興味深い級数を全て示す

146

定義

\begin{eqnarray} \beta_n:&=&\frac{_{2n}C_n}{2^{2n}}\ ,K(x):=\frac{\pi}{2}\sum_{n=0}^{\infty}\beta_n^2x^{2n}\\ A_n&=&(-1)^n\beta_n^3\sum_{n\lt m}\frac{(-1)^{m-1}(4m-1)}{(2m\beta_m)^3}\\ B_n&=&\frac{(-1)^n}{(2n+1)^3\beta_n^3}\sum_{m=0}^{n}(-1)^m(4m+1)\beta_m^3\\ C_n&=&\beta_n^4\left(\frac{7}2\zeta(3)+\sum_{m=1}^n\frac{4m-1}{(2m\beta_m)^4}\right)\\ D_n&=&\frac{1}{(2n+1)^4\beta_n^4}\sum_{m=0}^{n-1}(4n+1)\beta_n^4\\ E_n&=&\beta_n^2\left(2\beta(2)+\sum_{m=1}^n\frac{1}{(2m\beta_m)^2}\right)\\ F_n&=&\frac{1}{(2n+1)^2\beta_n^2}\sum_{m=0}^{n-1}\beta_n^2\\ V_n&=&\sum_{m=0}^n\beta_{m}^2\beta_{n-m}^2 \end{eqnarray}

始めに

3年程前からある記事非常に興味深い級数一覧の級数を全て示します.

最近僕が出した記事がこの記事の補題です.

以下最近の記事

先ず,今までの記事で得られた結果をある程度分類して羅列します.

FL展開

\begin{eqnarray} \frac1{\sqrt2}\frac{1}{\sqrt{1-x}}&=&\sum_{n=0}^{\infty}P_n(x)\\ \frac1\pi\frac{\kappa(x)}{\sqrt{1-x}}&=&\sum_{n=0}^{\infty}P_n(x)(2n+1)V_n \\ \frac{\pi}{4}\kappa(x)&=&\sum_{n=0}^{\infty}P_n(2x-1)\frac1{2n+1} \\ \frac1{4}\frac{\ln\frac1{1-x}}{\sqrt{1-x}}&=&\sum_{n=0}^{\infty}P_n(2x-1)\left(\frac1{2n+1}+2\sum_{k=0}^{n}\frac1{2k+1}\right)\\ \frac12\frac{\ln\frac1x}{1-x}&=&\sum_{n=0}^{\infty}P_{n}(2x-1)(2n+1)\sum_{k\gt n}\frac{(-1)^{k-1}}{k^2}\\ \frac12\frac{1}{\sqrt{1- \left| x \right| }}&=&\sum_{n=0}^{\infty}P_{2n}(x)\left(2\sum_{k=0}^n(-1)^k\beta_k-(-1)^n\beta_n\right)\\ \frac14\frac{\mathrm{sign}(x)}{\sqrt{1-\left| x \right|}}&=&\sum_{n=0}^{\infty}P_{2n+1}(x)\sum_{k=0}^n(-1)^k\beta_k\\ \end{eqnarray}

\begin{eqnarray} \frac{\sin^{-1}x}{x\sqrt{1-x^2}}&=&\sum_{n=0}^{\infty}P_{2n}(x)(-1)^n(4n+1)\beta_n\sum_{n\lt m}\frac{(-1)^{m-1}(4m-1)}{(2m\beta_m)^3}=\sum_{n=0}^{\infty}P_{2n}(x)\frac{4n+1}{\beta_n^2}A_n\\ \frac{\Gamma(\frac13)^3}{4\pi^2}\frac{1}{(x(1-x))^{\frac13}}&=&\sum_{n=0}^{\infty}P_{2n}(2x-1)(4n+1)\beta_n\frac{(\frac13)_n}{(\frac76)_n} \end{eqnarray}

\begin{eqnarray} \frac{2}{\pi}\frac1{\sqrt{1-x^2}}&=&\sum_{n=0}^{\infty}P_{2n}(x)(4n+1)\beta_n^2\\ \frac{\mathrm{sign}(x)}{\sqrt{1-x^2}}&=&\sum_{n=0}^{\infty}P_{2n+1}(x)\frac{4n+3}{(2n+1)^2\beta_n^2}\sum_{k=0}^{n}(-1)^k(4k+1)\beta_k^3=\sum_{n=0}^{\infty}P_{2n+1}(-1)^n(4n+3)(2n+1)\beta_nB_n\\ \frac{\pi}{4}\frac{\ln\frac1{1-x}}{\sqrt{1-x}}&=&\sum_{n=0}^{\infty}P_{2n}(x)(4n+1)\beta_n^2\left(2\ln2+\sum_{k=1}^{2n}\frac1k\right)+\sum_{n=0}^{\infty}P_{2n+1}(x)\frac{4n+3}{(2n+1)^2\beta_n^2}\sum_{k=0}^{n}(4k+1)\beta_k^4=\sum_{n=0}^{\infty}\left(P_{2n}(x)(4n+1)\beta_n^2\left(2\ln2+H_{2n}\right)+P_{2n+1}(x)(4n+3)(2n+1)^2\beta_n^2D_n\right)\\ \frac12\frac{\tanh^{-1}\sqrt{x}}{\sqrt{x(1-x)}}&=&\sum_{n=0}^{\infty}\left(P_{2n}(x)(4n+1)E_n+P_{2n+1}(x)F_n\right)\\ \end{eqnarray}

\begin{eqnarray} K(x)&=&\sum_{n=0}^{\infty}P_{2n}(x)(4n+1)A_n\\ \mathrm{sign}(x)K(x)&=&\sum_{n=0}^{\infty}P_{2n+1}(x)(4n+3)B_n\\ \frac{4}{\pi^2}K'(x)&=&\sum_{n=0}^{\infty}P_{2n}(x)(-1)^n(4n+1)\beta_n^3\\ \mathrm{sign}(x)K'(x)&=&\sum_{n=0}^{\infty}P_{2n+1}(x)\frac{(-1)^n(4n+3)}{(2n+1)^3\beta_n^3}\sum_{k=0}^{n}(4k+1)\beta_k^4=\sum_{n=0}^{\infty}P_{2n+1}(x)(-1)^n(4n+3)(2n+1)\beta_nD_n\\ \end{eqnarray}

\begin{eqnarray} \frac{2}{\pi}\kappa(x)\kappa(1-x)&=&\sum_{n=0}^{\infty}P_{2n}(2x-1)(4n+1)\beta_n^4\\ \frac{\pi^2}{4}\kappa(x)^2&=&\sum_{n=0}^{\infty}\left(P_{2n}(2x-1)(4n+1)C_n+P_{2n+1}(2x-1)(4n+3)D_n\right) \end{eqnarray}

