中央二項係数の4乗のFourier-Legendre展開
\begin{align}
\beta_n:=\frac{\Gamma(n+\frac12)}{\Gamma(\frac12)\Gamma(n+1)},\kappa(x):=\sum_{n=0}^{\infty}\beta_n^2x^n,\beta(2):=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}
\end{align}
$P_n(x):\mathrm{Legendre}$多項式
今回は次を示します.
\begin{align} \frac{\pi^2}{4}\kappa\left(x\right)^2=\sum_{n=0}^{\infty}(4n+1)\beta_n^4\left(\frac{7}{2}\zeta(3)+\sum_{m=1}^{n}\frac{4m-1}{(2m\beta_m)^4}\right)P_{2n}(2x-1)+\sum_{n=1}^{\infty}\frac{4n-1}{(2n\beta_n)^4}P_{2n-1}(2x-1)\sum_{k=0}^{n-1}(4k+1)\beta_k^4 \end{align}
この記事
に書かれている4乗のFourier-Legendre展開の計算をします.
\begin{align}
\left(\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}-\frac{d}{dx}(1-x^2)-x\right)P_n(x)=\frac{1}{2n+1}\left((n+1)^4P_{n+1}(x)-n^4P_{n-1}(x)\right)
\end{align}
について,$f(x)$をかけて0から1で積分します.
\begin{eqnarray}
(RHS)&=&\frac{1}{2n+1}\left((n+1)^4\int_0^1f(x)P_{n+1}(x)dx-n^4\int_0^1f(x)P_{n-1}(x)dx\right) \\
(LHS)&=&\int_0^1f(x)\left(-x-\frac{d}{dx}(1-x^2)+\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}\right)P_n(x)dx\\
&=&\int_0^1P_n(x)\left(-xf(x)+(1-x^2)\frac{d}{dx}f(x)\right)-\left((1-x^2)\frac{d}{dx}f(x)\right) \cdot \left(\frac{d}{dx}(1-x^2)\frac{d}{dx}P_n(x)\right)dx+\left[(1-x^2)f(x)\frac{d}{dx}(1-x^2)\frac{d}{dx}P_n(x)-(1-x^2)f(x)P_n(x)\right]_0^1 \\
&=&\int_0^1P_n(x)\left(x+\frac{d}{dx}(1-x^2)\right)f(x)+\left((1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)\right)\cdot\left(\frac{d}{dx}P_n(x)\right)dx+\left[(1-x^2)f(x)\left(\frac{d}{dx}(1-x^2)\frac{d}{dx}-1\right)P_n(x)-(1-x^2)^2\frac{d}{dx}f(x)\cdot \frac{d}{dx}P_n(x)\right]_0^1\\
&=&-\int_0^1P_n(x)\left(\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}-\frac{d}{dx}(1-x^2)-x\right)f(x)dx+\left[(1-x^2)f(x)\left(\frac{d}{dx}(1-x^2)\frac{d}{dx}-1\right)P_n(x)-(1-x^2)^2\frac{d}{dx}f(x)\cdot \frac{d}{dx}P_n(x)+(1-x^2)P_n(x)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)\right]_0^1
\end{eqnarray}
これにより,
\begin{align}
g(x):=\left(\frac{d}{dx}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}-\frac{d}{dx}(1-x^2)-x\right)f(x),D_n(x):=(1-x^2)f(x)\left(\frac{d}{dx}(1-x^2)\frac{d}{dx}-1\right)P_n(x)-(1-x^2)^2\frac{d}{dx}f(x)\cdot \frac{d}{dx}P_n(x)+(1-x^2)P_n(x)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)
\end{align}
として,
\begin{align}
\frac{1}{2n+1}\left((n+1)^4\int_0^1f(x)P_{n+1}(x)dx-n^4\int_0^1f(x)P_{n-1}(x)dx\right)=[D_n(x)]_0^1-\int_0^1g(x)P_n(x)dx
\end{align}
が得られると書いてあります.
これを用いて定理1を証明します.
$f(x)=\kappa(\frac{1-x}{2})^2,\kappa(\frac{1-x}{2})\kappa(\frac{1+x}{2}),\kappa(\frac{1+x}{2})^2$のとき$g(x)=0$です.
