ZhouによるLegendre関数の3つの積の積分3
今回はZhouによるLegendre関数の3つの積が入った積分
\begin{align}
\int_{-1}^1P_{\nu}(x)^2P_{\nu}(-x)(1-x^2)^{\frac{\nu-1}2}\,dx&=\frac 1{\pi}\left(\frac{\cos\frac{\pi\nu}2}{2^{\nu}}\right)^3\left(\frac{\Gamma\left(\frac{1+\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)}\right)^4
\end{align}
を示す.
\begin{align} &\int_0^1P_{\nu}(1-2t)(t(1-t))^{a-1}\,dt\\ &=\frac{\pi\Gamma(a)^2}{2^{2a-1}\Gamma\left(a+\frac{\nu+1}2\right)\Gamma\left(a-\frac{\nu}2\right)\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)} \end{align}
まず, 項別積分により
\begin{align}
&\int_0^1P_{\nu}(1-2t)(t(1-t))^{a-1}\,dt\\
&=\int_0^1\sum_{0\leq n}\frac{(-\nu,\nu+1)_n}{n!^2}t^{n+a-1}(1-t)^{a-1}\,dt\\
&=\frac{\Gamma(a)^2}{\Gamma(2a)}\F32{-\nu,\nu+1,a}{1,2a}{1}
\end{align}
となる. ここで,
Whippleの和公式
より,
\begin{align}
\F32{-\nu,\nu+1,a}{1,2a}{1}&=\frac{\pi\Gamma(2a)}{2^{2a-1}\Gamma\left(a+\frac{\nu+1}2\right)\Gamma\left(a-\frac{\nu}2\right)\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}
\end{align}
であるからこれを代入して示すべき等式を得る.
前の記事 の定理3の1つ目の式において$a\mapsto -\nu$と置き換えたものは以下のようになる.
$0< x<1$において
\begin{align}
&P_{\nu}(1-2t)P_{\nu}(2t-1)\\
&=\frac{\sin^2\pi\nu}{\pi^{\frac 52}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma(1+\nu+s)\Gamma\left(\frac 12+s\right)\Gamma(-s)^2}{\Gamma(s+1)}(4x(1-x))^s\,ds
\end{align}
が成り立つ.
これらを用いて以下を示す.
\begin{align}
\int_{-1}^1P_{\nu}(x)^2P_{\nu}(-x)(1-x^2)^{\frac{\nu-1}2}\,dx&=\frac 1{\pi}\left(\frac{\cos\frac{\pi\nu}2}{2^{\nu}}\right)^3\left(\frac{\Gamma\left(\frac{1+\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)}\right)^4
\end{align}
が成り立つ.
補題1, 補題2を用いると,
\begin{align}
&\int_{-1}^1P_{\nu}(x)^2P_{\nu}(-x)(1-x^2)^{\frac{\nu-1}2}\,dx\\
&=2^{\nu}\int_0^1P_{\nu}(1-2t)^2P_{\nu}(2t-1)(t(1-t))^{\frac{\nu-1}2}\,dt\qquad x\mapsto 1-2t\\
&=\frac{2^{\nu}\sin^2\pi\nu}{\pi^{\frac 52}}\int_0^1P_{\nu}(1-2t)(t(1-t))^{\frac{\nu-1}2}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma(1+\nu+s)\Gamma\left(\frac 12+s\right)\Gamma(-s)^2}{\Gamma(s+1)}(4t(1-t))^s\,ds\,dt\\
&=\frac{2^{\nu}\sin^2\pi\nu}{\pi^{\frac 52}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{4^s\Gamma(-\nu+s)\Gamma(1+\nu+s)\Gamma\left(\frac 12+s\right)\Gamma(-s)^2}{\Gamma(s+1)}\int_0^1P_{\nu}(1-2t)(t(1-t))^{s+\frac{\nu-1}2}\,dt\,ds\\
