ZhouによるLegendre関数の3つの積の積分2

前の記事 で示した系1, 定理2をまとめると以下のようになる.

Legendre多項式の3つの積の積分が
\begin{align} \int_{-1}^1P_l(x)P_m(x)P_n(x)\,dx&=\frac{\left(\frac 12\right)_{N-l}\left(\frac 12\right)_{N-m}\left(\frac 12\right)_{N-n}N!}{(N-l)!(N-m)!(N-n)!\left(\frac 12\right)_{N+1}}\qquad 2N=l+m+n \end{align}
と表されることは良く知られている. 特に, $n,m,l$を全て$2n$とすると,
\begin{align} \int_{-1}^1P_{2n}(x)^3\,dx&=\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}} \end{align}
を得る. 定理1はいずれもこれを一般化しており, 例えば1つ目の式は
\begin{align} \lim_{\mu\to 2n}\frac{2\sin\pi\mu}{\pi\mu(\mu+1)}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n}}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_{2n}}&=\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}} \end{align}
となることが直接的に確かめられる.

前の記事 における3つ目の等式の証明において
\begin{align} \int_{-1}^1\left.\frac{\partial}{\partial \nu}P_{\nu}(x)\right|_{\nu=n}P_n(x)^2\,dx \end{align}
が一番上の式を用いて計算できるということが重要だった. 前の記事においてはその計算は省略したので, 今回は上の積分をちゃんと計算するとともに, $\nu$に関する二階微分が入ったものについてもZhouの結果から計算できるものを計算しておきたいと思う. 以下
\begin{align} P^{(k)}_{\nu}(x):=\left(\frac{\partial}{\partial \nu}\right)^kP_{\nu}(x) \end{align}
とする. まず, 定理1の1つ目の等式より,
\begin{align} \int_{-1}^1P_n^{(1)}(x)P_n(x)^2\,dx&=\left.\frac{\partial}{\partial \mu}\frac{2\sin\pi\mu}{\pi\mu(\mu+1)}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_n}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_n}\right|_{\mu=n} \end{align}
$n$が奇数のとき, これは$0$である. $n\mapsto 2n$を偶数として,
\begin{align} &\int_{-1}^1P_{2n}^{(1)}(x)P_{2n}(x)^2\,dx\\ &=\left.\frac{\partial}{\partial \mu}\frac{2\sin\pi\mu}{\pi\mu(\mu+1)}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n}}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_{2n}}\right|_{\mu=2n}\\ &=-\left.\frac{\partial}{\partial \mu}\frac{\sin\pi\mu}{2\pi}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n}}{\left(\frac{\mu+1}2,-\frac{\mu}2\right)_{2n+1}}\right|_{\mu=2n}\\ &=-\left.\frac{\sin\pi\mu}{2\pi}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n}}{\left(\frac{\mu+1}2,-\frac{\mu}2\right)_{2n+1}}\left(\pi\cot\pi\mu+\frac 12\sum_{k=0}^{2n-1}\left(\frac 1{k+1+\frac{\mu}2}-\frac 1{k+\frac{1-\mu}2}\right)+\frac 12\sum_{k=0}^{2n}\left(\frac 1{k-\frac{\mu}2}-\frac 1{k+\frac{1+\mu}2}\right)\right)\right|_{\mu=2n}\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left(\frac 12\sum_{k=0}^{2n-1}\left(\frac 1{k+1+n}-\frac 1{k+\frac{1}2-n}\right)-\frac 12\sum_{k=0}^{2n}\frac 1{k+\frac{1}2+n}\right)\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{2n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac{1}2}\right) \end{align}
を得る.　次に定理1の1つ目の式の2階微分も計算してみる. まず, 奇数の場合