積分

\begin{eqnarray} \int_0^1\frac{K(x)}{\sqrt{1+x}}dx&=&\frac13\int_0^1\frac{K(x)}{\sqrt{1-x}}dx=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{48\sqrt{2}\pi}\\ \int_0^1\frac{K(x)}{\sqrt{1-x^2}}dx&=&\frac{\pi^3}{4\Gamma(\frac34)^4}=\frac{\Gamma(\frac14)^4}{16\pi}\\ \int_0^1\frac{K(x)^2}{\sqrt{1-x^2}}dx&=&2\int_0^1K(x)K'(x)dx=\frac{\pi^3}{4}\sum_{n=0}^{\infty}\beta_n^4\\ \int_0^1\frac{xK(x)^2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}dx&=&\frac{\pi^4}{4}\sum_{n=0}^{\infty}\beta_n^4\\ \int_0^1K(x)^2dx&=&\frac12\int_0^1K'(x)^2dx=\frac{\pi^4}{32}\sum_{n=0}^{\infty}(4n+1)\beta_n^6\\ \int_0^1xK(x)^3dx&=&\frac{6}{5}\int_0^1xK(x)K'(x)^2dx=\frac{\Gamma(\frac14)^8}{640\pi^2} \\ \int_0^1xK(x)^2dx&=&\frac74\zeta(3)\\ \int_0^1xK(x)K'(x)dx&=&\frac{\pi^3}{16}\\ \int_0^1x(2x^2-1)K(x)^4dx&=&\frac{93}{16}\zeta(5)\\ \int_0^1\frac{K'(x)}{\sqrt{1-x}}dx&=&\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{32\pi}+\frac12\sum_{n=0}^{\infty}\frac{(4n+1)\beta_{2n}}{(2n+1)^3\beta_n^2}\\ \int_0^1\frac{\kappa(x)^2}{(x(1-x))^{\frac13}}dx&=&\frac{\sqrt{3}\Gamma(\frac13)^9}{2^{3}2^{\frac13}\pi^5}\\ \int_0^1K(x)K'(x)^2dx&=&\frac12\int_0^1K'(x)^3dx=\frac{\Gamma(\frac14)^8}{256\pi^2}\\ \int_0^1\frac{\kappa(x)\kappa(1-x)}{x}\ln\frac1{1-x}dx&=&\frac{7\zeta(3)}{\pi}\\ \int_0^1\frac{\sin^{-1}x}{x\sqrt{1-x^2}}K(x)dx&=&\frac{7}{2}\zeta(3)\\ \int_0^1\frac{\tanh^{-1}x\tanh^{-1}\sqrt{1-x^2}}{x(1-x^2)}dx&=&\frac{\pi^2\ln2}{2} \end{eqnarray}

今回ほぼ全ての導出で用いる計算があります.
\begin{align} \frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}(2n+1)a_nP_n(x)\right)\cdot\left(\sum_{n=0}^{\infty}(2n+1)b_nP_{n}(x)\right)dx=\sum_{n=0}^{\infty}(2n+1)a_nb_n \end{align}
これについて理解できればこの記事を読むのには十分なのかもしれません.
以下,級数とその導出です.

1

\begin{align} \sum_{n=0}^{\infty}A_n=2\sum_{n=0}^{\infty}B_n=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{48\pi} \end{align}

元となっている積分は
\begin{align} \int_0^1\frac{K(x)}{\sqrt{1+x}}dx=\frac13\int_0^1\frac{K(x)}{\sqrt{1-x}}dx=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{48\sqrt{2}\pi} \end{align}
です.
\begin{eqnarray} K(x)&=&\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(4n-1)}{(2n\beta_n)^3}\sum_{m=0}^{n-1}(-1)^m(4m+1)\beta_m^3P_{2m}(x)=\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\\ K(x)\mathrm{sign}(x)&=&\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(4n-1)}{(2n\beta_n)^3}\sum_{m=0}^{n-1}(-1)^m(4m+1)\beta_m^3=\sum_{n=0}^{\infty}(4n+3)B_nP_{2n+1}(x)\\ \frac{1}{\sqrt{1-x}}&=&\sqrt{2}\sum_{n=0}^{\infty}P_n(x) \end{eqnarray}
なので,
\begin{eqnarray} \sum_{n=0}^{\infty}A_n&=&\frac{1}{2}\int_{-1}^1\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}P_n(x)\right)dx=\frac{1}2\int_{-1}^1K(x)\cdot\frac1{\sqrt2\sqrt{1-x}}dx\\ &=&\frac{1}{2\sqrt{2}}\int_0^1K(x)\left(\frac1{\sqrt{1-x}}+\frac1{\sqrt{1+x}}\right)dx=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{48\pi}\\ \sum_{n=0}^{\infty}B_n&=&\frac{1}{2}\int_{-1}^1\left(\sum_{n=0}^{\infty}(4n+3)B_nP_{2n+1}(x)\right)\cdot\left(\sum_{n=0}^{\infty}P_n(x)\right)dx=\frac{1}{2}\int_{-1}^{1}K(x)\mathrm{sign}(x)\cdot\frac1{\sqrt2\sqrt{1-x}}dx\\ &=&\frac{1}{2\sqrt2}\int_0^1K(x)\left(\frac{1}{\sqrt{1-x}}-\frac{1}{\sqrt{1+x}}\right)dx=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{96\pi} \end{eqnarray}
となります.

2

\begin{align} \sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_nA_n=\frac{\pi}{2} \end{align}

\begin{align} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(4n-1)}{(2n\beta_n)^3}\sum_{m=0}^{n-1}(4m+1)\beta_m^4=\frac{\pi}{2} \end{align}
ですが,$P_n(0)=\begin{cases} 0 　　　　(n=2m+1)\\ (-1)^n\beta_n (n=2m) \end{cases}$なので,
\begin{align} K(x)=\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x) \end{align}
からすぐにわかります.

