ZhouによるLegendre関数の3つの積の積分4

前の記事 で, ZhouによるLegendre関数の3つの積の積分
\begin{align} &\int_{-1}^1P_{\mu}(x)P_{\nu}(x)P_{\nu}(-x)\,dx\\ &=\frac{2}{\pi^2}\frac{\sin\pi\mu\sin\pi\nu}{\mu(\mu+1)}\left(\frac 1\nu\F43{1,\frac{1-\mu}2,\frac{\mu+2}2,-\nu}{\frac{2-\mu}2,\frac{\mu+3}2,1-\nu}1-\frac 1{\nu+1}\F43{1,\frac{1-\mu}2,\frac{\mu+2}2,\nu+1}{\frac{2-\mu}2,\frac{\mu+3}2,\nu+2}1\right)\\ &=\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{2\nu+1}\left(\frac 1{\mu}\F54{\frac 12,\frac 12,-\frac{\mu}2,-\nu,1+\nu}{1,\frac{2-\mu}2,\frac 12-\nu,\frac 32+\nu}1-\frac 1{\mu+1}\F54{\frac 12,\frac 12,\frac{1+\mu}2,-\nu,1+\nu}{1,\frac{3+\mu}2,\frac 12-\nu,\frac 32+\nu}1\right) \end{align}
を示した. 1つ目の表示は
\begin{align} &\frac{2}{\pi^2}\frac{\sin\pi\mu\sin\pi\nu}{\mu(\mu+1)}\left(\frac 1\nu\F43{1,\frac{1-\mu}2,\frac{\mu+2}2,-\nu}{\frac{2-\mu}2,\frac{\mu+3}2,1-\nu}1-\frac 1{\nu+1}\F43{1,\frac{1-\mu}2,\frac{\mu+2}2,\nu+1}{\frac{2-\mu}2,\frac{\mu+3}2,\nu+2}1\right)\\ &=\frac{2}{\pi^2}\frac{\sin\pi\mu\sin\pi\nu}{\mu(\mu+1)}\sum_{0\leq n}\left(\frac 1{\nu-n}-\frac 1{\nu+n+1}\right)\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_n}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_n}\\ &=\frac{2}{\pi^2}\frac{\sin\pi\mu\sin\pi\nu}{\mu(\mu+1)}\sum_{0\leq n}\frac{2n+1}{(\nu-n)(\nu+n+1)}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_n}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_n}\\ &=\frac{2}{\pi^2}\frac{\sin\pi\mu\sin\pi\nu}{\mu(\mu+1)\nu(\nu+1)}\F65{1,\frac 32,\frac{1-\mu}2,\frac{\mu+2}2,-\nu,\nu+1}{\frac 12,\frac{\mu+3}2,\frac{2-\mu}2,\nu+2,1-\nu}1 \end{align}
と表され, 2つ目の表示は
\begin{align} &\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{2\nu+1}\left(\frac 1{\mu}\F54{\frac 12,\frac 12,-\frac{\mu}2,-\nu,1+\nu}{1,\frac{2-\mu}2,\frac 12-\nu,\frac 32+\nu}1-\frac 1{\mu+1}\F54{\frac 12,\frac 12,\frac{1+\mu}2,-\nu,1+\nu}{1,\frac{3+\mu}2,\frac 12-\nu,\frac 32+\nu}1\right)\\ &=\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{2\nu+1}\sum_{0\leq n}\left(\frac 1{\mu-2n}-\frac 1{\mu+2n+1}\right)\frac{\left(\frac 12,\frac 12,\nu+1,-\nu\right)_n}{k!^2\left(\frac 12-\nu,\frac 32+\nu\right)_n}\\ &=\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{2\nu+1}\sum_{0\leq n}\frac{4n+1}{(\mu-2n)(\mu+2n+1)}\frac{\left(\frac 12,\frac 12,\nu+1,-\nu\right)_n}{k!^2\left(\frac 12-\nu,\frac 32+\nu\right)_n}\\ &=\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{(2\nu+1)\mu(\mu+1)}\F76{\frac 12,\frac 54,\frac 12,-\frac{\mu}2,\frac{\mu+1}2,\nu+1,-\nu}{\frac 14,\frac{2-\mu}2,\frac{\mu+3}2,\frac 12-\nu,\frac 32+\nu}1 \end{align}
となる. まとめると以下を得る.