\begin{align} &\int_{-1}^1P_{2n+1}^{(2)}(x)P_{2n+1}(x)^2\,dx\\ &=\left.\frac{\partial^2}{\partial \mu^2}\frac{2\sin\pi\mu}{\pi\mu(\mu+1)}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n+1}}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_{2n+1}}\right|_{\mu=2n+1}\\ &=\lim_{\mu\to 2n+1}\frac 2{(\mu-2n-1)^2}\frac{2\sin\pi\mu}{\pi\mu(\mu+1)}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n+1}}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_{2n+1}}\\ &=-2\lim_{\mu\to 2n+1}\frac 1{\mu-2n-1}\frac{1}{(2n+1)(n+1)}\frac{\left(n+\frac{3}2,\frac{1-\mu}2\right)_{2n+1}}{\left(n+2,\frac{1}2-n\right)_{2n+1}}\\ &=\frac{1}{(2n+1)(n+1)}\frac{\left(n+\frac{3}2\right)_{2n+1}(-n)_n n!}{(n+2)_{2n+1}\left(\frac{1}2-n\right)_{n}\left(\frac 12\right)_{n+1}}\\ &=\frac{n!^3\left(\frac{1}2\right)_{3n+2}}{2\left(\frac 12\right)_{n+1}^3\left(3n+2\right)!} \end{align}
となる. 次に, 偶数の場合, 先ほどの一階微分の計算も用いて
\begin{align} &\int_{-1}^1P_{2n}^{(2)}(x)P_{2n}(x)^2\,dx\\ &=\left.\frac{\partial^2}{\partial \mu^2}\frac{2\sin\pi\mu}{\pi\mu(\mu+1)}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n}}{\left(\frac{\mu+3}2,\frac{2-\mu}2\right)_{2n}}\right|_{\mu=2n}\\ &=-\left.\frac{\sin\pi\mu}{2\pi}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n}}{\left(\frac{\mu+1}2,-\frac{\mu}2\right)_{2n+1}}\left(\pi\cot\pi\mu+\frac 12\sum_{k=0}^{2n-1}\left(\frac 1{k+1+\frac{\mu}2}-\frac 1{k+\frac{1-\mu}2}\right)+\frac 12\sum_{k=0}^{2n}\left(\frac 1{k-\frac{\mu}2}-\frac 1{k+\frac{1+\mu}2}\right)\right)^2\right|_{\mu=2n}\\ &\qquad-\frac{\sin\pi\mu}{2\pi}\frac{\left(\frac{\mu+2}2,\frac{1-\mu}2\right)_{2n}}{\left(\frac{\mu+1}2,-\frac{\mu}2\right)_{2n+1}}\left.\frac{\partial}{\partial\mu}\left(\pi\cot\pi\mu+\frac 12\sum_{k=0}^{2n-1}\left(\frac 1{k+1+\frac{\mu}2}-\frac 1{k+\frac{1-\mu}2}\right)+\frac 12\sum_{k=0}^{2n}\left(\frac 1{k-\frac{\mu}2}-\frac 1{k+\frac{1+\mu}2}\right)\right)\right|_{\mu=2n}\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac{1}2}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left.\left(-\frac{\pi^2}{\sin^2\pi\mu}-\frac 14\sum_{k=0}^{2n-1}\left(\frac 1{\left(k+1+\frac{\mu}2\right)^2}+\frac 1{\left(k+\frac{1-\mu}2\right)^2}\right)+\frac 14\sum_{k=0}^{2n}\left(\frac 1{\left(k-\frac{\mu}2\right)^2}+\frac 1{\left(k+\frac{1+\mu}2\right)^2}\right)\right)\right|_{\mu=2n}\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac{1}2}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{\pi^2}{3}-\frac 14\sum_{k=0}^{2n-1}\left(\frac 1{\left(k+1+n\right)^2}+\frac 1{\left(k+\frac{1}2-n\right)^2}\right)+\frac 14\sum_{0\leq k\leq 2n,k\neq n}\frac 1{\left(k-n\right)^2}+\frac 14\sum_{k=0}^{2n}\frac 1{\left(k+\frac{1}2+n\right)^2}\right)\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac{1}2}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{4\pi^2}{3}+\sum_{k=0}^{3n}\frac 1{\left(k+\frac{1}2\right)^2}-\sum_{k=1}^{3n}\frac 1{k^2}-3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac{1}2\right)^2}+3\sum_{k=1}^n\frac 1{k^2}\right) \end{align}