3

\begin{align} 2\pi\sum_{n=0}^{\infty}(-1)^nA_n=\frac{\pi^2}{2}\sum_{n=0}^{\infty}(4n+1)\beta_n^2A_n=\frac{\pi}{2}\sum_{n=0}^{\infty}\beta_nE_n=\sum_{n=1}^{\infty}2n\beta_nF_{n-1}^2=2\sum_{n=0}^{\infty}\left(\frac{E_n}{4n+1}+\frac{F_n}{4n+3}\right)=\frac{\Gamma(\frac14)^4}{16} \end{align}

元となる積分は,
\begin{align} \int_0^1\frac{K(x)}{\sqrt{1-x^2}}dx=\frac{\pi^3}{4\Gamma(\frac34)^4}=\frac{\Gamma(\frac14)^4}{16\pi} \end{align}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}(-1)^nA_n&=&\frac{1}{2}\int_{-1}^{1}\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}\right)\cdot\left(\sum_{n=0}^{\infty}(-1)^nP_{2n}(x)\right)dx\\ &=&\int_0^1K(x)\cdot\left(\mathrm{Re}\sum_{n=0}^{\infty}P_n(x)i^n\right)dx =\int_0^1K(x)\cdot\left(\mathrm{Re}\frac{1}{\sqrt{-2xi}}\right)dx\\ &=&\frac{1}{2}\int_0^1K(x)\frac{dx}{\sqrt{x}} =\int_0^1K(x^2)dx=\int_0^1\frac{1}{1+x^2}K\left(\frac{2x}{1+x^2}\right)dx\\ &=&\frac{1}{2}\int_0^1\frac{K(t)}{\sqrt{1-t^2}}dt\ \left(\frac{2x}{1+x^2}\to t\right)\\ \\ \sum_{n=0}^{\infty}(4n+1)\beta_n^2A_n&=&\frac{1}{2}\int_{-1}^{1}\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+1)\beta_n^2P_{2n}(x)\right)dx\\ &=&\frac{2}{\pi}\int_0^1K(x)\cdot\frac1{\sqrt{1-x^2}}dx\\ \\ \end{eqnarray}
以降3つはmomentによる結果になっています.
\begin{eqnarray} \sum_{n=0}^{\infty}\beta_nE_n&=&\sum_{n=0}^{\infty}\beta_n\int_0^1x^{2n}K(x)dx=\int_0^1\frac{1}{\sqrt{1-x^2}}K(x)dx\\ \\ \sum_{n=1}^{\infty}2n\beta_nF_{n-1}^2&=&\sum_{n=1}^{\infty}2n\beta_nF_{n-1}\int_0^1x^{2n-1}K(x)dx=\int_0^1\frac{\tanh^{-1} x}{\sqrt{1-x^2}}K(x)dx\\ &=&\frac{\pi}{2}\sum_{n=0}^{\infty}\beta_n^2\int_0^1x^{2n}\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx\\ &=&\frac{\pi}{2}\sum_{n=0}^{\infty}\beta_n\int_0^1x^{2n}K(x)dx=\frac{\pi}{2}\int_0^1\frac{K(x)}{\sqrt{1-x^2}}dx\\ \\ \sum_{n=0}^{\infty}\left(\frac{E_n}{4n+1}+\frac{F_n}{4n+3}\right)&=&\sum_{n=0}^{\infty}\int_0^1\left(\frac{x^{2n}}{4n+1}+\frac{x^{2n+1}}{4n+3}\right)K(x)dx=\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^1x^nK(x)dx\\ &=&\int_0^1\frac{\tanh^{-1}\sqrt{x}}{\sqrt{x}}K(x)dx =2\int_0^1\tanh^{-1}x\ K(x^2)dx=\int_0^1\frac{\tanh^{-1}x}{1+x^2}K\left(\frac{2x}{1+x^2}\right)dx\\ &=&\frac{1}{2}\int_0^1\frac{\tanh^{-1}t}{\sqrt{1-t^2}}K(t)dt \end{eqnarray}

4

\begin{align} \pi\sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_n^3A_n=\sum_{n=0}^{\infty}(4n+1)\beta_n^2C_n=\frac{\pi^2}{2}\sum_{n=0}^{\infty}\beta_n^4=\frac{4}{\pi^2}\sum_{n=0}^{\infty}\left(C_n\left(\frac{1}{4n+1}+\sum_{k=0}^{2n-1}\frac{2}{2k+1}\right)+D_n\left(\frac1{4n+3}+\sum_{k=0}^{2n}\frac{2}{2k+1}\right)\right) \end{align}

元となる積分は,
\begin{eqnarray} \int_0^1\frac{K(x)^2}{\sqrt{1-x^2}}dx&=&2\int_0^1K(x)K'(x)dx=\frac{\pi^3}{4}\sum_{n=0}^{\infty}\beta_n^4\\ \\ \int_0^1\frac{xK(x)^2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}dx&=&\frac{\pi^4}{4}\sum_{n=0}^{\infty}\beta_n^4 \end{eqnarray}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_n^3A_n&=&\frac{1}{2}\int_{-1}^{1}\left(\sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_n^3P_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\right)dx\\ &=&\int_0^1\frac4{\pi^2}K'(x)\cdot K(x)dx=\frac{\pi}{2}\sum_{n=0}^{\infty}\beta_n^4\\ \\ \sum_{n=0}^{\infty}(4n+1)\beta_n^2C_n&=&\frac12\int_{-1}^{1}\left(\sum_{n=0}^{\infty}(4n+1)\beta_n^2P_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+1)C_nP_{2n}(x)\right)dx\\ &=&\frac12\int_{-1}^{1}\left(\sum_{n=0}^{\infty}(4n+1)\beta_n^2P_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+1)C_nP_{2n}(x)+\sum_{n=0}^{\infty}(4n+3)D_nP_{2n+1}(x)\right)dx\\ &=&\frac12\int_{-1}^{1}\frac{2}{\pi}\frac{1}{\sqrt{1-x^2}}\cdot\frac{\pi^2}{4}\kappa\left(\frac{1+x}{2}\right)^2dx\\ &=&\frac{\pi}{4}\int_0^1\frac{\kappa(x)^2}{\sqrt{x(1-x)}}dx=\frac2{\pi}\int_0^1\frac{K(x)^2}{\sqrt{1-x^2}}dx\\ \\ \end{eqnarray}
\begin{eqnarray} \sum_{n=0}^{\infty}\left(C_n\left(\frac{1}{4n+1}+\sum_{k=0}^{2n-1}\frac{2}{2k+1}\right)+D_n\left(\frac{1}{4n+3}+\sum_{k=0}^{2n}\frac2{2k+1}\right)\right)&=&\int_{0}^1\left(\sum_{n=0}^{\infty}\left((4n+1)C_nP_{2n}(2x-1)+(4n+3)D_nP_{2n+1}(2x-1)\right)\right)\cdot\left(\sum_{n=0}^{\infty}\left(\frac{1}{2n+1}+\sum_{k=0}^{n-1}\frac2{2k+1}\right)P_{n}(2x-1)\right)dx\\ &=&\int_{0}^1\frac{\pi^2}{4}\kappa\left(x\right)^2\cdot \frac{\ln\frac1{1-x}}{4\sqrt{1-x}}dx =\frac{1}{2}\int_0^1\frac{xK(x)^2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}dx \end{eqnarray}

