変形Bessel関数の4つの積の積分3
今回は変形Bessel関数の4つの積の積分
\begin{align}
\int_0^{\infty}I_{\mu}(t)I_{\nu}(t)K_{\mu}(t)K_{\nu}(t)\,dt\\
\int_0^{\infty}I_{\nu}(t)K_{\mu}(t)^2K_{\nu}(t)\,dt\\
\int_0^{\infty}K_{\mu}(t)^2K_{\nu}(t)^2\,dt
\end{align}
を${}_7F_6$を用いて表したいと思う.
前の記事
で用いた補題1は以下のようなものだった.
\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を$u\mapsto 2u, s=1$としたものは以下のようになる.
\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.
これらを用いて以下を示す.
\begin{align} &\int_0^{\infty}I_{\mu}(t)I_{\nu}(t)K_{\mu}(t)K_{\nu}(t)\,dt\\ &=\frac{\pi\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 32+\mu\right)}{8\nu\Gamma(1+\mu)^2}\F76{\frac 12+\mu,\frac{\mu}2+\frac 54,\frac 12,\frac 12,\frac 12+\mu+\nu,\frac 12+\mu-\nu,\frac 12+\mu}{\frac{\mu}2+\frac 14,1+\mu,1+\mu,1-\nu,1+\nu,1}1\\ &\qquad+\frac{\Gamma\left(\frac 32+\mu+2\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12+\nu\right)^2\Gamma\left(\frac 12+\mu+\nu\right)\Gamma(-\nu)}{8\Gamma(1+2\nu)\Gamma(1+\mu+\nu)^2\Gamma(1+\nu)\Gamma\left(\frac 12+\mu-\nu\right)}\\ &\qquad\cdot\F76{\frac 12+\mu+2\nu,\frac{\mu}2+\nu+\frac 54,\frac 12+\nu,\frac 12+\nu,\frac 12+\mu,\frac 12+\mu+\nu,\frac 12+\mu+\nu}{\frac{\mu}2+\nu+\frac 14,1+\mu+\nu,1+\mu+\nu,1+2\nu,1+\nu,1+\nu}1\\ &\int_0^{\infty}I_{\nu}(t)K_{\mu}(t)^2K_{\nu}(t)\,dt\\ &=\frac{\pi^2 \Gamma\left(\frac 32+\mu+\nu\right)\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12+\nu\right)^2}{8\Gamma(1+\mu+\nu)^2\Gamma(1+\nu)^2\cos\pi\mu}\\ &\qquad\cdot\F76{\frac 12+\mu+\nu,\frac 54+\frac{\mu}2+\frac{\nu}2,\frac 12,\frac 12,\frac 12+\mu,\frac 12+\mu,\frac 12+\mu+\nu}{\frac 14+\frac{\mu}2+\frac{\nu}2,1+\mu+\nu,1+\mu+\nu,1+\nu,1+\nu,1}1\\ &\int_0^{\infty}K_{\mu}(t)^2K_{\nu}(t)^2\,dt\\ &=\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma\left(\frac 32-\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12-\mu\right)\Gamma(\nu)\Gamma\left(\frac 12-\mu-\nu\right)\Gamma\left(\frac12+\mu-\nu\right)}{8\Gamma(1-\nu)\Gamma(1+\mu-\nu)\Gamma(1-\mu-\nu)}\\ &\qquad\cdot\F76{\frac 12-\nu,\frac 54-\frac{\nu}2,\frac 12-\nu,\frac 12+\mu,\frac 12-\mu,\frac 12,\frac 12}{\frac 14-\frac{\nu}2,1,1-\mu-\nu,1+\mu+\nu,1-\nu,1-\nu}1\\ &\qquad+\frac{\pi\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12-\mu+\nu\right)\Gamma\left(\frac 12+\nu\right)\Gamma\left(\frac 32+\nu\right)\Gamma(-\nu)\Gamma\left(\frac 12-\mu-\nu\right)\Gamma\left(\frac12+\mu-\nu\right)}{8\Gamma(1+\nu)\Gamma(1+\mu)\Gamma(1-\mu)}\\ &\qquad\cdot\F76{\frac 12+\nu,\frac 54+\frac{\nu}2,\frac 12+\nu,\frac 12+\mu+\nu,\frac 12-\mu+\nu,\frac 12,\frac 12}{\frac 14+\frac{\nu}2,1,1-\mu,1+\mu,1+\nu,1+\nu}1 \end{align}
$f(t)=I_{\mu}(t)K_{\mu}(t),g(t)=I_{\nu}(t)K_{\nu}(t)$に関して補題1, 補題2を用いて,
\begin{align}