$f(x)$はx=1で発散する場合があるので$[D_n(x)]_0^1=\lim_{x\to 1^-}D_n(x)-D_n(0)$として計算します.
\begin{eqnarray}
D_n(0)&=&f(0)(P_n^{\prime\prime}(0)-P_n(0))-f^{\prime}(0)P_n^{\prime}(0)+P_n(0)f^{\prime \prime}(0)\\
\lim_{x\to1^{-}}D_n(x)&=&\left(-2P_n^{\prime}(1)-P_n(1)\right)\left(\lim_{x\to1^{-}}(1-x^2)f(x)\right)-P_n^{\prime}(1)\left(\lim_{x\to1^-}(1-x^2)^2\frac{d}{dx}f(x)\right)+P_n(1)\left(\lim_{x\to1^-}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)\right)
\end{eqnarray}
以下,区別しやすくするため$f_1(x)=\kappa(\frac{1-x}{2})^2,f_2(x)=\kappa(\frac{1-x}{2})\kappa(\frac{1+x}{2}),f_3(x)=\kappa(\frac{1+x}{2})^2$とします.
先ず,3つの極限から計算しましょう.
$f(x)=\kappa(\frac{1-x}{2})^2$のときどれも0となります.
$f(x)=\kappa(\frac{1-x}{2})\kappa(\frac{1+x}{2})$のとき,
\begin{eqnarray}
\lim_{x\to1^-}(1-x^2)f(x)&=&\lim_{x\to0^+}4x(1-x)\kappa(x)\kappa(1-x)=\lim_{x\to0^+}x\cdot \frac{1}{\pi}\sum_{n=0}^{\infty}\beta_n^2x^{n}\left(4\ln2-4\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}-\ln x\right)=0\\
\lim_{x\to1^-}(1-x^2)^2\frac{d}{dx}f(x)&=&\lim_{x\to1^-}\frac{d}{dx}\left((1-x^2)^2f(x)\right)+4x(1-x^2)f(x)=\lim_{x\to1^-}\frac{d}{dx}(1-x^2)^2\kappa\left(\frac{1-x}{2}\right)\cdot\frac1\pi\sum_{n=0}^{\infty}\beta_n^2\left(\frac{1-x}{2}\right)^n\left(4\ln2-4\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}-\ln\frac{1-x}{2}\right)=\lim_{x\to1^-}\frac{d}{dx}(1-x^2)^2\cdot\frac1\pi\ln\frac{1}{1-x}=0\\
\lim_{x\to1^-}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)&=&\int_0^1\left(\frac{d}{dx}(1-x^2)+x\right)f(x)dx+f^{\prime\prime}(0)=f''(0)+\left[(1-x^2)f(x)\right]_0^1+\int_0^1xf(x)dx=f''(0)-f(0)+\int_0^1x\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)dx
\end{eqnarray}
途中,$\kappa(1-x)=\frac1\pi\sum_{n=0}^{\infty}\beta_n^2x^n\left(4\ln2-4\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}-\ln x\right)$を用いました.
最後の計算について,過去に書いた
K^２のmomentの計算
を参考して,
\begin{eqnarray}
2\int_0^1x\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)dx&=&\int_0^1\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)dx\\
&=&\int_0^1\left(\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)dx+\int_0^1\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2dx\\
&=&\frac{8\Gamma(\frac34)^4}{\pi^3}+\sum_{n=0}^{\infty}\frac{\beta_n^3}{n+1}\\
&=&\frac{\pi}{\Gamma(\frac{3}4)^4}+\frac{4\Gamma(\frac34)^4}{\pi^3}
\end{eqnarray}
と,
\begin{eqnarray}
f_2(0)=\frac{\pi}{\Gamma(\frac34)^4},f_2''(0)=\frac{\pi}{2\Gamma(\frac34)^4}-\frac{2\Gamma(\frac34)^4}{\pi^3}
\end{eqnarray}
から,
\begin{align}
\lim_{x\to1^-}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)=0
\end{align}
がわかり,いづれも0であることがわかります.