5

\begin{align} \frac{\pi^4}{32}\sum_{n=0}^{\infty}(4n+1)\beta_n^6=\sum_{n=0}^{\infty}(4n+1)A_n^2=\frac{2}{3}\sum_{n=0}^{\infty}C_n=2\sum_{n=0}^{\infty}D_n=\frac{\pi}{2}\sum_{n=0}^{\infty}\beta_n^2E_n \end{align}

元となる積分は,
\begin{align} \int_0^1K(x)^2dx=\frac12\int_0^1K'(x)^2dx=\frac{\pi^4}{32}\sum_{n=0}^{\infty}(4n+1)\beta_n^6 \end{align}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}(4n+1)A_n^2&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\right)dx\\ &=&\frac12\int_{-1}^1K(x)\cdot K(x)dx=\int_0^1K(x)^2dx\\ \\ \sum_{n=0}^{\infty}C_n&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}\left((4n+1)C_nP_{2n}(x)+(4n+3)P_{2n+1}(x)\right)\right)\cdot\left(\sum_{n=0}^{\infty}P_{2n}(x)\right)dx\\ &=&\frac12\int_{-1}^1\frac{\pi^2}{4}\kappa\left(\frac{1+x}{2}\right)^2\cdot \frac12\left(\frac{1}{\sqrt{2(1-x)}}+\frac{1}{\sqrt{2(1+x)}}\right)dx\\ &=&\frac{\pi^2}{16}\int_0^1\kappa(x)^2\left(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{x}}\right)dx=\frac{1}2\int_0^1K'(x)^2+K(x)^2\ dx\\ &=&\frac32\int_0^1K(x)^2dx\\ \\ \sum_{n=0}^{\infty}D_n&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}\left((4n+1)C_nP_{2n}(x)+(4n+3)P_{2n+1}(x)\right)\right)\cdot\left(\sum_{n=0}^{\infty}P_{2n+1}(x)\right)dx\\ &=&\frac12\int_{-1}^1\frac{\pi^2}{4}\kappa\left(\frac{1+x}{2}\right)^2\cdot \frac12\left(\frac{1}{\sqrt{2(1-x)}}-\frac{1}{\sqrt{2(1+x)}}\right)dx\\ &=&\frac12\int_0^1K'(x)^2-K(x)^2 dx\\ &=&\frac12\int_0^1K(x)^2dx \\ \end{eqnarray}
最後はmomentを用いた計算になり,
\begin{eqnarray} \sum_{n=0}^{\infty}\beta_n^2E_n&=&\int_0^1K(x)\sum_{n=0}^{\infty}\beta_n^2x^{2n}dx =\frac{2}{\pi}\int_0^1K(x)^2dx \end{eqnarray}
となります.

6

\begin{align} \sum_{n=0}^{\infty}\frac{C_n}{4n+1}=11\sum_{n=0}^{\infty}\frac{D_n}{4n+3}=\frac{11}{96}\pi^3\sum_{n=0}^{\infty}\frac{\beta_n^4}{4n+1}=\frac{11}{7680}\frac{\Gamma(\frac14)^8}{\pi^2} \end{align}

元となる積分は
\begin{align} \int_0^1xK(x)^3dx=\frac{\Gamma(\frac14)^8}{640\pi^2}\ ,\ \int_0^1xK(x)K'(x)^2dx=\frac{\Gamma(\frac14)^8}{768\pi^2} \end{align}
です.
今回は用いるFL展開を書いておきます.
\begin{eqnarray} \sum_{n=0}^{\infty}(4n+1)C_nP_{2n}(2x-1)&=&\frac{1}{2}\left(K(\sqrt{x})^2+K'(\sqrt{x})^2\right)\\ \sum_{n=0}^{\infty}(4n+3)D_nP_{2n+1}(2x-1)&=&\frac{1}{2}\left(K(\sqrt{x})^2-K'(\sqrt{x})^2\right)\\ \sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(2x-1)&=&\frac{8}{\pi^3}K(\sqrt{x})K'(\sqrt{x})\\ \sum_{n=0}^{\infty}\frac{1}{2n+1}P_n(2x-1)&=&\frac12K(\sqrt{x}) \end{eqnarray}
これらを用いて計算します.
\begin{eqnarray} \sum_{n=0}^{\infty}\frac{C_n}{4n+1}&=&\int_0^1\left(\sum_{n=0}^{\infty}(4n+1)C_nP_{2n}(2x-1)\right)\cdot\left(\sum_{n=0}^{\infty}\frac1{2n+1}P_{n}(2x-1)\right)dx\\ &=&\int_0^1\frac12\left(K(\sqrt{x})^2+K'(\sqrt{x})^2\right)\cdot \frac{1}{2}K(\sqrt{x})dx\\ &=&\frac12\int_0^1xK(x)^3+xK(x)K'(x)^2\ dx\\ \\ \sum_{n=0}^{\infty}\frac{D_n}{4n+3}&=&\int_0^1\left(\sum_{n=0}^{\infty}(4n+3)D_nP_{2n+1}(2x-1)\right)\cdot\left(\sum_{n=0}^{\infty}\frac1{2n+1}P_{n}(2x-1)\right)dx\\ &=&\int_0^1\frac12\left(K(\sqrt{x})^2-K(\sqrt{x})^2\right)\cdot\frac12K(\sqrt{x})dx\\ &=&\frac12\int_0^1xK(x)^3-xK(x)K'(x)^2\ dx\\ \\ \end{eqnarray}
また,
\begin{align} \sum_{n=0}^{\infty}\frac{\beta_{n}^4}{4n+1}=\frac{\Gamma(\frac14)^8}{96\pi^5} \end{align}
についてはガンマ関数で表せる二項係数の4乗が付いた級数を参照
係数は$\frac{11}{96}$ではなく$\frac{11}{80}$のようです.