\int_0^{\infty}I_{\mu}(t)I_{\nu}(t)K_{\mu}(t)K_{\nu}(t)\,dt&=\frac 1{2\pi i}\frac 1{8\pi}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)^2\Gamma\left(u+\mu\right)\Gamma\left(\frac{1}2-u\right)^2\Gamma\left(\frac 12-\nu-u\right)}{\Gamma\left(1+\mu-u\right)\Gamma\left(\nu+\frac 12+u\right)}\,du
\end{align}
となる. ここで,
前の記事
で示したnon-terminating Whippleの変換公式のMellin-Barnes積分類似より,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(u\right)^2\Gamma\left(u+\mu\right)\Gamma\left(\frac{1}2-u\right)^2\Gamma\left(\nu+\frac 12-u\right)}{\Gamma\left(1+\mu-u\right)\Gamma\left(\nu+\frac 12+u\right)}\,du\\
&=\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12+u\right)^2\Gamma\left(\frac 12+u+\mu\right)\Gamma\left(-u\right)^2\Gamma\left(\nu-u\right)}{\Gamma\left(\frac 12+\mu-u\right)\Gamma\left(\nu+1+u\right)}\,du\\
&=\frac 1{2\pi i}\frac{\pi}{\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12+\mu-\nu\right)}\\
&\qquad\cdot\int_{-i\infty}^{i\infty}\frac{\left(2s+\mu+\frac 12\right)\Gamma\left(\frac 12+\mu+\nu+s\right)\Gamma\left(\frac 12+\mu-\nu+s\right)\Gamma\left(\frac 12+s\right)^2\Gamma\left(\frac 12+\mu+s\right)^2\Gamma(\nu-s)\Gamma(-s)}{\Gamma(1+\nu+s)\Gamma(1+\mu+s)^2\Gamma(1+s)}\,ds
\end{align}
ここで,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\left(2s+\mu+\frac 12\right)\Gamma\left(\frac 12+\mu+\nu+s\right)\Gamma\left(\frac 12+\mu-\nu+s\right)\Gamma\left(\frac 12+s\right)^2\Gamma\left(\frac 12+\mu+s\right)^2\Gamma(\nu-s)\Gamma(-s)}{\Gamma(1+\nu+s)\Gamma(1+\mu+s)^2\Gamma(1+s)}\,ds\\
&=\sum_{0\leq n}\frac{(-1)^n\left(2n+\mu+\frac 12\right)\Gamma\left(\frac 12+\mu+\nu+n\right)\Gamma\left(\frac 12+\mu-\nu+n\right)\Gamma\left(\frac 12+n\right)^2\Gamma\left(\frac 12+\mu+n\right)^2\Gamma(\nu-n)}{n!^2\Gamma(1+\nu+n)\Gamma(1+\mu+n)^2}\\
&\qquad+\sum_{0\leq n}\frac{(-1)^n\left(2n+2\nu+\mu+\frac 12\right)\Gamma\left(\frac 12+\mu+2\nu+n\right)\Gamma\left(\frac 12+\mu+n\right)\Gamma\left(\frac 12+\nu+n\right)^2\Gamma\left(\frac 12+\mu+\nu+n\right)^2\Gamma(-\nu-n)}{n!\Gamma(1+2\nu+n)\Gamma(1+\mu+\nu+n)^2\Gamma(1+\nu+n)}\\
&=\frac{\pi\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12+\mu-\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 32+\mu\right)}{\nu\Gamma(1+\mu)^2}\\
&\qquad\cdot\F76{\frac 12+\mu,\frac{\mu}2+\frac 54,\frac 12,\frac 12,\frac 12+\mu+\nu,\frac 12+\mu-\nu,\frac 12+\mu}{\frac{\mu}2+\frac 14,1+\mu,1+\mu,1-\nu,1+\nu,1}1\\
&\qquad+\frac{\Gamma\left(\frac 32+\mu+2\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12+\nu\right)^2\Gamma\left(\frac 12+\mu+\nu\right)^2\Gamma(-\nu)}{\Gamma(1+2\nu)\Gamma(1+\mu+\nu)^2\Gamma(1+\nu)}\\
&\qquad\cdot\F76{\frac 12+\mu+2\nu,\frac{\mu}2+\nu+\frac 54,\frac 12+\nu,\frac 12+\nu,\frac 12+\mu,\frac 12+\mu+\nu,\frac 12+\mu+\nu}{\frac{\mu}2+\nu+\frac 14,1+\mu+\nu,1+\mu+\nu,1+2\nu,1+\nu,1+\nu}1