$f(x)=\kappa(\frac{1+x}{2})^2$のとき,
\begin{eqnarray}
\lim_{x\to1^-}(1-x^2)f(x)&=&\lim_{x\to0^+}4x(1-x)\kappa(1-x)^2=\lim_{x\to0^+}4x\cdot \frac1{\pi^2}\ln^2x=0\\
\lim_{x\to1^-}(1-x^2)^2\frac{d}{dx}f(x)&=&\lim_{x\to1^-}\frac{d}{dx}(1-x^2)^2f(x)=\lim_{x\to1^-}\frac{d}{dx}(1-x^2)^2\cdot\frac1{\pi^2}\ln^2(1-x)=0
\\
\lim_{x\to1^-}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}f(x)&=&f''(0)-f(0)+\int_0^1x\kappa\left(\frac{1+x}{2}\right)^2dx
\end{eqnarray}
最後の計算について,先程と同様にして,
\begin{eqnarray}
2\int_0^1x\kappa\left(\frac{1+x}{2}\right)^2dx&=&\int_0^1\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2dx\\
&=&\int_0^1\left(\kappa\left
(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)dx+\int_0^1\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2dx\\
&=&\frac{16}{\pi^2}+\sum_{n=0}^{\infty}\frac{\beta_n^3}{n+1}=\frac{16}{\pi^2}+\frac{\pi}{\Gamma(\frac34)^4}-\frac{4\Gamma(\frac34)^4}{\pi^3}
\end{eqnarray}
と,
\begin{align}
f_3(0)=\frac{\pi}{\Gamma(\frac34)^4},f_3''(0)=\frac{\pi}{2\Gamma(\frac34)^4}+\frac{2\Gamma(\frac34)^4}{\pi^3}
\end{align}
から,
\begin{align}
\lim_{x\to1^-}(1-x^2)\frac{d}{dx}(1-x^2)\frac{d}{dx}f_3(x)=\frac{8}{\pi^2}
\end{align}
がわかりました.
よって,
\begin{align}
\lim_{x\to1^-}D_n(x)=\begin{cases}
0 \ \ \ \ \ (f(x)=\kappa(\frac{1-x}{2})^2,\kappa(\frac{1-x}{2})\kappa(\frac{1+x}{2}))\\
\frac{8}{\pi^2}\ \ (f(x)=\kappa(\frac{1+x}{2})^2)
\end{cases}
\end{align}
となります.
次に,$D_n(0)$を計算します.
\begin{align}
\sum_{n=0}^{\infty}P_n(x)t^n=\frac{1}{\sqrt{1-2xt+t^2}} \Rightarrow \sum_{n=0}^{\infty}\left(\frac{d}{dx}P_n(x)\right)t^n=\frac{t}{(1-2xt+t^2)^{3/2}} \Rightarrow \sum_{n=0}^{\infty}\left(\frac{d^2}{dx^2}P_n(x)\right)t^n=\frac{t^2}{(1-2xt+t^2)^{5/2}}
\end{align}
ですから,$x=0$としたときの両辺の$t^n$の係数比較から
\begin{align}
P_n(0)=\begin{cases}
0\ \ (n=2m+1)\\
(-1)^m\beta_m\ (n=2m)
\end{cases} ,P'_n(0)=\begin{cases}
(-1)^m(2m+1)\beta_m\ (n=2m+1)\\
0 \ (n=2m)
\end{cases},P''(0)=\begin{cases}
0\ (n=2m+1)\\
(-1)^m2m(2m+1)\beta_m\ (n=2m)
\end{cases}
\end{align}
がわかります.
\begin{align}
f_1(0)=\frac{\pi}{\Gamma(\frac{3}{4})^4},f_1'(0)=-\frac{2}{\pi},f_1''(0)=\frac{\pi}{2\Gamma(\frac34)^4}+\frac{2\Gamma(\frac34)^4}{\pi^3}\\
f_2(0)=f_1(0),f_2'(0)=0,f_2''(0)=\frac{\pi}{2\Gamma(\frac34)^4}-\frac{2\Gamma(\frac34)^4}{\pi^3}\\
f_3(0)=f_2(0),f_3'(0)=-f_1'(0),f_3''(0)=f_1''(0)
\end{align}
なので,
\begin{align}
D_n(0)=\begin{cases}
-f'(0)\cdot (-1)
^m(2m+1)\beta_m\ \ (n=2m+1)\\
f(0)\cdot(-1)^m4m^2\beta_m+f''(0)\cdot(-1)^m\beta_m\ (n=2m)
\end{cases}
\end{align}
より,$f(x)=f_1(x)$のとき
\begin{align}
D_n(0)=\begin{cases}
\frac{2}{\pi}\cdot(-1)^m(2m+1)\beta_m\ \ (n=2m+1)\\
\left(4m^2\frac{\pi}{\Gamma(\frac34)^4}+\frac{\pi}{2\Gamma(\frac34)^4}+\frac{2\Gamma(\frac34)^4}{\pi^3}\right)\cdot(-1)^m\beta_m\ (n=2m)
\end{cases}
\end{align}
$f(x)=f_2(x)$のとき,
\begin{align}
D_n(0)=\begin{cases}
0\ \ \ (n=2m+1)\\
\left(4m^2\frac{\pi}{\Gamma(\frac34)^4}+\frac{\pi}{2\Gamma(\frac34)^4}-\frac{2\Gamma(\frac34)^4}{\pi^3}\right)\cdot(-1)^m\beta_m\ (n=2m)
\end{cases}
\end{align}
$f(x)=f_3(x)$のとき,
\begin{align}
D_n(0)=\begin{cases}
-\frac2\pi\cdot(-1)^m(2m+1)\beta_m\ \ (n=2m+1)\\
\left(4m^2\frac{\pi}{\Gamma(\frac{3}{4})^4}+\frac{\pi}{2\Gamma(\frac34)^4}+\frac{2\Gamma(\frac34)^4}{\pi^3}\right)\cdot(-1)^m\beta_m\ (n=2m)
\end{cases}
\end{align}
以上で$[D_n(x)]_0^1$の計算を終えました.
