Bessel関数の積のMellin変換
第1種, 第2種のBessel関数は
\begin{align}
J_{\nu}(z)&:=\sum_{0\leq m}\frac{(-1)^m}{m!\Gamma(m+\nu+1)}\left(\frac x2\right)^{2m+\nu}\\
Y_{\nu}(z)&:=\frac{J_{\nu}(z)\cos\nu\pi-J_{-\nu}(z)}{\sin\nu\pi}
\end{align}
によって定義される.
前の記事
でBessel関数, 変形Bessel関数のMellin変換を計算した. 今回はBessel関数の2つの積のMellin変換を計算したいと思う.
\begin{align} \int_0^{\infty}t^{s-1}J_{\mu}(t)J_{\nu}(t)\,dt=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)} \end{align}
前の記事
で示したBessel関数の積の級数表示と
Ramanujan's master theorem
より
\begin{align}
\int_0^{\infty}t^{s-1}J_{\mu}(t)J_{\nu}(t)\,dt&=\int_0^{\infty}t^{\mu+\nu+s-1}\sum_{0\leq m}\frac{(-t^2)^m\Gamma(\mu+\nu+2m+1)}{m!2^{\mu+\nu+2m}\Gamma(\mu+\nu+m+1)\Gamma(\mu+m+1)\Gamma(\nu+m+1)}\,dt\\
&=\frac 12\int_0^{\infty}t^{\frac{\mu+\nu+s}2-1}\sum_{0\leq m}\frac{(-t)^m\Gamma(\mu+\nu+2m+1)}{m!2^{\mu+\nu+2m}\Gamma(\mu+\nu+m+1)\Gamma(\mu+m+1)\Gamma(\nu+m+1)}\,dt\\
&=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)}
\end{align}
と示される.
\begin{align} \int_0^{\infty}t^{s-1}J_{\mu}(t)Y_{\nu}(t)\,dt&=-\frac{2^{s-1}\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)\cos\frac{\pi(\mu-\nu+s)}2}{\pi\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\mu+\nu-s}2\right)} \end{align}
定理1より,
\begin{align}
&\int_0^{\infty}t^{s-1}J_{\mu}(t)Y_{\nu}(t)\,dt\\
&=\frac{\cos\nu\pi}{\sin\nu\pi}\int_0^{\infty}t^{s-1}J_{\mu}(t)J_{\nu}(t)\,dt-\frac 1{\sin\nu\pi}\int_0^{\infty}t^{s-1}J_{\mu}(t)J_{-\nu}(t)\,dt\\
&=\frac{\cos\nu\pi}{\sin\nu\pi}\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)}-\frac 1{\sin\nu\pi}\frac{2^{s-1}\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1-\frac{\nu+\mu+s}2\right)}\\
&=\frac 1{\sin\nu\pi}\frac{2^{s-1}\pi\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)\Gamma\left(1-\frac{\nu+\mu+s}2\right)}\\
&\qquad\cdot \left(\frac{\cos\nu\pi}{\sin\frac{\pi(\mu+\nu+s)}2}-\frac 1{\sin\frac{\pi(\mu-\nu+s)}{2}}\right)\\
&=-\frac 1{\sin\nu\pi}\frac{2^{s-1}\pi\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)\Gamma\left(1-\frac{\nu+\mu+s}2\right)}\\
&\qquad\cdot \frac{\sin\nu\pi\cos\frac{\pi(\mu-\nu+s)}2}{\sin\frac{\pi(\mu+\nu+s)}2\sin\frac{\pi(\mu-\nu+s)}{2}}\\
&=-\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma(1-s)\cos\frac{\pi(\mu-\nu+s)}2}{\pi\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}2\right)}
\end{align}
と示される.
定理2において$\mu,\nu$を入れ替えたものとの差を考えると
\begin{align}
&\int_0^{\infty}t^{s-1}(J_{\mu}(t)Y_{\nu}(t)-J_{\nu}(t)Y_{\mu}(t))\,dt\\
&=\frac{2^{s-1}\Gamma\left(\frac{\nu-\mu+s}2\right)\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)\cos\frac{\pi(\nu-\mu+s)}2}{\pi\Gamma\left(1+\frac{\nu-\mu-s}2\right)\Gamma\left(1+\frac{\mu+\nu-s}2\right)}-\frac{2^{s-1}\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)\cos\frac{\pi(\mu-\nu+s)}2}{\pi\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\mu+\nu-s}2\right)}\\
&=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}2\right)}\left(\frac{\cos\frac{\pi(\nu-\mu+s)}2}{\sin\frac{\pi(\nu-\mu+s)}2}-\frac{\cos\frac{\pi(\mu-\nu+s)}2}{\sin\frac{\pi(\mu-\nu+s)}2}\right)\\
&=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{\nu-\mu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma(1-s)\sin\pi(\mu-\nu)}{\pi^2\Gamma\left(1+\frac{\mu+\nu-s}2\right)}
\end{align}
となる. つまり以下の系を得る.