7

\begin{align} \sum_{n=0}^{\infty}\left((2n+1)A_n+(2n+2)A_{n+1}\right)B_n=\frac{7}{4}\zeta(3) \end{align}

元となる積分は,
\begin{align} \int_0^1xK(x)^2dx=\frac74\zeta(3) \end{align}
です.
\begin{align} (2n+1)xP_n(x)=(n+1)P_{n+1}(x)+nP_{n-1}(x) \end{align}
を用います.
\begin{eqnarray} \sum_{n=0}^{\infty}\left((2n+1)A_n+(2n+2)A_{n+1}\right)B_{n}&=&\frac12\int_{-1}^{1}\left(\sum_{n=0}^{\infty}(4n+1)xP_{2n}(x)A_n\right)\cdot\left(\sum_{n=0}^{\infty}(4n+3)B_{n}P_{2n+1}(x)\right)dx\\ &=&\frac12\int_{-1}^1xK(x)\cdot\mathrm{sign}(x)K(x)dx\\ &=&\int_0^1xK(x)^2dx \end{eqnarray}

8

\begin{align} \sum_{n=0}^{\infty}\left((2n+1)A_n+(2n+2)A_{n+1}\right)(-1)^n(2n+1)\beta_nD_n=2\beta(3) \end{align}

元となる積分は,
\begin{align} \int_0^1xK(x)K'(x)dx=\frac{\pi^3}{16} \end{align}
です.
今回は,
\begin{align} \sum_{n=0}^{\infty}(4n+3)(-1)^n(2n+1)\beta_nD_nP_{2n+1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(4n+3)}{(2n+1)^3\beta_n^3}\sum_{k=0}^{n}(4k+1)\beta_k^4 P_{2n+1}(x)=\mathrm{sign}(x)K'(x) \end{align}
を用います.
\begin{eqnarray} \sum_{n=0}^{\infty}\left((2n+1)A_n+(2n+2)A_{n+1}\right)(-1)^n(2n+1)\beta_nD_n&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}(4n+1)xP_{2n}(x)A_n\right)\cdot\left(\sum_{n=0}^{\infty}(4n+3)(-1)^n(2n+1)\beta_nP_{2n+1}(x)\right)dx\\ &=&\frac12\int_{-1}^1xK(x)\cdot\mathrm{sign}(x)K'(x)dx\\ &=&\int_0^1xK(x)K'(x)dx \end{eqnarray}

9

\begin{align} \sum_{n=0}^{\infty}\left((2n+1)C_n+(2n+2)C_{n+1}\right)D_n=\frac{93}{16}\zeta(5) \end{align}

元となる積分は,
\begin{align} \int_0^1x(2x^2-1)K(x)^4dx=\frac{93}{16}\zeta(5) \end{align}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}\left((2n+1)C_n+(2n+2)C_{n+1}\right)D_n&=&\int_{0}^1\left(\sum_{n=0}^{\infty}(4n+1)(2x-1)P_{2n}(2x-1)C_n\right)\left(\sum_{n=0}^{\infty}(4n+3)D_nP_{2n+1}(2x-1)\right)dx\\ &=&\int_0^1\frac12(2x-1)\left(K(\sqrt{x})^2+K'(\sqrt{x})^2\right)\cdot\frac12\left(K(\sqrt{x})^2-K'(\sqrt{x})^2\right)dx\\ &=&\int_0^1x(2x^2-1)K(x)^4dx \end{eqnarray}

10

\begin{align} \sum_{n=0}^{\infty}\left((4n+1)C_n^2-(4n+3)D_n^2\right)=\frac{\pi^6}{64}\sum_{n=0}^{\infty}(4n+1)\beta_n^8 \end{align}

\begin{eqnarray} \sum_{n=0}^{\infty}\left((4n+1)C_n^2-(4n+3)D_n^2\right)&=&\int_0^1\left(\sum_{n=0}^{\infty}\left((4n+1)C_nP_{2n}(2x-1)+(4n+3)D_nP_{2n+1}(2x-1)\right)\right)\cdot\left(\sum_{n=0}^{\infty}\left((4n+1)C_nP_{2n}(2x-1)-(4n+3)D_nP_{2n+1}(2x-1)\right)\right)dx\\ &=&\int_0^1K(\sqrt{x})^2\cdot K'(\sqrt{x})^2dx\\ &=&\int_0^1\left(K(\sqrt{x})K'(\sqrt{x})\right)^2dx\\ &=&\int_0^1\left(\frac{\pi^3}{8}\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(2x-1)\right)^2dx\\ &=&\frac{\pi^6}{64}\sum_{n=0}^{\infty}(4n+1)\beta_n^8 \end{eqnarray}

11

\begin{align} \pi^2\sum_{n=0}^{\infty}(-1)^n\beta_n^3\left(2\sum_{k=0}^{n}(-1)^k\beta_k-(-1)^n\beta_n\right)=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{16\pi}+\sum_{n=0}^{\infty}\frac{(4n+1)\beta_{2n}}{(2n+1)^3\beta_n^2} \end{align}

元となる積分は,
\begin{align} \int_0^1\frac{K'(x)}{\sqrt{1-x}}dx=\frac{\Gamma(\frac18)^2\Gamma(\frac38)^2}{32\pi}+\frac12\sum_{n=0}^{\infty}\frac{(4n+1)\beta_{2n}}{(2n+1)^3\beta_n^2} \end{align}
です.
用いるFL展開は,
\begin{eqnarray} \sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_n^3P_{2n}(x)&=&\frac{4}{\pi^2}K'(x)\\ \sum_{n=0}^{\infty}\left(2\sum_{k=0}^n(-1)^k\beta_k-(-1)^n\beta_n\right)P_{2n}(x)&=&\frac{1}{2\sqrt{1- \left| x \right| }} \end{eqnarray}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}(-1)^n\beta_n^3\left(2\sum_{k=0}^{n}(-1)^k\beta_k-(-1)^n\beta_n\right)&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_n^3P_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}\left(2\sum_{k=0}^n(-1)^k\beta_k-(-1)^n\beta_n\right)P_{2n}(x)\right)dx\\ &=&\frac12\int_{-1}^1\frac4{\pi^2}K'(x)\cdot\frac{1}{2\sqrt{1- \left| x \right| }}dx\\ &=&\frac{2}{\pi^2}\int_0^1\frac{K'(x)}{\sqrt{1-x}}dx \end{eqnarray}