\begin{eqnarray}
(n+1)^4\int_0^1f(x)P_{n+1}(x)dx-n^4\int_0^1f(x)P_{n-1}(x)dx=(2n+1)[D_n(x)]_0^1
\end{eqnarray}
について,偶奇に分けて計算すると,
\begin{eqnarray}
\int_0^1f(x)P_{2n-1}(x)dx&=&\frac{1}{(2n\beta_n)^4}\sum_{k=0}^{n-1}(4k+1)\beta_k^4[D_{2k}(x)]_0^1\\
\int_0^1f(x)P_{2n}(x)dx&=&\beta_n^4\left(\int_0^1f(x)dx+\sum_{k=1}^{n}\frac{4k-1}{(2k\beta_k)^4}[D_{2k-1}(x)]_0^1\right)
\end{eqnarray}
ですから,$\int_0^1f(x)dx$の計算が必要です.
\begin{eqnarray}
\int_0^1f_1(x)dx&=&\int_0^1\kappa\left(\frac{1-x}{2}\right)^2dx=\frac12\int_0^1\frac{1}{\sqrt{1-x}}\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2dx\\
&=&\frac{1}{2}\sum_{n=0}^{\infty}\beta_n^3\int_0^1\frac{x^{n}}{\sqrt{1-x}}dx=\sum_{n=0}^{\infty}\frac{\beta_n^2}{2n+1}\\
&=&\frac{2}{\pi}\int_0^1K(x)dx=\frac{4\beta(2)}{\pi}\\
\int_0^1f_2(x)dx&=&\int_0^1\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)dx=\int_{-1}^0\kappa\left(\frac{1+x}{2}\right)\kappa\left(\frac{1-x}{2}\right)dx\\
&=&\frac12\int_{-1}^{1}\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)dx=\int_0^1\kappa(x)\kappa(1-x)dx\\
&=&\frac\pi2\\
\int_0^1f_3(x)dx&=&\int_0^1\kappa\left(\frac{1+x}{2}\right)^2dx=\frac{1}2\int_0^1\frac{1}{\sqrt{1-x}}\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2dx\\
&=&\frac12\int_0^1\frac{1}{\sqrt{1-x}}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)dx+\frac12\int_0^1\frac{1}{\sqrt{1-x}}\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2dx\\
&=&\frac12\sum_{n=0}^{\infty}\beta_n\int_0^1x^{n}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)dx+\int_0^1\kappa\left(\frac{1-x}{2}\right)^2dx\\
&=&\frac12\sum_{n=0}^{\infty}\beta_n\cdot\frac{2}{\pi^2}\frac{1}{(n+\frac12)^3\beta_n^3}+\frac{4\beta(2)}{\pi}=\frac{1}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{(n+\frac12)^3\beta_n^2}+\frac{4\beta(2)}{\pi}\\
&=&\frac{28\zeta(3)}{\pi^2}-\frac{4\beta(2)}{\pi}
\end{eqnarray}
なので,以下を得ます.