\end{align}
となる. よって,
\begin{align}
&\int_0^{\infty}I_{\mu}(t)I_{\nu}(t)K_{\mu}(t)K_{\nu}(t)\,dt\\
&=\frac{1}{8\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12+\mu-\nu\right)}\\
&\qquad\bigg(\frac{\pi\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12+\mu-\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 32+\mu\right)}{\nu\Gamma(1+\mu)^2}\\
&\qquad\cdot\F76{\frac 12+\mu,\frac{\mu}2+\frac 54,\frac 12,\frac 12,\frac 12+\mu+\nu,\frac 12+\mu-\nu,\frac 12+\mu}{\frac{\mu}2+\frac 14,1+\mu,1+\mu,1-\nu,1+\nu,1}1\\
&\qquad+\frac{\Gamma\left(\frac 32+\mu+2\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12+\nu\right)^2\Gamma\left(\frac 12+\mu+\nu\right)^2\Gamma(-\nu)}{\Gamma(1+2\nu)\Gamma(1+\mu+\nu)^2\Gamma(1+\nu)}\\
&\qquad\cdot\F76{\frac 12+\mu+2\nu,\frac{\mu}2+\nu+\frac 54,\frac 12+\nu,\frac 12+\nu,\frac 12+\mu,\frac 12+\mu+\nu,\frac 12+\mu+\nu}{\frac{\mu}2+\nu+\frac 14,1+\mu+\nu,1+\mu+\nu,1+2\nu,1+\nu,1+\nu}1\bigg)\\
&=\frac{\pi\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 32+\mu\right)}{8\nu\Gamma(1+\mu)^2}\F76{\frac 12+\mu,\frac{\mu}2+\frac 54,\frac 12,\frac 12,\frac 12+\mu+\nu,\frac 12+\mu-\nu,\frac 12+\mu}{\frac{\mu}2+\frac 14,1+\mu,1+\mu,1-\nu,1+\nu,1}1\\
&\qquad+\frac{\Gamma\left(\frac 32+\mu+2\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12+\nu\right)^2\Gamma\left(\frac 12+\mu+\nu\right)\Gamma(-\nu)}{8\Gamma(1+2\nu)\Gamma(1+\mu+\nu)^2\Gamma(1+\nu)\Gamma\left(\frac 12+\mu-\nu\right)}\\
&\qquad\cdot\F76{\frac 12+\mu+2\nu,\frac{\mu}2+\nu+\frac 54,\frac 12+\nu,\frac 12+\nu,\frac 12+\mu,\frac 12+\mu+\nu,\frac 12+\mu+\nu}{\frac{\mu}2+\nu+\frac 14,1+\mu+\nu,1+\mu+\nu,1+2\nu,1+\nu,1+\nu}1
\end{align}
を得る. 次に, $f(t)=K_{\mu}(t)^2,g(t)=I_{\nu}(t)K_{\nu}(t)$に関して補題1, 補題2を用いて,
\begin{align}
&\int_0^{\infty}I_{\nu}(t)K_{\mu}(t)^2K_{\nu}(t)\,dt=\frac 1{2\pi i}\frac 18\int_{-i\infty}^{i\infty}\frac{\Gamma(u)^2\Gamma(u-\mu)\Gamma(u+\mu)\Gamma\left(\frac 12-u\right)\Gamma\left(\frac 12+\nu-u\right)}{\Gamma\left(\frac 12+u\right)\Gamma\left(\frac 12+\nu+u\right)}\,du
\end{align}
ここで,
non-terminating Whippleの変換公式のMellin-Barnes積分表示
より,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(u)^2\Gamma(u-\mu)\Gamma(u+\mu)\Gamma\left(\frac 12-u\right)\Gamma\left(\frac 12+\nu-u\right)}{\Gamma\left(\frac 12+u\right)\Gamma\left(\frac 12+\nu+u\right)}\,du\\
&=\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12+u\right)^2\Gamma\left(\frac 12+u-\mu\right)\Gamma\left(\frac 12+u+\mu\right)\Gamma\left(-u\right)\Gamma\left(\nu-u\right)}{\Gamma\left(1+u\right)\Gamma\left(1+\nu+u\right)}\,du\\
&=\frac{\pi \Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 32+\mu+\nu\right)\Gamma\left(\frac 12-\mu\right)\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12+\nu\right)^2}{\Gamma(1+\mu+\nu)^2\Gamma(1+\nu)^2}\\