&=\frac{2^{\nu}\sin^2\pi\nu}{\pi^{\frac 52}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{4^s\Gamma(-\nu+s)\Gamma(1+\nu+s)\Gamma\left(\frac 12+s\right)\Gamma(-s)^2}{\Gamma(s+1)}\frac{\pi\Gamma\left(s+\frac{\nu+1}2\right)^2}{2^{2s+\nu}\Gamma\left(s+\nu+1\right)\Gamma\left(s+\frac 12\right)\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\,ds\\
&=\frac{\sin^2\pi\nu}{\pi^{\frac 32}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma\left(s+\frac{\nu+1}2\right)^2\Gamma(-s)^2}{\Gamma(s+1)}\,ds\\
\end{align}
ここで,
Barnesの第2補題
より,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma\left(s+\frac{\nu+1}2\right)^2\Gamma(-s)^2}{\Gamma(s+1)}\,ds\\
&=\frac{\Gamma(-\nu)^2\Gamma\left(\frac{\nu+1}2\right)^4}{\Gamma(1+\nu)\Gamma\left(\frac{1-\nu}2\right)^2}
\end{align}
であるから, これを代入すると
\begin{align}
&\int_{-1}^1P_{\nu}(x)^2P_{\nu}(-x)(1-x^2)^{\frac{\nu-1}2}\,dx\\
&=\frac{\sin^2\pi\nu}{\pi^{\frac 32}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\frac{\Gamma(-\nu)^2\Gamma\left(\frac{\nu+1}2\right)^4}{\Gamma(1+\nu)\Gamma\left(\frac{1-\nu}2\right)^2}\\
&=\frac{\sqrt{\pi}\Gamma\left(\frac{\nu+1}2\right)^4}{\Gamma(1+\nu)^3\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)^3}\\
&=\frac{\pi^2\Gamma\left(\frac{\nu+1}2\right)}{2^{3\nu}\Gamma\left(1+\frac{\nu}2\right)^4\Gamma\left(\frac{1-\nu}2\right)^3}\\
&=\frac{\cos^3\frac{\pi\nu}2\Gamma\left(\frac{\nu+1}2\right)^4}{2^{3\nu}\pi\Gamma\left(1+\frac{\nu}2\right)^4}\\
\end{align}
となって示すべき等式が得られる.
上の証明はZhouの2025年の証明を参考にしたものである. Zhouの2014年の論文においては上の結果の類似として
\begin{align}
\int_{-1}^1P_{\nu}(x)^3(1-x^2)^{\frac{\nu-1}2}\,dx&=\frac {3-2\cos\pi\nu}{\pi}\left(\frac{\cos\frac{\pi\nu}2}{2^{\nu}}\right)^3\left(\frac{\Gamma\left(\frac{1+\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)}\right)^4
\end{align}
も示されている. その証明の方針は有限Hilbert変換のParseval型の等式を用いて
\begin{align}
\int_{-1}^1P_{\nu}(x)^3(1-x^2)^{\frac{\nu-1}2}\,dx&=(3-2\cos\pi\nu)\int_{-1}^1P_{\nu}(x)^2P_{\nu}(-x)(1-x^2)^{\frac{\nu-1}2}\,dx
\end{align}
を示すというもので, それは比較的簡潔な証明である.
追記
\begin{align}
\int_{-1}^1P_{\nu}(x)^3(1-x^2)^{\frac{\nu-1}2}\,dx&=\frac {3-2\cos\pi\nu}{\pi}\left(\frac{\cos\frac{\pi\nu}2}{2^{\nu}}\right)^3\left(\frac{\Gamma\left(\frac{1+\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)}\right)^4
\end{align}
も上の積分と同様に示せるようなので, それを書いておく.
前の記事
で示した定理3の2つ目の式は以下のようになる.
$0< x<1$おいて,
\begin{align}
&P_{\nu}(2x-1)^2+P_{\nu}(1-2x)^2\\
&=-\frac{2\sin^3\pi\nu}{\pi^{\frac 52}}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma(\nu+1+s)\Gamma(-s)^3}{\Gamma\left(\frac 12-s\right)}(4x(1-x))^s\,ds
\end{align}
これを用いると以下が示される.
