2つの超球関数の積の[0,1]における積分

前の記事 で超球関数の二次変換公式
\begin{align} P_{\nu}^{\left(a,-\frac 12\right)}(2x-1)&=\frac{\Gamma\left(\nu+\frac 12\right)\Gamma\left(a+\frac 12\right)}{\sqrt{\pi}\Gamma\left(a+\nu+\frac 12\right)}C_{2\nu}^{\left(a+\frac 12\right)}(\sqrt x)\\ Q_{\nu}^{\left(a,-\frac 12\right)}(2x-1)&=\frac{\Gamma\left(\nu+\frac 12\right)\Gamma\left(a+\frac 12\right)}{\sqrt{\pi}\Gamma\left(a+\nu+\frac 12\right)}D_{2\nu}^{\left(a+\frac 12\right)}(\sqrt x) \end{align}
を示した. 前の記事 でJacobi関数の2つの積の積分を計算したので, 今回はこれらを用いて, 2つの超球関数の積の[0,1]における積分を求めたいと思う. まず, 上の式は$0< x<1$に対して,
\begin{align} C_{2\nu}^{(a)}(x)&=\frac{\sqrt{\pi}\Gamma\left(a+\nu\right)}{\Gamma\left(\nu+\frac 12\right)\Gamma\left(a\right)}P_{\nu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1)\\ D_{2\nu}^{(a)}(x)&=\frac{\sqrt{\pi}\Gamma\left(a+\nu\right)}{\Gamma\left(\nu+\frac 12\right)\Gamma\left(a\right)}Q_{\nu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1) \end{align}
と書き換えられる. 前の記事 の定理2において, $a\mapsto a-\frac 12,b=-\frac 12,x\mapsto 2x^2-1$とすると,
\begin{align} &\int_0^1P_{\mu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1)P_{\nu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1)(1-x^2)^{a-\frac 12}\,dx\\ &=\frac{\sin\pi\mu\sin\pi\nu}{2\pi(\nu-\mu)(a+\mu+\nu)}\left(\frac{\Gamma(-\nu)\Gamma\left(a+\mu+\frac 12\right)}{\Gamma\left(\frac 12-\nu\right)\Gamma\left(a+\mu\right)}-\frac{\Gamma(-\mu)\Gamma\left(a+\nu+\frac 12\right)}{\Gamma\left(\frac 12-\mu\right)\Gamma\left(a+\nu\right)}\right)\\ &\int_0^1P_{\mu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1)(P_{\nu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1)\cot\pi\nu-Q_{\nu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1))(1-x^2)^{a-\frac 12}\,dx\\ &=\frac{\Gamma\left(a+\mu+\frac 12\right)\Gamma\left(\nu+\frac 12\right)\sin\pi\mu}{2\pi^2(\nu-\mu)(a+\mu+\nu)}\left(\frac{\Gamma(-\nu)}{\Gamma\left(a+\mu\right)}-\frac{\Gamma(-\mu)}{\Gamma\left(a+\nu\right)}\right)\\ &\int_0^1(P_{\mu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1)\cot\pi\mu-Q_{\mu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1))(P_{\nu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1)\cot\pi\nu-Q_{\nu}^{\left(a-\frac 12,-\frac 12\right)}(2x^2-1))(1-x^2)^{a-\frac 12}\,dx\\ &=\frac{1}{2\pi(\nu-\mu)(a+\mu+\nu)\cos\pi a}\left(\frac{\Gamma(-\nu)\Gamma\left(\mu+\frac 12\right)}{\Gamma\left(\frac 12-a-\nu\right)\Gamma\left(a+\mu\right)}-\frac{\Gamma(-\mu)\Gamma\left(\nu+\frac 12\right)}{\Gamma\left(\frac 12-a-\mu\right)\Gamma\left(a+\nu\right)}\right) \end{align}
となるから, これを超球関数で書き換えると
\begin{align} &\int_0^1C_{2\mu}^{(a)}(x)C_{2\nu}^{(a)}(x)(1-x^2)^{a-\frac 12}\,dx\\ &=\frac{\sin\pi\mu\sin\pi\nu}{2(\nu-\mu)(a+\mu+\nu)}\frac{\Gamma(a+\mu)\Gamma\left(a+\nu\right)}{\Gamma\left(a\right)^2\Gamma\left(\mu+\frac 12\right)\Gamma\left(\nu+\frac 12\right)}\left(\frac{\Gamma(-\nu)\Gamma\left(a+\mu+\frac 12\right)}{\Gamma\left(\frac 12-\nu\right)\Gamma\left(a+\mu\right)}-\frac{\Gamma(-\mu)\Gamma\left(a+\nu+\frac 12\right)}{\Gamma\left(\frac 12-\mu\right)\Gamma\left(a+\nu\right)}\right)\\ &=\frac{\sin\pi\mu\sin\pi\nu\cos\pi\mu\cos\pi\nu\left(\Gamma(-\nu)\Gamma\left(a+\nu\right)\Gamma\left(\frac 12-\mu\right)\Gamma\left(a+\mu+\frac 12\right)-\Gamma(-\mu)\Gamma\left(a+\mu\right)\Gamma\left(\frac 12-\nu\right)\Gamma\left(a+\nu+\frac 12\right)\right)}{2\pi^2(\nu-\mu)(a+\mu+\nu)\Gamma\left(a\right)^2}\\ &=\frac{\sin 2\pi\mu\sin 2\pi\nu\left(\Gamma(-\nu)\Gamma\left(a+\nu\right)\Gamma\left(\frac 12-\mu\right)\Gamma\left(a+\mu+\frac 12\right)-\Gamma(-\mu)\Gamma\left(a+\mu\right)\Gamma\left(\frac 12-\nu\right)\Gamma\left(a+\nu+\frac 12\right)\right)}{8\pi^2(\nu-\mu)(a+\mu+\nu)\Gamma\left(a\right)^2}\\ &\int_0^1C_{2\mu}^{(a)}(x)(C_{2\nu}^{(a)}(x)\cot\pi\nu-D_{2\nu}^{(a)}(x))(1-x^2)^{a-\frac 12}\,dx\\ &=\frac{\Gamma\left(a+\mu+\frac 12\right)\Gamma\left(\nu+\frac 12\right)\sin\pi\mu}{2\pi(\nu-\mu)(a+\mu+\nu)}\frac{\Gamma(a+\mu)\Gamma\left(a+\nu\right)}{\Gamma\left(a\right)^2\Gamma\left(\mu+\frac 12\right)\Gamma\left(\nu+\frac 12\right)}\left(\frac{\Gamma(-\nu)}{\Gamma\left(a+\mu\right)}-\frac{\Gamma(-\mu)}{\Gamma\left(a+\nu\right)}\right)\\ &=\frac{\sin\pi\mu\Gamma\left(a+\mu+\frac 12\right)(\Gamma(-\nu)\Gamma\left(a+\nu\right)-\Gamma(-\mu)\Gamma(a+\mu))}{2\pi(\nu-\mu)(a+\mu+\nu)\Gamma\left(a\right)^2\Gamma\left(\mu+\frac 12\right)}\\ &\int_0^1(C_{2\mu}^{(a)}(x)\cot\pi\mu-D_{2\mu}^{(a)}(x))(C_{2\nu}^{(a)}(x)\cot\pi\nu-D_{2\nu}^{(a)}(x))(1-x^2)^{a-\frac 12}\,dx\\ &=\frac{1}{2(\nu-\mu)(a+\mu+\nu)\cos\pi a}\frac{\Gamma(a+\mu)\Gamma\left(a+\nu\right)}{\Gamma\left(a\right)^2\Gamma\left(\mu+\frac 12\right)\Gamma\left(\nu+\frac 12\right)}\left(\frac{\Gamma(-\nu)\Gamma\left(\mu+\frac 12\right)}{\Gamma\left(\frac 12-a-\nu\right)\Gamma\left(a+\mu\right)}-\frac{\Gamma(-\mu)\Gamma\left(\nu+\frac 12\right)}{\Gamma\left(\frac 12-a-\mu\right)\Gamma\left(a+\nu\right)}\right)\\ &=\frac{1}{2(\nu-\mu)(a+\mu+\nu)\cos\pi a\Gamma\left(a\right)^2}\left(\frac{\Gamma(-\nu)\Gamma\left(a+\nu\right)}{\Gamma\left(\nu+\frac 12\right)\Gamma\left(\frac 12-a-\nu\right)}-\frac{\Gamma(-\mu)\Gamma(a+\mu)}{\Gamma\left(\mu+\frac 12\right)\Gamma\left(\frac 12-a-\mu\right)}\right) \end{align}
となる. つまり, 以下を得る.