を得る. 次に2つの1階微分が入った
\begin{align} \int_{-1}^1P_n^{(1)}(x)^2P_n(x)\,dx\\ \int_{-1}^1P_n^{(1)}(x)P_n^{(1)}(-x)P_n(x)\,dx \end{align}
を求めたい. まず, 下の積分から考える. $P_n^{(1)}(x)P_n^{(1)}(-x)$は偶関数だから, $n\mapsto 2n$として偶数の場合だけを考えればよい. 2つ目の式を2階微分して$\nu=2n$とすると
\begin{align} &\int_{-1}^1P_{2n}(x)(P_{2n}^{(2)}(x)P_{2n}(-x)+P_{2n}(x)P_{2n}^{(2)}(-x)+2P_{2n}^{(1)}(x)P_{2n}^{(1)}(-x))\,dx\\ &=\left.\frac{\partial^2}{\partial \nu^2}\frac{2\cos\pi\nu}{2\nu+1}\frac{\left(\frac 12\right)_{n}^2}{n!^2}\frac{(\nu+1,-\nu)_{n}}{\left(\nu+\frac 32,\frac 12-\nu\right)_{n}}\right|_{\nu=2n}\\ &=\frac{\left(\frac 12\right)_{n}^2}{n!^2}\left.\frac{\partial^2}{\partial \nu^2}\frac{(\nu+1,-\nu)_{n}\cos\pi\nu}{\left(\nu+\frac 12\right)_{n+1}\left(\frac 12-\nu\right)_{n}}\right|_{\nu=2n}\\ &=\frac{\left(\frac 12\right)_{n}^2}{n!^2}\left.\frac{\partial}{\partial \nu}\frac{(\nu+1,-\nu)_{n}\cos\pi\nu}{\left(\nu+\frac 12\right)_{n+1}\left(\frac 12-\nu\right)_{n}}\left(-\pi\tan\pi\nu+\sum_{k=0}^{n-1}\left(\frac 1{k+\nu+1}-\frac 1{k-\nu}+\frac 1{k+\frac 12-\nu}\right)-\sum_{k=0}^{n}\frac 1{k+\frac 12+\nu}\right)\right|_{\nu=2n}\\ &=\frac{\left(\frac 12\right)_{n}^2}{n!^2}\left.\frac{(\nu+1,-\nu)_{n}\cos\pi\nu}{\left(\nu+\frac 12\right)_{n+1}\left(\frac 12-\nu\right)_{n}}\left(-\pi\tan\pi\nu+\sum_{k=0}^{n-1}\left(\frac 1{k+\nu+1}-\frac 1{k-\nu}+\frac 1{k+\frac 12-\nu}\right)-\sum_{k=0}^{n}\frac 1{k+\frac 12+\nu}\right)^2\right|_{\nu=2n}\\ &\qquad+\frac{\left(\frac 12\right)_{n}^2}{n!^2}\left.\frac{(\nu+1,-\nu)_{n}\cos\pi\nu}{\left(\nu+\frac 12\right)_{n+1}\left(\frac 12-\nu\right)_{n}}\left(-\frac{\pi^2}{\cos^2\pi\nu}-\sum_{k=0}^{n-1}\left(\frac 1{(k+\nu+1)^2}+\frac 1{(k-\nu)^2}-\frac 1{\left(k+\frac 12-\nu\right)^2}\right)+\sum_{k=0}^{n}\frac 1{\left(k+\frac 12+\nu\right)^2}\right)\right|_{\nu=2n}\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=0}^{n-1}\left(\frac 1{k+2n+1}-\frac 1{k-2n}+\frac 1{k+\frac 12-2n}\right)-\sum_{k=0}^{n}\frac 1{k+\frac 12+2n}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left(-\pi^2-\sum_{k=0}^{n-1}\left(\frac 1{(k+2n+1)^2}+\frac 1{(k-2n)^2}-\frac 1{\left(k+\frac 12-2n\right)^2}\right)+\sum_{k=0}^{n}\frac 1{\left(k+\frac 12+2n\right)^2}\right)\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left(-\pi^2-\sum_{k=n+1}^{3n}\frac 1{k^2}+\sum_{k=n}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right) \end{align}
よって,