12

\begin{align} \sum_{n=0}^{\infty}\frac{4n+3}{(2n+1)^2\beta_n^2}B_n=\pi\sum_{n=0}^{\infty}\frac{\beta_n^2}{(4n+1)^2}=\frac{\pi}{8}\sum_{n=0}^{\infty}\beta_n^3\left(\frac{\pi^2}{2}+\sum_{k=1}^{n}\frac1{k^2\beta_k}\right) \end{align}

\begin{eqnarray} \sum_{n=0}^{\infty}\frac{4n+3}{(2n+1)^2\beta_n^2}B_n&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}\frac{4n+3}{(2n+1)^2\beta_n^2}P_{2n+1}(x)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+3)B_nP_{2n+1}(x)\right)dx\\ &=&\frac12\int_{-1}^1\frac{\sin^{-1}x}{\sqrt{1-x^2}}\cdot\mathrm{sign}(x)K(x)dx\\ &=&\int_0^1\frac{\sin^{-1}x}{\sqrt{1-x^2}}K(x)dx \end{eqnarray}
ですが,
\begin{align} \int_0^1x^{2n}\frac{\sin^{-1}x}{\sqrt{1-x^2}}dx=\beta_n\left(\frac{\pi^2}{8}+\sum_{k=1}^n\frac1{(2k)^2\beta_k}\right) \end{align}
を用いれば,
\begin{align} \sum_{n=0}^{\infty}\frac{4n+3}{(2n+1)^2\beta_n^2}B_n=\frac{\pi}{8}\sum_{n=0}^{\infty}\beta_n^3\left(\frac{\pi^2}{2}+\sum_{k=1}^n\frac1{k^2\beta_k}\right) \end{align}
がわかります.また,
\begin{eqnarray} \int_0^1K(x)\frac{\sin^{-1}x}{\sqrt{1-x^2}}dx&=&\int_0^1K(x)\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n\beta_n}dx\\ &=&\sum_{n=1}^{\infty}\frac1{(2n\beta_n)^3}\sum_{k=0}^{n-1}\beta_k^2\\ &=&\int_0^1K'(x)\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n\beta_n}\sum_{k=0}^{n-1}\beta_k^2dx\\ &=&\int_0^1K'(x)\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx=\int_0^1\frac{K'(x)}{\sqrt{1-x^2}}dx\int_0^x\frac{dy}{1-y^2}\\ &=&\int_0^1\frac{dy}{y\sqrt{1-y^2}}\int_0^y\frac{K(x)}{\sqrt{1-x^2}}dx \ \left(u\to\sqrt{1-u^2}\right)\\ &=&\int_0^1\frac{dy}{y}\int_0^y\frac{2}{1+x^2}K\left(\frac{2x}{1+x^2}\right)dx\ \left(u\to\frac{2u}{1+u^2}\right)\\&=&2\int_0^1\frac{dy}{y}\int_0^yK(x^2)dx\\ &=&\pi\sum_{n=0}^{\infty}\frac{\beta_n^2}{(4n+1)^2} \end{eqnarray}
と計算できます.

13

\begin{align} \sum_{n=0}^{\infty}(4n+1)\beta_n\frac{(\frac13)_n}{(\frac76)_n}C_n=\frac{\sqrt{3}\Gamma(\frac13)^{12}}{2^72^{\frac13}\pi^5} \end{align}

元となる積分は,
\begin{align} \int_0^1\frac{\kappa(x)^2}{(x(1-x))^{\frac13}}dx=\frac{\sqrt{3}\Gamma(\frac13)^9}{2^{3}2^{\frac13}\pi^5} \end{align}
です.用いるFL展開は,
\begin{eqnarray} \sum_{n=0}^{\infty}(4n+1)\beta_n\frac{(\frac13)_n}{(\frac76)_n}P_{2n}(2x-1)=\frac{\Gamma(\frac13)^3}{4\pi^2}\frac{1}{(x(1-x))^{\frac13}} \end{eqnarray}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}(4n+1)\beta_n\frac{(\frac13)_n}{(\frac76)_n}C_n&=&\int_{0}^1\left(\sum_{n=0}^{\infty}(4n+1)\beta_n\frac{(\frac13)_n}{(\frac76)_n}P_{2n}(2x-1)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+1)C_nP_{2n}(2x-1)\right)dx\\ &=&\int_0^1\frac{\Gamma(\frac13)^3}{4\pi^2}\frac{1}{(x(1-x))^{\frac13}}\cdot\frac{\pi^2}{4}\kappa(x)^2dx\\ &=&\frac{\Gamma(\frac13)^3}{16}\int_0^1\frac{\kappa(x)^2}{(x(1-x))^{\frac13}}dx \end{eqnarray}

14

\begin{align} \frac{\pi^3}{4}\sum_{n=0}^{\infty}(4n+1)\beta_n^4V_{2n}=\sum_{n=0}^{\infty}\left((4n+1)C_nV_{2n}+(4n+3)D_{n}V_{2n+1}\right)=\frac{\Gamma(\frac14)^8}{32\pi^4} \end{align}

元となる積分は,
\begin{align} \int_0^1K(x)K'(x)^2dx=\frac{\Gamma(\frac14)^8}{256\pi^2} \ ,\ \int_0^1K'(x)^3dx=\frac{\Gamma(\frac14)^8}{128\pi^2} \end{align}
です.

\begin{eqnarray} \sum_{n=0}^{\infty}(4n+1)\beta_n^4V_{2n}&=&\int_0^1\left(\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(2x-1)\right)\cdot\left(\sum_{n=0}^{\infty}(2n+1)V_nP_n(2x-1)\right)dx\\ &=&\int_0^1\frac{2}{\pi}\kappa(x)\kappa(1-x)\cdot\frac1\pi\frac{\kappa(x)}{\sqrt{1-x}}dx\\ &=&\frac{2^5}{\pi^5}\int_0^1K(x)K'(x)^2dx\\ \end{eqnarray}
\begin{eqnarray} \sum_{n=0}^{\infty}\left((4n+1)C_nV_{2n}+(4n+3)D_{n}V_{2n+1}\right)&=&\int_0^1\left(\sum_{n=0}^{\infty}\left((4n+1)C_nP_{2n}(2x-1)+(4n+3)D_nP_{2n+1}(2x-1)\right)\right)\cdot\left(\sum_{n=0}^{\infty}(2n+1)V_nP_{n}(2x-1)\right)dx\\ &=&\int_0^1\frac{\pi^2}{4}\kappa(x)^2\cdot\frac{1}{\pi}\frac{\kappa(x)}{\sqrt{1-x}}dx\\ &=&\frac{4}{\pi^2}\int_0^1K'(x)^3dx \end{eqnarray}