この${}_6F_5$は
\begin{align} \F65{1,\frac 32,\frac{1-\mu}2,\frac{\mu+2}2,-\nu,\nu+1}{\frac 12,\frac{\mu+3}2,\frac{2-\mu}2,\nu+2,1-\nu}1&=\F76{1,\frac 32,\frac{1-\mu}2,\frac{\mu+2}2,-\nu,\nu+1,1}{\frac 12,\frac{\mu+3}2,\frac{2-\mu}2,\nu+2,1-\nu,1}1 \end{align}
と書けるから, 定理1の表示はいずれもvery-well-poisedな${}_7F_6$である. 上の2つの表示が等しいことは ${}_7F_6$の二項変換公式
\begin{align} &\F76{a,1+\frac a2,b,c,d,e,f}{\frac a2,1+a-b,1+a-c,1+a-d,1+a-e,1+a-f}1\\ &=\frac{\Gamma(1+a-e)\Gamma(1+a-f)\Gamma(1+w)\Gamma(1+w-e-f)}{\Gamma(1+a)\Gamma(1+a-e-f)\Gamma(1+w-e)\Gamma(1+w-f)}\\ &\qquad\cdot\F76{w,1+\frac w2,w+b-a,w+c-a,w+d-a,e,f}{\frac w2,1+a-b,1+a-c,1+a-d,1+w-e,1+w-f}1\qquad w=1+2a-b-c-d \end{align}
において, $e,f$を$-\nu,\nu+1$と選ぶと示される. また, 定理1の2つ目の表示において, $e,f$を$\frac 12,-\nu$と選ぶと
\begin{align} &\F76{\frac 12,\frac 54,\frac 12,-\frac{\mu}2,\frac{\mu+1}2,-\nu,\nu+1}{\frac 14,1,\frac{2-\mu}2,\frac{\mu+3}2,\frac 32+\nu,\frac 12-\nu}1\\ &=\frac{\Gamma\left(\frac 32+\nu\right)\Gamma\left(\frac 32-\nu\right)}{\Gamma\left(\frac 32\right)\Gamma(1+\nu)\Gamma\left(\frac 32\right)\Gamma(1-\nu)}\F76{\frac{1}2-\nu,\frac 54-\frac{\nu}2,1,\frac 12,\frac{\mu+1}2-\nu,-\frac{\mu}2-\nu,-\nu}{\frac 14-\frac{\nu}2,\frac 12-\nu,1-\nu,\frac{2-\mu}2,\frac{\mu+3}2,\frac 32}1\\ &=\frac{(2\nu+1)(1-2\nu)\sin\pi\nu}{\pi\nu\cos\pi\nu}\F65{\frac 54-\frac{\nu}2,1,\frac 12,\frac{\mu+1}2-\nu,-\frac{\mu}2-\nu,-\nu}{\frac 14-\frac{\nu}2,1-\nu,\frac{2-\mu}2,\frac{\mu+3}2,\frac 32}1 \end{align}
だから以下の表示を得る.

Mellin-Barnes積分表示

\begin{align} &\int_{-1}^1P_{\mu}(x)P_{\nu}(x)P_{\nu}(-x)\,dx\\ &=\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{(2\nu+1)\mu(\mu+1)}\F76{\frac 12,\frac 54,\frac 12,-\frac{\mu}2,\frac{\mu+1}2,-\nu,\nu+1}{\frac 14,1,\frac{2-\mu}2,\frac{\mu+3}2,\frac 32+\nu,\frac 12-\nu}1 \end{align}

定理1の2つ目の表示にnon-terminating Whippleの変換公式のMellin-Barnes積分表示を用いると
\begin{align} &\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{(2\nu+1)\mu(\mu+1)}\F76{\frac 12,\frac 54,\frac 12,-\frac{\mu}2,\frac{\mu+1}2,-\nu,\nu+1}{\frac 14,1,\frac{2-\mu}2,\frac{\mu+3}2,\frac 32+\nu,\frac 12-\nu}1\\ &=\frac 2{\pi}\frac{\sin\pi\mu\cos\pi\nu}{(2\nu+1)\mu(\mu+1)}\frac{\Gamma\left(\frac{2-\mu}2\right)\Gamma\left(\frac{\mu+3}2\right)\Gamma\left(\frac 32+\nu\right)\Gamma\left(\frac 12-\nu\right)}{\Gamma\left(\frac 12\right)^2\Gamma\left(\frac 32\right)\Gamma(-\nu)^2\Gamma(\nu+1)^2}\\ &\qquad\cdot \frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12+s\right)\Gamma(1+s)\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(-s)^2}{\Gamma\left(\frac{2-\mu}2+s\right)\Gamma\left(\frac{\mu+3}2+s\right)}\,ds\\ &=-\frac{\Gamma\left(\frac{-\mu}2\right)\Gamma\left(\frac{\mu+1}2\right)\sin\pi\mu\sin^2\pi\nu}{2\pi^{\frac 72}}\\ &\qquad\cdot \frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12+s\right)\Gamma(1+s)\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(-s)^2}{\Gamma\left(\frac{2-\mu}2+s\right)\Gamma\left(\frac{\mu+3}2+s\right)}\,ds\\ \end{align}
つまり以下を得る.