\begin{eqnarray}
\int_0^1f_1(x)P_{2n-1}(x)dx&=&\frac{1}{(2n\beta_n)^4}\sum_{k=0}^{n-1}(-1)^{k-1}(4k+1)\beta_k^5\left(4m^2\frac{\pi}{\Gamma(\frac34)^4}+\frac{\pi}{2\Gamma(\frac34)^4}+\frac{2\Gamma(\frac34)^4}{\pi^3}\right)\\
\int_0^1f_1(x)P_{2n}(x)dx&=&\beta_n^4\left(\frac{4\beta(2)}{\pi}+\frac{2}{\pi}\sum_{k=1}^{n}\frac{(-1)^{k}(4k-1)}{(2k\beta_k)^3}\right)\\
\int_0^1f_2(x)P_{2n-1}(x)dx&=&\frac{1}{(2n\beta_n)^4}\sum_{k=0}^{n-1}(-1)^{k-1}(4k+1)\beta_k^5\left(4m^2\frac{\pi}{\Gamma(\frac34)^4}+\frac{\pi}{2\Gamma(\frac34)^4}-\frac{2\Gamma(\frac34)^4}{\pi^3}\right)\\
\int_0^1f_2(x)P_{2n}(x)dx&=&\beta_n^4\cdot\frac{\pi}{2}\\
\int_0^1f_3(x)P_{2n-1}(x)dx&=&\frac{1}{(2n\beta_n)^4}\sum_{k=0}^{n-1}(-1)^{k-1}(4k+1)\beta_k^5\left(4m^2\frac{\pi}{\Gamma(\frac34)^4}+\frac{\pi}{2\Gamma(\frac34)^4}+\frac{2\Gamma(\frac34)^4}{\pi^3}\right)+\frac{1}{(2n\beta_n)^4}\sum_{k=0}^{n-1}(4k+1)\beta_{k}^4\cdot\frac8{\pi^2}\\
\int_0^1f_3(x)P_{2n}(x)dx&=&\beta_n^4\left(\frac{28\zeta(3)}{\pi^2}-\frac{4\beta(2)}{\pi}-\frac{2}{\pi}\sum_{k=1}^{n}\frac{(-1)^{k-1}(4k-1)}{(2k\beta_k)^3}+\frac{8}{\pi^2}\sum_{k=1}^{n}\frac{4k-1}{(2k\beta_k)^4}
\right)
\end{eqnarray}
右辺がに綺麗になるように,足し引きすれば次を得ます.
\begin{eqnarray}
\int_0^1\left(\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)-\kappa\left(\frac{1-x}{2}\right)^2\right)P_{2n-1}(x)dx&=&\frac{4\Gamma(\frac34)^4}{\pi^3}\frac{1}{(2n\beta_n)^4}\sum_{k=0}^
{n-1}(-1)^k(4k+1)\beta_k^5\\
\int_0^1\left(\kappa\left(\frac{1+x}{2}\right)^2-\kappa\left(\frac{1-x}{2}\right)^2\right)P_{2n-1}(x)dx&=&\frac{8}{\pi^2}\frac{1}{(2n\beta_n)^4}\sum_{k=0}^{n-1}(4k+1)\beta_k^4\\
\int_0^1\kappa\left(\frac{1-x}{2}\right)^2P_{2n}(x)dx&=&\frac2\pi\beta_n^4\left(2\beta(2)-\sum_{k=1}^{n}\frac{(-1)^{k-1}(4k-1)}{(2k\beta_k)^3}\right)\\
\int_0^1\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)P_{2n}(x)dx&=&\frac{\pi}{2}\beta_n^4\\
\int_0^1\left(\kappa\left(\frac{1+x}{2}\right)^2+\kappa\left(\frac{1-x}{2}\right)^2\right)P_{2n}(x)dx&=&\frac{8}{\pi^2}\beta_n^4\left(\frac{7}{2}\zeta(3)+\sum_{k=1}^{n}\frac{4k-1}{(2k\beta_k)^4}\right)
\end{eqnarray}
左辺について,0から1の積分に変形することを目指します.