\begin{align} &\int_{-1}^1P_{2n}(x)P_{2n}^{(1)}(x)P_{2n}^{(1)}(-x)\,dx\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{2n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{2n!^3\left(\frac 12\right)_{3n+1}}\left(-\pi^2-\sum_{k=n+1}^{3n}\frac 1{k^2}+\sum_{k=n}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &\qquad-\int_{-1}^1P_{2n}^{(2)}(x)P_{2n}(x)^2\,dx\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{2n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{2n!^3\left(\frac 12\right)_{3n+1}}\left(-\pi^2-\sum_{k=n+1}^{3n}\frac 1{k^2}+\sum_{k=n}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &\qquad-\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac{1}2}\right)^2\\ &\qquad-\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{4\pi^2}{3}+\sum_{k=0}^{3n}\frac 1{\left(k+\frac{1}2\right)^2}-\sum_{k=1}^{3n}\frac 1{k^2}-3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac{1}2\right)^2}+3\sum_{k=1}^n\frac 1{k^2}\right)\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{2\pi^2}3-\sum_{k=1}^{3n}\frac 1{k^2}+\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}-\sum_{k=1}^n\frac 1{k^2}+\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}\right) \end{align}
が得られる. 次に, 定理1の3つ目の式を$\nu$に関して2階微分して$\nu=2n+1$とすると,
\begin{align} &\int_{-1}^1(2P_n^{(2)}(x)P_n(x)P_n(-x)+P_n(x)^2P_n^{(2)}(-x)+2P_n^{(1)}(x)^2P_n(-x)+4P_n(x)P_n^{(1)}(x)P_n^{(1)}(-x))\,dx\\ &=\left.\frac{\partial^2}{\partial \nu^2}\frac{1+2\cos\pi\nu}3\frac{\pi\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+2}2\right)}{\Gamma\left(\frac{1-\nu}2\right)^2\Gamma\left(\frac{\nu+2}2\right)^3\Gamma\left(\frac{3\nu+3}2\right)}\right|_{\nu=n} \end{align}
まず, $n$が奇数のとき, $n\mapsto 2n+1$として,
\begin{align} &-\int_{-1}^1(P_{2n+1}^{(2)}(x)P_{2n+1}(x)^2+2P_{2n+1}^{(1)}(x)^2P_{2n+1}(x))\,dx\\ &=\left.\frac{\partial^2}{\partial \nu^2}\frac{1+2\cos\pi\nu}3\frac{\pi\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+2}2\right)}{\Gamma\left(\frac{1-\nu}2\right)^2\Gamma\left(\frac{\nu+2}2\right)^3\Gamma\left(\frac{3\nu+3}2\right)}\right|_{\nu=2n+1}\\ &=-\frac{2}3\lim_{\nu\to 2n+1}\frac{\left(\frac{1-\nu}2\right)_{n+1}^2}{(\nu-2n-1)^2}\frac{\pi n!\Gamma\left(3n+\frac 52\right)}{\Gamma\left(n+\frac 32\right)^3(3n+2)!}\\ &=-\frac{n!^3\left(\frac 12\right)_{3n+2}}{6\left(\frac 12\right)_{n+1}^3(3n+2)!} \end{align}
となる. よって,
\begin{align} &\int_{-1}^1P_{2n+1}^{(1)}(x)^2P_{2n+1}(x)\,dx\\ &=\frac{1}{12}\frac{n!^3\left(\frac 12\right)_{3n+2}}{\left(\frac 12\right)_{n+1}^3(3n+2)!}-\frac 12\int_{-1}^1P_{2n+1}^{(2)}(x)P_{2n+1}(x)^2\,dx\\ &=\frac{1}{12}\frac{n!^3\left(\frac 12\right)_{3n+2}}{\left(\frac 12\right)_{n+1}^3(3n+2)!}-\frac{1}{4}\frac{n!^3\left(\frac{1}2\right)_{3n+2}}{\left(\frac 12\right)_{n+1}^3\left(3n+2\right)!}\\ &=-\frac{n!^3\left(\frac{1}2\right)_{3n+2}}{6\left(\frac 12\right)_{n+1}^3\left(3n+2\right)!} \end{align}