特に, $\mu\to\nu$とすると, 以下を得る.

Legendre関数の場合

第1種, 第2種のLegendre関数をそれぞれ
\begin{align} P_{\nu}(x)&:=C_{\nu}^{\left(\frac 12\right)}(x)\\ Q_{\nu}(x)&:=D_{\nu}^{\left(\frac 12\right)}(x) \end{align}
と定義する. 定理1において$a\mapsto \frac 12$とすると,以下を得る.

系1において, $a\mapsto \frac 12$すると,
\begin{align} &\int_0^1P_{2\nu}(x)^2\,dx\\ &=\frac{\sin^2 2\pi\nu\Gamma(-\nu)\Gamma\left(\frac 12-\nu\right)\Gamma\left(\frac 12+\nu\right)\Gamma\left(\nu+1\right)\left(\psi\left(\frac 12-\nu\right)-\psi\left(\nu+1\right)-\psi(-\nu)+\psi\left(\frac 12+\nu\right)\right)}{4\pi^3(4\nu+1)}\\ &=\frac{\sin2\pi\nu\left(\psi\left(\nu+1\right)+\psi(-\nu)-\psi\left(\frac 12-\nu\right)-\psi\left(\frac 12+\nu\right)\right)}{2\pi(4\nu+1)}\\ &\int_0^1P_{2\nu}(x)(P_{2\nu}(x)\cot\pi\nu-Q_{2\nu}(x))\,dx\\ &=\frac{\sin\pi\nu\Gamma\left(\nu+1\right)\Gamma(-\nu)\left(\psi\left(\frac 12+\nu\right)-\psi(-\nu)\right)}{\pi^2(4\nu+1)}\\ &=\frac{\psi(-\nu)-\psi\left(\frac 12+\nu\right)}{\pi(4\nu+1)}\\ &\int_0^1(P_{2\nu}(x)\cot\pi\nu-Q_{2\nu}(x))^2\,dx\\ &=\frac{\psi'(-\nu)-\psi'\left(\frac 12+\nu\right)}{\pi^2(4\nu+1)} \end{align}
つまり以下を得る.

特に, 1番上の式において, $\nu\mapsto -\frac 14$とすると,
\begin{align} \int_0^1P_{-\frac 12}(x)^2\,dx&=\frac{\psi'\left(\frac 14\right)-\psi'\left(\frac 34\right)}{4\pi}\\ &=\frac{4\beta(2)}{\pi} \end{align}
つまり,
\begin{align} \int_0^{\frac 12}\F21{\frac 12,\frac 12}1{x}^2\,dx=&\frac{2\beta(2)}{\pi} \end{align}
を得る. ここで,
\begin{align} \beta(2):=\sum_{0\leq n}\frac{(-1)^n}{(2n+1)^2} \end{align}
はCatalanの定数である.