu)\Gamma(-2\nu)}{\sin\pi\nu\Gamma\left(\frac 12-\nu\right)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1-\frac{\pi^3}{\sin^2\pi\nu\sin 2\pi\nu}\frac{\Gamma(2\nu)}{\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)\Gamma(1-\nu)}\\
&\qquad+\frac{\pi^2\Gamma(1-3\nu)\Gamma(2\nu)}{\sin\pi\nu\sin 2\pi\nu\Gamma\left(\frac 12+\nu\right)^2\Gamma(1-\nu)}\\
&=\frac{2\sqrt{\pi}\Gamma(1-\nu)\Gamma\left(\nu+\frac 12\right)}{\sin^2\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\left(\frac 1{\sin 3\pi\nu}-\frac 1{\sin\pi\nu}\right)\frac{\pi^2\Gamma(\nu)\Gamma(2\nu)}{\sin 2\pi\nu\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}
\end{align}
となる. よってこれを代入して,
\begin{align}
\int_0^{\infty}t^{1-2\nu}K_{\nu}(t)^4\,dt&=\frac{\pi^{\frac 32}\Gamma(1-\nu)\Gamma\left(\nu+\frac 12\right)}{4\sin^2\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\left(\frac 1{\sin 3\pi\nu}-\frac 1{\sin\pi\nu}\right)\frac{\pi^3\Gamma(\nu)\Gamma(2\nu)}{8\sin 2\pi\nu\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}
\end{align}
を得る.
$K_{\nu}$の定義には
\begin{align}
K_{\nu}(x)&=\frac{\pi}2\frac{I_{-\nu}(x)-I_{\nu}(x)}{\sin\pi\nu}
\end{align}
と$\frac{\pi}2$が掛かっているので, その分を補正すると定理1は
\begin{align}
\left(\frac 2{\pi}\right)^2\int_0^{\infty}t^{1-2\nu}I_{\nu}(t)^2K_{\nu}(t)^2\,dt&=\frac{2\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{\pi^{\frac 52}\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
\left(\frac 2{\pi}\right)^3\int_0^{\infty}t^{1-2\nu}I_{\nu}(t)K_{\nu}(t)^3\,dt&=-\frac{3\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{\pi^{\frac 52}\sin\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1+\frac{\Gamma(\nu)\Gamma(2\nu)}{\pi\sin 2\pi\nu\Gamma\left(\nu+\frac 12\right)^2\Gamma(3\nu)}\\
\left(\frac 2{\pi}\right)^4\int_0^{\infty}t^{1-2\nu}K_{\nu}(t)^4\,dt&=\frac{4\Gamma\left(\nu+\frac 12\right)\Gamma(1-\nu)}{\pi^{\frac 52}\sin^2\pi\nu\Gamma(2\nu+1)}\F32{1,1-\nu,\frac 12+\nu}{\frac 32,2\nu+1}1\\
&\qquad+\left(\frac 1{\sin 3\pi\nu}-\frac 1{\sin\pi\nu}\right)\frac{2\Gamma(\nu)\Gamma(2\nu)}{\pi\sin 2\pi\nu\Gamma\left(\frac 12+\nu\right)^2\Gamma(3\nu)}
\end{align}
と書くことができる. 特に, ${}_3F_2$の係数に規則性が見える.