\begin{eqnarray}
\int_0^1\left(\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)-\kappa\left(\frac{1-x}{2}\right)^2\right)P_{2n-1}(x)dx&=&-\int_{-1}^0\left(\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)-\kappa\left(\frac{1+x}{2}\right)^2\right)P_{2n-1}(x)dx\\
&=&\frac12\int_{-1}^{1}\mathrm{sign}(x)\left(\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)-\kappa\left(\frac{1- \left| x \right| }{2}\right)^2\right)P_{2n-1}(x)dx\\
\int_0^1\left(\kappa\left(\frac{1+x}{2}\right)^2-\kappa\left(\frac{1-x}{2}\right)^2\right)P_{2n-1}(x)dx&=&\int_0^1\kappa\left(\frac{1+x}{2}\right)^2P_{2n-1}(x)dx+\int_{-1}^0\kappa\left(\frac{1+x}{2}\right)^2P_{2n-1}(x)dx\\
&=&\int_{-1}^{1}\kappa\left(\frac{1+x}{2}\right)^2P_{2n-1}(x)dx\\
\int_0^1\kappa\left(\frac{1-x}{2}\right)^2P_{2n}(x)dx&=&\int_{-1}^{0}\kappa\left(\frac{1+x}{2}\right)^2P_{2n}(x)dx\\
&=&\frac{1}{2}\int_{-1}^1\kappa\left(\frac{1- \left| x \right| }{2}\right)^2P_{2n}(x)dx\\
\int_0^1\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)P_{2n}(x)dx&=&\frac{1}{2}\int_{-1}^{1}\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)P_{2n}(x)dx\\
\int_0^1\left(\kappa\left(\frac{1+x}{2}\right)^2+\kappa\left(\frac{1-x}{2}\right)^2\right)P_{2n}(x)dx&=&\int_{-1}^{1}\kappa\left(\frac{1+x}{2}\right)^2P_{2n}(x)dx
\end{eqnarray}
これにより,以下のFL展開が得られました.
\begin{eqnarray} \mathrm{sign}(x)\left(\kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)-\kappa\left(\frac{1- \left| x \right| }{2}\right)^2\right)&=&\frac{4\Gamma(\frac34)^4}{\pi^3}\sum_{n=1}^{\infty}\frac{4n-1}{(2n\beta_n)^4}P_{2n-1}(x)\sum_{k=0}^{n-1}(-1)^k(4k+1)\beta_k^5\\ \kappa\left(\frac{1- \left| x \right| }{2}\right)^2&=&\frac{2}{\pi}\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(x)\left(2\beta(2)-\sum_{k=1}^{n}\frac{(-1)^{k-1}(4k-1)}{(2k\beta_k)^3}\right) =\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(4n-1)}{(2n\beta_n)^3}\sum_{k=0}^{n-1}(4k+1)\beta_k^4P_{2k}(x) \\ \kappa\left(\frac{1-x}{2}\right)\kappa\left(\frac{1+x}{2}\right)&=&\frac{\pi}{2}\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(x) \\ \kappa\left(\frac{1+x}{2}\right)^2&=&\frac{4}{\pi^2}\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(x)\left(\frac{7}{2}\zeta(3)+\sum_{k=1}^{n}\frac{4k-1}{(2k\beta_k)^4}\right)+\frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{4n-1}{(2n\beta_n)^4}P_{2n-1}(x)\sum_{k=0}^{n-1}(4k+1)\beta_k^4 \end{eqnarray}
定理1より次がわかります.
\begin{align}
\sum_{n=0}^{\infty}\frac{\beta_n^4}{4n+1}\left(\frac{7}{2}\zeta(3)+\sum_{k=1}^{n}\frac{4k-1}{(2k\beta_k)^4}\right)=\frac{11}{480}\frac{\pi^4}{\Gamma(\frac34)^8}
\end{align}
導出
\begin{align} \kappa(x)^2=\frac{4}{\pi^2}\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(2x-1)\left(\frac{7}{2}\zeta(3)+\sum_{k=1}^{n}\frac{4k-1}{(2k\beta_k)^4}\right)+\sum_{n=1}^{\infty}a_nP_{2n-1}(2x-1) \end{align}
と,
\begin{align} \kappa(x)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{2n+1}P_n(2x-1) \end{align}
より,
\begin{eqnarray} \int_0^1\kappa(x)^2\cdot(\kappa(x)+\kappa(1-x))dx&=&\frac{32}{\pi^3}\int_0^1\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}(4n+1)\beta_n^4\frac{1}{4m+1}P_{2n}(2x-1)P_{2m}(2x-1)\left(\frac{7}{2}\zeta(3)+\sum_{k=1}^{n}\frac{4k-1}{(2k\beta_k)^4}\right)dx\\ &=&\frac{32}{\pi^3}\sum_{n=0}^{\infty}\frac{\beta_n^4}{4n+1}\left(\frac{7}{2}\zeta(3)+\sum_{k=1}^n\frac{4k-1}{(2k\beta_k)^4}\right) \end{eqnarray}
であり,
\begin{align} 5\int_0^1\kappa(x)^3dx=6\int_0^1\kappa(x)^2\kappa(1-x)dx=2\frac{\pi}{\Gamma(\frac34)^8} \end{align}
なので,級数の値がわかります.
これは 非常に興味深い級数一覧 と 閉形式で書ける／書けそうな級数まとめ に書かれている式です.
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