&\qquad\cdot\F76{\frac 12+\mu+\nu,\frac 54+\frac{\mu}2+\frac{\nu}2,\frac 12,\frac 12,\frac 12+\mu,\frac 12+\mu,\frac 12+\mu+\nu}{\frac 14+\frac{\mu}2+\frac{\nu}2,1+\mu+\nu,1+\mu+\nu,1+\nu,1+\nu,1}1\\
&=\frac{\pi^2 \Gamma\left(\frac 32+\mu+\nu\right)\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12+\nu\right)^2}{\Gamma(1+\mu+\nu)^2\Gamma(1+\nu)^2\cos\pi\mu}\\
&\qquad\cdot\F76{\frac 12+\mu+\nu,\frac 54+\frac{\mu}2+\frac{\nu}2,\frac 12,\frac 12,\frac 12+\mu,\frac 12+\mu,\frac 12+\mu+\nu}{\frac 14+\frac{\mu}2+\frac{\nu}2,1+\mu+\nu,1+\mu+\nu,1+\nu,1+\nu,1}1
\end{align}
となるから, これを代入して, 2つ目の等式を得る. 次に, $f(t)=K_{\mu}(t)^2,g(t)=K_{\nu}(t)^2$に関して補題1, 補題2を用いて,
\begin{align}
&\int_0^{\infty}K_{\mu}(t)^2K_{\nu}(t)^2\,dt\\
&=\frac 1{2\pi i}\frac{\pi}8\int_{-i\infty}^{i\infty}\frac{\Gamma(u)\Gamma(u-\mu)\Gamma(u+\mu)\Gamma\left(\frac 12-u\right)\Gamma\left(\frac 12+\nu-u\right)\Gamma\left(\frac 12-\nu-u\right)}{\Gamma\left(\frac 12+u\right)\Gamma(1-u)}\,du
\end{align}
ここで,
前の記事
で示したnon-terminating Whippleの変換公式のMellin-Barnes積分類似より,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(u)\Gamma(u-\mu)\Gamma(u+\mu)\Gamma\left(\frac 12-u\right)\Gamma\left(\frac 12+\nu-u\right)\Gamma\left(\frac 12-\nu-u\right)}{\Gamma\left(\frac 12+u\right)\Gamma(1-u)}\,du\\
&=\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12+u\right)\Gamma\left(u+\frac 12-\mu\right)\Gamma\left(u+\frac 12+\mu\right)\Gamma\left(-u\right)\Gamma\left(\nu-u\right)\Gamma\left(-\nu-u\right)}{\Gamma\left(1+u\right)\Gamma\left(\frac 12-u\right)}\,du\\
&=\frac{\Gamma\left(\frac 12-\mu-\nu\right)\Gamma\left(\frac12+\mu-\nu\right)}{\pi}\\
&\qquad\cdot\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\left(2s+\frac 12-\nu\right)\Gamma\left(\frac 12-\nu+s\right)^2\Gamma\left(\frac 12+\mu+s\right)\Gamma\left(\frac 12-\mu+s\right)\Gamma\left(\frac 12+s\right)\Gamma(\nu-s)\Gamma(-s)}{\Gamma(1+s)\Gamma(1-\nu+s)\Gamma(1+\mu-\nu+s)\Gamma(1-\mu-\nu+s)}\,ds
\end{align}
ここで,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\left(2s+\frac 12-\nu\right)\Gamma\left(\frac 12-\nu+s\right)^2\Gamma\left(\frac 12+\mu+s\right)\Gamma\left(\frac 12-\mu+s\right)\Gamma\left(\frac 12+s\right)^2\Gamma(\nu-s)\Gamma(-s)}{\Gamma(1+s)\Gamma(1-\nu+s)\Gamma(1+\mu-\nu+s)\Gamma(1-\mu-\nu+s)}\,ds\\
&=\sum_{0\leq n}\frac{(-1)^n\left(2n+\frac 12-\nu\right)\Gamma\left(\frac 12-\nu+n\right)^2\Gamma\left(\frac 12+\mu+n\right)\Gamma\left(\frac 12-\mu+n\right)\Gamma\left(\frac 12+n\right)^2\Gamma(\nu-n)}{n!^2\Gamma(1-\nu+n)\Gamma(1+\mu-\nu+n)\Gamma(1-\mu-\nu+n)}\\
&\qquad+\sum_{0\leq n}\frac{(-1)^n\left(2n+\frac 12+\nu\right)\Gamma\left(\frac 12+n\right)^2\Gamma\left(\frac 12+\mu+\nu+n\right)\Gamma\left(\frac 12-\mu+\nu+n\right)\Gamma\left(\frac 12+\nu+n\right)^2\Gamma(-\nu-n)}{n!^2\Gamma(1+\nu+n)\Gamma(1+\mu+n)\Gamma(1-\mu+n)}\\