次に,
\begin{align} \int_{-1}^1(P_{\mu}(x)+P_{\mu}(-x))P_{\nu}(x)^2\,dx \end{align}
を考える. 前の記事 の用いた補題4, 補題1を用いると
\begin{align} &\int_{-1}^1(P_{\mu}(x)+P_{\mu}(-x))P_{\nu}(x)^2\,dx\\ &=\int_{-1}^1P_{\mu}(x)(P_{\nu}(x)^2+P_{\nu}(-x)^2)\,dx\\ &=2\int_0^1P_{\mu}(1-2t)(P_{\nu}(2t-1)^2+P_{\nu}(1-2t)^2)\,dt\qquad x\mapsto 1-2t\\ &=-\frac{4\sin^3\pi\nu}{\pi^{\frac 52}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{4^s\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(-s)^3}{\Gamma\left(\frac 12-s\right)}\int_0^1P_{\mu}(1-2t)(t(1-t))^s\,dt\,ds\\ &=-\frac{4\sin^3\pi\nu}{\pi^{\frac 52}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{4^s\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(-s)^3}{\Gamma\left(\frac 12-s\right)}\frac{\pi\Gamma(s+1)^2}{2^{2s+1}\Gamma\left(s+\frac{\mu+3}2\right)\Gamma\left(s+\frac{2-\mu}2\right)\Gamma\left(1+\frac{\mu}2\right)\Gamma\left(\frac{1-\mu}2\right)}\,ds\\ &=-\frac{2\sin^3\pi\nu}{\pi^{\frac 32}\Gamma\left(1+\frac{\mu}2\right)\Gamma\left(\frac{1-\mu}2\right)}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(s+1)^2\Gamma(-s)^3}{\Gamma\left(\frac 12-s\right)\Gamma\left(s+\frac{\mu+3}2\right)\Gamma\left(s+\frac{2-\mu}2\right)}\,ds\\ &=\frac{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(\frac{\mu+1}2\right)\sin^3\pi\nu\sin\pi\mu}{\pi^{\frac 72}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(s+1)^2\Gamma(-s)^3}{\Gamma\left(\frac 12-s\right)\Gamma\left(s+\frac{\mu+3}2\right)\Gamma\left(s+\frac{2-\mu}2\right)}\,ds \end{align}
となる. つまり以下を得る.

この右辺は積分路の右側の極に関して展開すると,
\begin{align} &\frac{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(\frac{\mu+1}2\right)\sin\pi\mu\sin^3\pi\nu}{\pi^{\frac 72}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(s+1)^2\Gamma(-s)^3}{\Gamma\left(\frac 12-s\right)\Gamma\left(s+\frac{\mu+3}2\right)\Gamma\left(s+\frac{2-\mu}2\right)}\,ds\\ &=-\frac{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(\frac{\mu+1}2\right)\sin\pi\mu\sin^3\pi\nu}{\pi^{\frac 92}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\frac 12+s\right)\Gamma(-\nu+s)\Gamma(\nu+1+s)}{\Gamma(s+1)\Gamma\left(s+\frac{\mu+3}2\right)\Gamma\left(s+\frac{2-\mu}2\right)}\frac{\pi^3\cos\pi s}{\sin^3\pi s}\,ds\\ &=\frac{\Gamma\left(-\frac{\mu}2\right)\Gamma\left(\frac{\mu+1}2\right)\sin\pi\mu\sin^3\pi\nu}{2\pi^{\frac 92}}\sum_{0\leq n}\left.\frac{\partial^2}{\partial s^2}\frac{\Gamma\left(\frac 12+s\right)\Gamma(-\nu+s)\Gamma(\nu+1+s)}{\Gamma(s+1)\Gamma\left(s+\frac{\mu+3}2\right)\Gamma\left(s+\frac{2-\mu}2\right)}\right|_{s=n}\\ &=\frac{2\sin\pi\mu\sin^2\pi\nu}{\pi^3\mu(\mu+1)}\left.\frac{\partial^2}{\partial s^2}\frac{c_{\mu,\nu}(s)}{c_{\mu,\nu}(0)}\F43{1,\frac 12+s,-\nu+s,\nu+1+s}{s+1,s+\frac{\mu+3}2,s+\frac{2-\mu}2}1\right|_{s=0}\\ c_{\mu,\nu}(s)&:=\frac{\Gamma\left(\frac 12+s\right)\Gamma(-\nu+s)\Gamma(\nu+1+s)}{\Gamma(s+1)\Gamma\left(s+\frac{\mu+3}2\right)\Gamma\left(s+\frac{2-\mu}2\right)} \end{align}
表される. Zhouの論文にはこの形で書かれている. 定理3も全く同様に
\begin{align} &\int_{-1}^1P_{\mu}(x)P_{\nu}(x)P_{\nu}(-x)\,dx\\ &=\frac{2\sin\pi\mu\sin\pi\nu}{\pi^2\mu(\mu+1)}\left.\frac{\partial}{\partial s}\frac{c_{\mu,\nu}(s)}{c_{\mu,\nu}(0)}\F43{1,\frac 12+s,-\nu+s,\nu+1+s}{s+1,s+\frac{\mu+3}2,s+\frac{2-\mu}2}1\right|_{s=0} \end{align}
と表すこともできる. 定理3の場合とは異なり, 定理4の右辺に現れるMellin-Barnes積分はnon-terminating Whippleの変換公式が使える形にはなっていない. そのため, 定理4の左辺がvery-well-poised ${}_7F_6$で表されるかどうかは分からないようである. 個人的には, それは一般にはいくつかのvery-well-poised ${}_7F_6$の和では表されないのではないかと予想している.