次に, $n\mapsto 2n$として,

\begin{align} &\int_{-1}^1(3P_{2n}^{(2)}(x)P_{2n}(x)^2+2P_{2n}^{(1)}(x)^2P_{2n}(x)+4P_{2n}(x)P_{2n}^{(1)}(x)P_{2n}^{(1)}(-x))\,dx\\ &=\left.\frac{\partial^2}{\partial \nu^2}\frac{1+2\cos\pi\nu}3\frac{\pi\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+2}2\right)}{\Gamma\left(\frac{1-\nu}2\right)^2\Gamma\left(\frac{\nu+2}2\right)^3\Gamma\left(\frac{3\nu+3}2\right)}\right|_{\nu=2n}\\ &=\frac{\partial}{\partial \nu}\frac{1+2\cos\pi\nu}3\frac{\pi\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+2}2\right)}{\Gamma\left(\frac{1-\nu}2\right)^2\Gamma\left(\frac{\nu+2}2\right)^3\Gamma\left(\frac{3\nu+3}2\right)}\\ &\qquad\cdot\left.\left(-\frac{2\pi\sin\pi\nu}{1+2\cos\pi\nu}+\frac 12\left(\psi\left(\frac{\nu+1}2\right)+3\psi\left(\frac{3\nu+2}2\right)+2\psi\left(\frac{1-\nu}2\right)-3\psi\left(\frac{\nu+2}2\right)-3\psi\left(\frac{3\nu+3}2\right)\right)\right)\right|_{\nu=2n}\\ &=\frac{1+2\cos\pi\nu}3\frac{\pi\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+2}2\right)}{\Gamma\left(\frac{1-\nu}2\right)^2\Gamma\left(\frac{\nu+2}2\right)^3\Gamma\left(\frac{3\nu+3}2\right)}\\ &\qquad\cdot\left.\left(-\frac{2\pi\sin\pi\nu}{1+2\cos\pi\nu}+\frac 12\left(\psi\left(\frac{\nu+1}2\right)+3\psi\left(\frac{3\nu+2}2\right)+2\psi\left(\frac{1-\nu}2\right)-3\psi\left(\frac{\nu+2}2\right)-3\psi\left(\frac{3\nu+3}2\right)\right)\right)^2\right|_{\nu=2n}\\ &\qquad+\frac{1+2\cos\pi\nu}3\frac{\pi\Gamma\left(\frac{\nu+1}2\right)\Gamma\left(\frac{3\nu+2}2\right)}{\Gamma\left(\frac{1-\nu}2\right)^2\Gamma\left(\frac{\nu+2}2\right)^3\Gamma\left(\frac{3\nu+3}2\right)}\\ &\qquad\cdot\left.\left(-\frac{\partial}{\partial \nu}\frac{2\pi\sin\pi\nu}{1+2\cos\pi\nu}+\frac 14\left(\psi'\left(\frac{\nu+1}2\right)+9\psi'\left(\frac{3\nu+2}2\right)-2\psi'\left(\frac{1-\nu}2\right)-3\psi'\left(\frac{\nu+2}2\right)-9\psi'\left(\frac{3\nu+3}2\right)\right)\right)\right|_{\nu=2n}\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\psi\left(\frac{1}2+n\right)+3\psi\left(3n+1\right)+2\psi\left(\frac{1}2-n\right)-3\psi\left(n+1\right)-3\psi\left(3n+\frac{3}2\right)\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{2\pi^2}{3}+\frac 14\left(\psi'\left(\frac{1}2+n\right)+9\psi'\left(3n+1\right)-2\psi'\left(\frac{1}2-n\right)-3\psi'\left(n+1\right)-9\psi'\left(3n+\frac{3}2\right)\right)\right)\\ &=\frac{9\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{20\pi^2}{3}-3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}+3\sum_{k=1}^n\frac 1{k^2}-9\sum_{k=1}^{3n}\frac 1{k^2}+9\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right) \end{align}
よって,