\begin{align} \int_{-1}^1P_{\nu}(x)^3(1-x^2)^{\frac{\nu-1}2}\,dx&=\frac {3-2\cos\pi\nu}{\pi}\left(\frac{\cos\frac{\pi\nu}2}{2^{\nu}}\right)^3\left(\frac{\Gamma\left(\frac{1+\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)}\right)^4 \end{align}
\begin{align}
&\int_{-1}^1(P_{\nu}(x)^2+P_{\nu}(-x)^2)P_{\nu}(x)(1-x^2)^{\frac{\nu-1}2}\,dx\\
&=2^{\nu}\int_0^1(P_{\nu}(2t-1)^2+P_{\nu}(1-2t)^2)P_{\nu}(1-2t)(t(1-t))^{\frac{\nu-1}2}\,dt\qquad x\mapsto 1-2t\\
&=-2^{\nu}\frac{2\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_{\nu}(1-2t)(t(1-t))^{s+\frac{\nu-1}2}\,dt\,ds\\
&=-2^{\nu}\frac{2\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\left(s+\frac{\nu+1}2\right)^2}{2^{2s+\nu}\Gamma\left(s+\nu+1\right)\Gamma\left(s+\frac 12\right)\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\,ds\\
&=-\frac{2\sin^3\pi\nu}{\pi^{\frac 52}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\Gamma(-\nu+s)\Gamma\left(s+\frac{\nu+1}2\right)^2\Gamma(-s)^3\cos\pi s\,ds
\end{align}
ここで,
Barnesの第2補題
より,
\begin{align}
&\frac 1{2\pi i}\int_{-i\infty}^{i\infty}\Gamma(-\nu+s)\Gamma\left(s+\frac{\nu+1}2\right)^2\Gamma(-s)^3\cos\pi s\,ds\\
&=\frac 1{2\pi i\sin\pi\nu}\int_{-i\infty}^{i\infty}\Gamma(-\nu+s)\Gamma\left(s+\frac{\nu+1}2\right)^2\Gamma(-s)^3(\sin\pi s\cos\pi\nu-\sin\pi(s-\nu))\,ds\\
&=-\frac 1{2\pi i}\frac{\pi\cos\pi\nu}{\sin\pi\nu}\int_{-i\infty}^{i\infty}\frac{\Gamma(-\nu+s)\Gamma\left(s+\frac{\nu+1}2\right)^2\Gamma(-s)^2}{\Gamma(s+1)}\,ds\\
&\qquad-\frac 1{2\pi i}\frac{\pi}{\sin\pi\nu}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(s+\frac{\nu+1}2\right)^2\Gamma(-s)^3}{\Gamma(1+\nu-s)}\,ds\\
&=-\frac{\pi\cos\pi\nu}{\sin\pi\nu}\frac{\Gamma(-\nu)^2\Gamma\left(\frac{\nu+1}2\right)^4}{\Gamma(1+\nu)\Gamma\left(\frac{1-\nu}2\right)^2}-\frac 1{2\pi i}\frac{\pi}{\sin\pi\nu}\int_{-i\infty}^{i\infty}\frac{\Gamma\left(s+\frac{\nu+1}2\right)^3\Gamma(-s)^2}{\Gamma\left(\frac{3\nu+3}2+s\right)}\,ds\\
&=-\frac{\pi\cos\pi\nu}{\sin\pi\nu}\frac{\Gamma(-\nu)^2\Gamma\left(\frac{\nu+1}2\right)^4}{\Gamma(1+\nu)\Gamma\left(\frac{1-\nu}2\right)^2}-\frac{\pi}{\sin\pi\nu}\frac{\Gamma\left(\frac{\nu+1}2\right)^6}{\Gamma(\nu+1)^3}\\
&=-\frac{\pi}{\sin^3\pi\nu}\left(\cos\pi\nu\cos^2\frac{\pi\nu}2+\sin^2\pi\nu\right)\frac{\Gamma\left(\frac{\nu+1}2\right)^6}{\Gamma(\nu+1)^3}\\
&=-\frac{\pi\cos^2\frac{\pi\nu}2}{\sin^3\pi\nu}\left(\cos\pi\nu+4\sin^2\frac{\pi\nu}2\right)\frac{\Gamma\left(\frac{\nu+1}2\right)^6}{\Gamma(\nu+1)^3}\\
&=-\frac{\pi\cos^2\frac{\pi\nu}2}{\sin^3\pi\nu}\left(2-\cos\pi\nu\right)\frac{\Gamma\left(\frac{\nu+1}2\right)^6}{\Gamma(\nu+1)^3}
\end{align}
となるから, これを代入して
\begin{align}
&\int_{-1}^1(P_{\nu}(x)^2+P_{\nu}(-x)^2)P_{\nu}(x)(1-x^2)^{\frac{\nu-1}2}\,dx\\
&=\left(4-2\cos\pi\nu\right)\frac{\cos^2\frac{\pi\nu}2}{\pi^{\frac 32}\Gamma\left(1+\frac{\nu}2\right)\Gamma\left(\frac{1-\nu}2\right)}\frac{\Gamma\left(\frac{\nu+1}2\right)^6}{\Gamma(\nu+1)^3}\\
&=\frac {4-2\cos\pi\nu}{\pi}\left(\frac{\cos\frac{\pi\nu}2}{2^{\nu}}\right)^3\left(\frac{\Gamma\left(\frac{1+\nu}2\right)}{\Gamma\left(1+\frac{\nu}2\right)}\right)^4
\end{align}
を得る. ここから定理3の等式を引くと示すべき等式を得る.