15

\begin{align} \sum_{n=0}^{\infty}\left((4n+1)\beta_n^2(2\ln2+H_{2n})C_n+(4n+3)(2n+1)^2\beta_n^2D_n^2\right)\overset{?}{=}\sum_{n=0}^{\infty}\beta_n^4\left(2\ln2+H_{2n}\right) \end{align}

\begin{eqnarray} \sum_{n=0}^{\infty}\left((4n+1)\beta_n^2(2\ln2+H_{2n})C_n+(4n+3)(2n+1)^2\beta_n^2D_n^2\right)&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}\left((4n+1)\beta_n^2\left(2\ln2+H_{2n}\right)P_{2n}(x)+(4n+3)(2n+1)^2\beta_n^2D_nP_{2n+1}(x)\right)\right)\cdot\left(\sum_{n=0}^{\infty}\left((4n+1)C_nP_{2n}(x)+(4n+3)D_nP_{2n+1}(x)\right)\right)dx\\ &=&\frac12\int_{-1}^1\frac2{\pi}\frac{\ln\frac1{1-x}}{\sqrt{1-x^2}}\cdot\frac{\pi^2}{4}\kappa\left(\frac{1+x}{2}\right)^2dx\\ &=&\frac{\pi}{4}\int_0^1\frac{\ln\frac1{2(1-x)}}{\sqrt{x(1-x)}}\kappa(x)^2dx \end{eqnarray}

となりますが,
\begin{align} \int_0^1\frac{\ln\frac1{2(1-x)}}{\sqrt{x(1-x)}}\kappa(x)^2dx\overset{?}{=}\frac{4}{\pi}\sum_{n=0}^{\infty}\beta_n^4(2\ln2+H_{2n}) \end{align}
は$H_{n}$が数値計算に向かないこともあり,正確なものかは判断できませんでした.
\begin{eqnarray} \int_0^1\frac{\ln\frac1{2(1-x)}}{\sqrt{x(1-x)}}\kappa(x)^2dx&=&\int_{-1}^1\frac{\ln\frac{1}{1-x}}{\sqrt{1-x^2}}\kappa\left(\frac{1+x}2\right)^2dx\ \left(x\to\frac{1+x}{2}\right)\\ &=&\int_0^1\frac{\ln\frac{1}{1-x}}{\sqrt{1-x^2}}\left(\kappa\left(\frac{1+x}{2}\right)^2-\kappa\left(\frac{1-x}{2}\right)^2\right)+\frac{\ln\frac1{1-x^2}}{\sqrt{1-x^2}}\kappa\left(\frac{1-x}{2}\right)^2\ dx\\ &=&\frac12\int_0^1\frac{\ln\frac1{1-\sqrt{1-x}}}{\sqrt{x(1-x)}}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)+\frac{\ln\frac{1}{x}}{\sqrt{x(1-x)}}\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2 \ dx\\ &=&\frac12\int_0^1\frac{\ln\frac{x}{1-\sqrt{1-x}}}{\sqrt{x(1-x)}}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)-\frac{\ln x}{\sqrt{x(1-x)}}\left(\kappa\left(\frac{1+\sqrt{1-x}}{2}\right)^2-\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2\right)+\frac{\ln\frac{1}{x}}{\sqrt{x(1-x)}}\kappa\left(\frac{1-\sqrt{1-x}}{2}\right)^2dx\\ &=&\frac12\sum_{n=0}\beta_n\int_0^1(\ln2+H_{2n}-H_n-\ln x)x^{n-\frac12}\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{\ln\frac1x}{\sqrt{x(1-x)}}\sum_{n=0}^{\infty}\beta_n^3x^ndx\\ &=&\pi\sum_{n=0}^{\infty}\beta_n^4\left(\ln2+H_{2n}-H_n-3\frac{\beta_n'}{\beta_n}\right)-\frac{\pi}2\sum_{n=0}^{\infty}\beta_n^3\beta_n'\ \left(\beta_n':=\beta_n\left(\psi\left(n+\frac12\right)-\psi(n+1)\right)\right)\\ &=&2\pi\sum_{n=0}^{\infty}\beta_n^4\left(4\ln2-3H_{2n}+7H_n\right) \end{eqnarray}
と計算できると思います.ほかの方法は思いつきませんでした.

16

\begin{align} \sum_{n=0}^{\infty}(4n+3)(2n+1)^2\beta_n^2D_{n}^2=\frac{\pi^4}{16}\sum_{n=0}^{\infty}(4n+1)\beta_n^6 \end{align}

\begin{eqnarray} \sum_{n=0}^{\infty}(4n+3)(2n+1)^2\beta_n^2D_n^2&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}(-1)^n(4n+3)(2n+1)\beta_nD_nP_{2n+1}(x)\right)^2dx\\ &=&\frac12\int_{-1}^1\left(\mathrm{sign}(x)K'(x)\right)^2dx\\ &=&\frac12\int_{-1}^1\left(K'(x)\right)^2dx\\ &=&\frac12\int_{-1}^1\left(\frac{\pi^2}{4}\sum_{n=0}^{\infty}(-1)^n(4n+1)\beta_n^3P_{2n}(x)\right)^2dx\\ &=&\frac{\pi^4}{16}\sum_{n=0}^{\infty}(4n+1)\beta_n^6 \end{eqnarray}

17

\begin{align} \sum_{n=0}^{\infty}(4n+1)\beta_n^4\sum_{2n\lt m}\frac{(-1)^{m-1}}{m^2}=\frac{7\zeta(3)}{\pi^2} \end{align}

元となる積分は,
\begin{align} \int_0^1\frac{\kappa(x)\kappa(1-x)}{x}\ln\frac1{1-x}dx=\frac{7\zeta(3)}{\pi} \end{align}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}(4n+1)\beta_n^4\sum_{2n\lt m}\frac{(-1)^{m-1}}{m^2}&=&\int_0^1\left(\sum_{n=0}^{\infty}(4n+1)\beta_n^4P_{2n}(2x-1)\right)\cdot\left(\sum_{n=0}^{\infty}(2n+1)P_n(x)\sum_{n\lt m}\frac{(-1)^{m-1}}{m^2}\right)dx\\ &=&\int_0^1\frac2\pi\kappa(x)\kappa(1-x)\cdot\frac12\frac{\ln\frac1x}{1-x}dx\\ &=&\frac1\pi\int_0^1\frac{\kappa(x)\kappa(1-x)}{x}\ln\frac{1}{1-x}dx \end{eqnarray}

18

\begin{align} \sum_{n=0}^{\infty}\frac{4n+1}{\beta_n^2}A_n^2=\frac72\zeta(3) \end{align}