\begin{align} &\int_0^{\infty}t^{s-1}(J_{\mu}(t)Y_{\nu}(t)-J_{\nu}(t)Y_{\mu}(t))\,dt\\ &=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{\nu-\mu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma(1-s)\sin\pi(\mu-\nu)}{\pi^2\Gamma\left(1+\frac{\mu+\nu-s}2\right)} \end{align}
次に, 第2種Bessel関数の2つの積のMellin変換を求める.
\begin{align} &\int_0^{\infty}t^{s-1}Y_{\mu}(t)Y_{\nu}(t)\,dt\\ &=\frac{2^{s-2}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)\Gamma(1-s)}{\pi^3}\\ &\qquad\cdot\bigg(2\cos\frac{\pi(\mu+\nu-s)}2\sin\pi s+\sin\frac{\pi(\mu+\nu-s)}2(\cos\pi s-\cos\pi(\mu-\nu))\bigg) \end{align}
定理2より,
\begin{align}
&\int_0^{\infty}t^{s-1}Y_{\mu}(t)Y_{\nu}(t)\,dt\\
&=\frac{\cos\mu\pi}{\sin\mu\pi}\int_0^{\infty}t^{s-1}J_{\mu}(t)Y_{\nu}(t)\,dt-\frac{1}{\sin\mu\pi}\int_0^{\infty}t^{s-1}J_{-\mu}(t)Y_{\nu}(t)\,dt\\
&=-\frac{\cos\mu\pi}{\sin\mu\pi}\frac{2^{s-1}\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)\cos\frac{\pi(\mu-\nu+s)}2}{\pi\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\mu+\nu-s}2\right)}\\
&\qquad+\frac{1}{\sin\mu\pi}\frac{2^{s-1}\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)\cos\frac{\pi(\mu-\nu+s)}2}{\pi\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\mu+\nu-s}2\right)}\\
&=\frac{2^{s-1}\Gamma(1-s)}{\pi\sin\mu\pi}\left(-\frac{\cos\mu\pi\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{\mu+\nu+s}2\right)\cos\frac{\pi(\mu-\nu+s)}2}{\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\mu+\nu-s}2\right)}+\frac{\Gamma\left(\frac{-\mu-\nu+s}2\right)\Gamma\left(\frac{\nu-\mu+s}2\right)\cos\frac{\pi(-\mu-\nu+s)}2}{\Gamma\left(1-\frac{\mu+\nu+s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)}\right)\\
&=\frac{2^{s-1}\pi\Gamma(1-s)}{\sin\mu\pi\Gamma\left(1+\frac{\mu-\nu-s}2\right)\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1-\frac{\mu+\nu+s}2\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)}\\
&\qquad\cdot\left(-\frac{\cos\mu\pi\cos\frac{\pi(\mu-\nu+s)}2}{\sin\frac{\pi(\mu+\nu+s)}2\sin\frac{\pi(\mu-\nu+s)}2}+\frac{\cos\frac{\pi(-\mu-\nu+s)}2}{\sin\frac{\pi(-\mu-\nu+s)}2\sin\frac{\pi(\nu-\mu+s)}2}\right)\\
&=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)\Gamma(1-s)}{\pi^3\sin\mu\pi}\\
&\qquad\cdot\bigg(-\cos\mu\pi\cos\frac{\pi(\mu-\nu+s)}2\sin\frac{\pi(-\mu-\nu+s)}2\sin\frac{\pi(\nu-\mu+s)}2\\
&\qquad\qquad+\cos\frac{\pi(-\mu-\nu+s)}2\sin\frac{\pi(\mu+\nu+s)}2\sin\frac{\pi(\mu-\nu+s)}2\bigg)\\
\end{align}
ここで,
\begin{align}
&-\cos\mu\pi\cos\frac{\pi(\mu-\nu+s)}2\sin\frac{\pi(-\mu-\nu+s)}2\sin\frac{\pi(\nu-\mu+s)}2\\
&\qquad+\cos\frac{\pi(-\mu-\nu+s)}2\sin\frac{\pi(\mu+\nu+s)}2\sin\frac{\pi(\mu-\nu+s)}2\\
&=\frac 12\cos\mu\pi(\sin\mu\pi+\sin\pi(\nu-s))\sin\frac{\pi(\nu-\mu+s)}2\\