\begin{align} &\int_{-1}^1P_{2n}^{(1)}(x)^2P_{2n}(x)\,dx\\ &=\frac{9\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{20\pi^2}{3}-3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}+3\sum_{k=1}^n\frac 1{k^2}-9\sum_{k=1}^{3n}\frac 1{k^2}+9\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &\qquad-\frac 32\int_{-1}^1P_{2n}^{(2)}(x)P_{2n}(x)^2\,dx\\ &\qquad-2\int_{-1}^1P_{2n}(x)P_{2n}^{(1)}(x)P_{2n}^{(1)}(-x)\,dx\\ &=\frac{9\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{20\pi^2}{3}-3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}+3\sum_{k=1}^n\frac 1{k^2}-9\sum_{k=1}^{3n}\frac 1{k^2}+9\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &\qquad-\frac{3\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac{1}2}\right)^2\\ &\qquad-\frac{3\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{4\pi^2}{3}+\sum_{k=0}^{3n}\frac 1{\left(k+\frac{1}2\right)^2}-\sum_{k=1}^{3n}\frac 1{k^2}-3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac{1}2\right)^2}+3\sum_{k=1}^n\frac 1{k^2}\right)\\ &\qquad-\frac{\left(\frac 12\right)_n^3(3n)!}{2n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad-\frac{\left(\frac 12\right)_n^3(3n)!}{2n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{2\pi^2}3-\sum_{k=1}^{3n}\frac 1{k^2}+\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}-\sum_{k=1}^n\frac 1{k^2}+\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{20\pi^2}{3}-3\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}+3\sum_{k=1}^n\frac 1{k^2}-9\sum_{k=1}^{3n}\frac 1{k^2}+9\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &\qquad-\frac{\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(-4\pi^2+3\sum_{k=0}^{3n}\frac 1{\left(k+\frac{1}2\right)^2}-3\sum_{k=1}^{3n}\frac 1{k^2}-9\sum_{k=0}^{n-1}\frac 1{\left(k+\frac{1}2\right)^2}+9\sum_{k=1}^n\frac 1{k^2}\right)\\ &\qquad-\frac{\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(-\frac{8\pi^2}3-4\sum_{k=1}^{3n}\frac 1{k^2}+4\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}-4\sum_{k=1}^n\frac 1{k^2}+4\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{8n!^3\left(\frac 12\right)_{3n+1}}\left(2\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}-2\sum_{k=1}^n\frac 1{k^2}-2\sum_{k=1}^{3n}\frac 1{k^2}+2\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right)\\ &=\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=n+1}^{3n}\frac 1k-\sum_{k=n}^{3n}\frac 1{k+\frac 12}\right)^2\\ &\qquad+\frac{\left(\frac 12\right)_n^3(3n)!}{4n!^3\left(\frac 12\right)_{3n+1}}\left(\sum_{k=0}^{n-1}\frac 1{\left(k+\frac 12\right)^2}-\sum_{k=1}^n\frac 1{k^2}-\sum_{k=1}^{3n}\frac 1{k^2}+\sum_{k=0}^{3n}\frac 1{\left(k+\frac 12\right)^2}\right) \end{align}
となる. ここまでの計算をまとめると以下のようになる.

ここにおいて, 奇数次の場合で$0$になるようなものは省略した.