元となる積分は,
\begin{align} \int_0^1\frac{\sin^{-1}x}{x\sqrt{1-x^2}}K(x)dx=\frac{7}{2}\zeta(3) \end{align}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}\frac{4n+1}{\beta_n^2}A_n^2&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}\frac{4n+1}{\beta_n^2}A_nP_{2n}(x)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\right)dx\\ &=&\frac12\int_{-1}^1\frac{\sin^{-1}x}{x\sqrt{1-x^2}}\cdot K(x)dx\\ &=&\int_0^1\frac{\sin^{-1}x}{x\sqrt{1-x^2}}K(x)dx \end{eqnarray}

19

\begin{align} \sum_{n=0}^{\infty}\left((4n+1)E_n^2-(4n+3)F_n^2\right)=\frac{\pi^2\ln2}{4} \end{align}

元となる積分は,
\begin{align} \int_0^1\frac{\tanh^{-1}x\tanh^{-1}\sqrt{1-x^2}}{x(1-x^2)}dx=\frac{\pi^2\ln2}{2} \end{align}
です.
\begin{eqnarray} \sum_{n=0}^{\infty}\left((4n+1)E_n^2-(4n+3)F_n^2\right)&=&\int_{0}^1\left(\sum_{n=0}^{\infty}\left((4n+1)E_nP_{2n}(2x-1)+(4n+3)F_nP_{2n+1}(2x-1)\right)\right)\cdot\left(\sum_{n=0}^{\infty}\left((4n+1)E_nP_{2n}(2x-1)-(4n+3)P_{2n+1}(2x-1)\right)\right)dx\\ &=&\int_0^1\frac{\tanh^{-1}\sqrt{x}}{2\sqrt{x(1-x)}}\cdot\frac{\tanh^{-1}\sqrt{1-x}}{2\sqrt{x(1-x)}}dx\\ &=&\frac12\int_0^1\frac{\tanh^{-1}x\tanh^{-1}\sqrt{1-x^2}}{x(1-x^2)}dx \end{eqnarray}

$\mathrm{CONJECTURE\ 1}$

\begin{align} \sum_{n=0}^{\infty}(4n+1)A_n^2=\sum_{n=0}^{\infty}(4n+3)B_n^2 \end{align}

\begin{eqnarray} \sum_{n=0}^{\infty}(4n+1)A_n^2&=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}(4n+1)A_nP_{2n}(x)\right)^2dx\\ &=&\frac12\int_{-1}^1K(x)^2dx\\ &=&\frac12\int_{-1}^1\left(\mathrm{sign}(x)K(x)\right)^2dx\\ &=&\frac12\int_{-1}^1\left(\sum_{n=0}^{\infty}(4n+3)B_nP_{2n+1}(x)\right)^2dx\\ &=&\sum_{n=0}^{\infty}(4n+3)B_n^2 \end{eqnarray}

$\mathrm{CONJECTURE\ 2}$

\begin{align} \sum_{n=0}^{\infty}(4n+1)C_n^2=5\sum_{n=0}^{\infty}(4n+3)D_n^2 \end{align}

\begin{eqnarray} \frac{\pi^4}{2^4}\int_0^1\kappa(x)^4dx&=&\sum_{n=0}^{\infty}(4n+1)C_n^2+\sum_{n=0}^{\infty}(4n+3)D_n^2\\ \frac{\pi^4}{2^4}\int_0^1\kappa(x)^2\kappa(1-x)^2dx&=&\sum_{n=0}^{\infty}(4n+1)C_n^2-\sum_{n=0}^{\infty}(4n+3)D_n^2\\ \int_0^1\kappa(x)^4dx&=&\frac32\int_0^1\kappa(x)^2\kappa(1-x)^2dx \end{eqnarray}
からわかります.三つ目の式については, 公式書付にある,
\begin{eqnarray} \mathrm{Re}\int_0^1\frac{x^{n-2m}}{(1-t^2x^2)\sqrt{1-x^2}}(K(x)+iK'(x))^ndx&=&\frac{\pi}{2}\frac{t^{2m}}{\sqrt{1-t^2}}K(t)^n \end{eqnarray}
から,
\begin{eqnarray} \int_0^1\frac{x^3-x}{(1-t^2x^2)\sqrt{1-x^2}}(K(x)^3-3K(x)K'(x)^2)dx&=&\frac{\pi}2\frac{1-t^2}{\sqrt{1-t^2}}K(t)^3\\ \int_0^1\frac{x}{(1-t^2x^2)\sqrt{1-x^2}}K(x)dx&=&\frac{\pi}2\frac{K(t)}{\sqrt{1-t^2}} \end{eqnarray}
であり,これを用いれば,
\begin{eqnarray} \int_0^1tK(t)^4dt&=&\frac2{\pi}\int_0^1\frac{t}{\sqrt{1-t^2}}K(t)\cdot\frac{\pi}2\frac{1-t^2}{\sqrt{1-t^2}}K(t)^3dt\\ &=&\frac2\pi\int_0^1\frac{t}{\sqrt{1-t^2}}K(t)\int_0^1\frac{x^3-x}{(1-t^2x^2)\sqrt{1-x^2}}(K(x)^3-3K(x)K'(x)^2)dxdt\\ &=&\frac2\pi\int_0^1\frac{x^3-x}{\sqrt{1-x^2}}(K(x)^3-3K(x)K'(x)^2)\int_0^1\frac{t}{(1-x^2t^2)\sqrt{1-t^2}}K(t)dtdx\\ &=&\int_0^1x(3K(x)^2K'(x)^2-K(x)^4)dx \end{eqnarray}
と示すことができます.

$\mathrm{CONJECTURE\ 3}$

\begin{align} \sum_{n=0}^{\infty}2n\beta_n^4D_{n-1}+\sum_{n=0}^{\infty}(2n+1)\beta_n^4D_n=\frac{\pi^2}{8} \end{align}

\begin{eqnarray} \sum_{n=0}^{\infty}\left((2n+2)\beta_{n+1}^4+(2n+1)\beta_n^4\right)D_n&=&\int_0^1\left(\sum_{n=0}^{\infty}(4n+1)\beta_n^4(2x-1)P_{2n}(2x-1)\right)\cdot\left(\sum_{n=0}^{\infty}(4n+3)D_nP_{2n+1}(2x-1)\right)dx\\ &=&\int_0^1\frac{2}{\pi}(2x-1)\kappa(x)\kappa(1-x)\cdot\frac{\pi^2}{4}\kappa(x)^2dx\\ &=&\frac{\pi}{2}\int_0^1x\left(\kappa(x)^3\kappa(1-x)-\kappa(x)\kappa(1-x)^3\right)dx\\ &=&\frac{\pi^2}{8}\ \end{eqnarray}