&\qquad+\frac 12(\sin\mu\pi-\sin\pi(\nu-s))\sin\frac{\pi(\mu+\nu+s)}2\\
&=\frac 12\sin\mu\pi\left(\cos\pi\mu\sin\frac{\pi(\nu-\mu+s)}2+\sin\frac{\pi(\mu+\nu+s)}2\right)\\
&\qquad+\frac 12\sin\pi(\nu-s)\left(\cos\pi\mu\sin\frac{\pi(\nu-\mu+s)}2-\sin\frac{\pi(\mu+\nu+s)}{2}\right)\\
&=\frac 12\sin\mu\pi\left(\cos\pi\mu\sin\frac{\pi(\nu-\mu+s)}2+\sin\frac{\pi(\mu+\nu+s)}2\right)\\
&\qquad-\frac 12\sin\mu\pi\sin\pi(\nu-s)\cos\frac{\pi(\nu-\mu+s)}2\\
&=\frac 12\sin\mu\pi\left(\cos\pi\mu\sin\frac{\pi(\nu-\mu+s)}2-\sin\pi(\nu-s)\cos\frac{\pi(\nu-\mu+s)}2+\sin\frac{\pi(\mu+\nu+s)}2\right)
\end{align}
であり,
\begin{align}
&\cos\pi\mu\sin\frac{\pi(\nu-\mu+s)}2-\sin\pi(\nu-s)\cos\frac{\pi(\nu-\mu+s)}2\\
&=\frac 12\sin\frac{\pi(\mu+\nu+s)}2+\frac 12\sin\frac{\pi(\nu-3\mu+s)}2+\frac 12\sin\frac{\pi(\mu-3\nu+s)}2-\frac 12\sin\frac{\pi(\mu+\nu-3s)}2\\
&=\cos\frac{\pi(\mu+\nu-s)}2\sin\pi s-\sin\frac{\pi(\mu+\nu-s)}2\cos\pi(\mu-\nu)
\end{align}
となるから,
\begin{align}
&\int_0^{\infty}t^{s-1}Y_{\mu}(t)Y_{\nu}(t)\,dt\\
&=\frac{2^{s-2}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)\Gamma(1-s)}{\pi^3}\\
&\qquad\cdot\bigg(\cos\frac{\pi(\mu+\nu-s)}2\sin\pi s-\sin\frac{\pi(\mu+\nu-s)}2\cos\pi(\mu-\nu)+\sin\frac{\pi(\mu+\nu+s)}2\bigg)\\
&=\frac{2^{s-2}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)\Gamma(1-s)}{\pi^3}\\
&\qquad\cdot\bigg(2\cos\frac{\pi(\mu+\nu-s)}2\sin\pi s+\sin\frac{\pi(\mu+\nu-s)}2(\cos\pi s-\cos\pi(\mu-\nu))\bigg)
\end{align}
を得る.
定理3の右辺の三角関数は積の形にはできなさそうであるが, 2つの項で書くと
\begin{align}
&\frac{2^{s-2}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)\Gamma(1-s)}{\pi^3}\\
&\qquad\cdot\bigg(2\cos\frac{\pi(\mu+\nu-s)}2\sin\pi s+\sin\frac{\pi(\mu+\nu-s)}2(\cos\pi s-\cos\pi(\mu-\nu))\bigg)\\
&=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)}{\pi^2\Gamma(s)}\cos\frac{\pi(\mu+\nu-s)}2\\
&\qquad-\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)\Gamma(1-s)}{\pi^3}\sin\frac{\pi(\mu+\nu-s)}2\sin\frac{\pi(s-\mu+\nu)}2\sin\frac{\pi(s+\nu-\mu)}2\\
&=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)}{\pi^2\Gamma(s)}\cos\frac{\pi(\mu+\nu-s)}2\\
&\qquad+\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma(1-s)}{\Gamma\left(1+\frac{\mu+\nu-s}2\right)\Gamma\left(1+\frac{\mu-\nu-s}{2}\right)\Gamma\left(1+\frac{\nu-\mu-s}2\right)}
\end{align}
となり, 各項はシンプルな形でまとまっている. 特に2つ目の項が定理1の右辺と一致していることに注目すると以下のようにも書ける
\begin{align} &\int_0^{\infty}t^{s-1}(Y_{\mu}(t)Y_{\nu}(t)-J_{\mu}(t)J_{\nu}(t))\,dt\\ &=\frac{2^{s-1}\Gamma\left(\frac{\mu+\nu+s}2\right)\Gamma\left(\frac{-\mu+\nu+s}2\right)\Gamma\left(\frac{\mu-\nu+s}2\right)\Gamma\left(\frac{-\mu-\nu+s}2\right)}{\pi^2\Gamma(s)}\cos\frac{\pi(\mu+\nu-s)}2 \end{align}