&=\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma\left(\frac 32-\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12-\mu\right)\Gamma(\nu)}{\Gamma(1-\nu)\Gamma(1+\mu-\nu)\Gamma(1-\mu-\nu)}\\
&\qquad\cdot\F76{\frac 12-\nu,\frac 54-\frac{\nu}2,\frac 12-\nu,\frac 12+\mu,\frac 12-\mu,\frac 12,\frac 12}{\frac 14-\frac{\nu}2,1,1-\mu-\nu,1+\mu+\nu,1-\nu,1-\nu}1\\
&\qquad+\frac{\pi\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12-\mu+\nu\right)\Gamma\left(\frac 12+\nu\right)\Gamma\left(\frac 32+\nu\right)\Gamma(-\nu)}{\Gamma(1+\nu)\Gamma(1+\mu)\Gamma(1-\mu)}\\
&\qquad\cdot\F76{\frac 12+\nu,\frac 54+\frac{\nu}2,\frac 12+\nu,\frac 12+\mu+\nu,\frac 12-\mu+\nu,\frac 12,\frac 12}{\frac 14+\frac{\nu}2,1,1-\mu,1+\mu,1+\nu,1+\nu}1
\end{align}
であるから, これを代入して,
\begin{align}
&\int_0^{\infty}K_{\mu}(t)^2K_{\nu}(t)^2\,dt\\
&=\frac{\Gamma\left(\frac 12-\mu-\nu\right)\Gamma\left(\frac12+\mu-\nu\right)}{8}\bigg(\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma\left(\frac 32-\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12-\mu\right)\Gamma(\nu)}{\Gamma(1-\nu)\Gamma(1+\mu-\nu)\Gamma(1-\mu-\nu)}\\
&\qquad\cdot\F76{\frac 12-\nu,\frac 54-\frac{\nu}2,\frac 12-\nu,\frac 12+\mu,\frac 12-\mu,\frac 12,\frac 12}{\frac 14-\frac{\nu}2,1,1-\mu-\nu,1+\mu+\nu,1-\nu,1-\nu}1\\
&\qquad+\frac{\pi\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12-\mu+\nu\right)\Gamma\left(\frac 12+\nu\right)\Gamma\left(\frac 32+\nu\right)\Gamma(-\nu)}{\Gamma(1+\nu)\Gamma(1+\mu)\Gamma(1-\mu)}\\
&\qquad\cdot\F76{\frac 12+\nu,\frac 54+\frac{\nu}2,\frac 12+\nu,\frac 12+\mu+\nu,\frac 12-\mu+\nu,\frac 12,\frac 12}{\frac 14+\frac{\nu}2,1,1-\mu,1+\mu,1+\nu,1+\nu}1\bigg)\\
&=\frac{\pi\Gamma\left(\frac 12-\nu\right)\Gamma\left(\frac 32-\nu\right)\Gamma\left(\frac 12+\mu\right)\Gamma\left(\frac 12-\mu\right)\Gamma(\nu)\Gamma\left(\frac 12-\mu-\nu\right)\Gamma\left(\frac12+\mu-\nu\right)}{8\Gamma(1-\nu)\Gamma(1+\mu-\nu)\Gamma(1-\mu-\nu)}\\
&\qquad\cdot\F76{\frac 12-\nu,\frac 54-\frac{\nu}2,\frac 12-\nu,\frac 12+\mu,\frac 12-\mu,\frac 12,\frac 12}{\frac 14-\frac{\nu}2,1,1-\mu-\nu,1+\mu+\nu,1-\nu,1-\nu}1\\
&\qquad+\frac{\pi\Gamma\left(\frac 12+\mu+\nu\right)\Gamma\left(\frac 12-\mu+\nu\right)\Gamma\left(\frac 12+\nu\right)\Gamma\left(\frac 32+\nu\right)\Gamma(-\nu)\Gamma\left(\frac 12-\mu-\nu\right)\Gamma\left(\frac12+\mu-\nu\right)}{8\Gamma(1+\nu)\Gamma(1+\mu)\Gamma(1-\mu)}\\
&\qquad\cdot\F76{\frac 12+\nu,\frac 54+\frac{\nu}2,\frac 12+\nu,\frac 12+\mu+\nu,\frac 12-\mu+\nu,\frac 12,\frac 12}{\frac 14+\frac{\nu}2,1,1-\mu,1+\mu,1+\nu,1+\nu}1
\end{align}
を得る.
定理3に現れた${}_7F_6$は全て異なっているが, それらが変換公式でどれぐらい移り合うのかについては今後考えていきたいところである.
