Proposition8
\begin{align} \beta_n:=2^{-2n}\binom{2n}{n},A:=\sum_{n\geq0}\beta_n^3=\frac{\pi}{\Gamma(\frac34)^4} \end{align}
\begin{align}
\int_0^1K(x)^2\tanh^{-1}xdx=\frac{\pi^4}{32}\sum_{n=0}^{\infty}\beta_n^4
\end{align}
\begin{align}
\int_0^1K\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)^2\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx=\frac{\pi^4}{16}\sum_{n=0}^{\infty}\beta_n^4-\frac{\pi}{32A}\sum_{n=1}^{\infty}\frac{1}{n^3\beta_n^3}\sum_{k=0}^{n-1}\beta_k^4
\end{align}
\begin{align}
\sum_{k=0}^{n}\frac{\beta_k^2}{n-k+\frac12}=\frac{1}{(n+\frac12)^2\beta_n^2}\sum_{k=0}^{n}\left(2k+\frac12\right)\beta_k^4
\end{align}
より,
\begin{align}
K(x)\tanh^{-1}x=\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{x^{2n-1}}{(2n)^2\beta_n^2}\sum_{k=0}^{n-1}\left(4k+1\right)\beta_k^4
\end{align}
なので,
\begin{eqnarray}
\int_0^1K(x)^2\tanh^{-1}xdx&=&\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{1}{(2n)^2\beta_n^2}\sum_{k=0}^{n-1}(4k+1)\beta_k^4\int_0^1x^{2n-1}K(x)dx\\
&=&\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{1}{(2n)^4\beta_n^4}\left(\sum_{k=0}^{n-1}\beta_k^2\right)\sum_{k=0}^{n-1}(4k+1)\beta_k^4
\end{eqnarray}
となる.
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n^2\beta_n^2}\left(\frac1{n+m}+\frac{1}{n-m-\frac12}\right)\sum_{k=0}^{n-1}\left(2k+\frac12\right)\beta_k^4=\pi^2\beta_m^2
\end{align}
と,
\begin{align}
\sum_{n=0}^{\infty}\frac{\beta_n^2}{n+m}=\sum_{n=0}^{\infty}\frac{\beta_n^2}{m-n-\frac12}=\frac{1}{\pi m^2\beta_m^2}\sum_{k=0}^{m-1}\beta_k^2
\end{align}
より,
\begin{eqnarray}
\sum_{n\gt0,m\geq0}\frac{1}{n^2\beta_n^2}\beta_m^2\left(\frac{1}{n+m}+\frac{1}{n-m-\frac12}\right)\sum_{k=0}^{n-1}\left(2k+\frac12\right)\beta_k^4&=&\\
\sum_{n=0}^{\infty}\frac{1}{n^2\beta_n^2}\left(\sum_{k=0}^{n-1}\left(2k+\frac12\right)\beta_k^4\right)\sum_{m=0}^{\infty}\beta_m^2\left(\frac{1}{m+n}+\frac{1}{n-m-\frac12}\right)&=&\sum_{m=0}^{\infty}\beta_m^2\sum_{n=1}^{\infty}\frac{1}{n^2\beta_n^2}\left(\frac1{n+m}+\frac{1}{n-m-\frac12}\right)\sum_{k=0}^{n-1}\left(2k+\frac12\right)\beta_k^4\\
2\sum_{n=0}^{\infty}\frac{1}{\pi n^4\beta_n^4}\left(\sum_{k=0}^{n-1}\beta_k^2\right)\sum_{k=0}^{n-1}\left(2k+\frac12\right)\beta_k^4&=&\pi^2\sum_{m=0}^{\infty}\beta_m^4
\end{eqnarray}
なので,
\begin{eqnarray}
\int_0^1K(x)^2\tanh^{-1}xdx&=&\frac{\pi}{16}\sum_{n=1}^{\infty}\frac{1}{n^4\beta_n^4}\left(\sum_{k=0}^{n-1}\beta_k^2\right)\sum_{k=0}^{n-1}\left(2k+\frac12\right)\beta_k^4\\
&=&\frac{\pi^4}{32}\sum_{n=0}^{\infty}\beta_n^4
\end{eqnarray}
です.
\begin{eqnarray}
\sum_{n=0}^{\infty}\frac{\beta_n^3}{n+m}&=&\frac{1}{m^3\beta_m^3}\sum_{k=0}^{m-1}\left(\frac{A}{2}\left(k+\frac14\right)\beta_k^3-\frac1{2\pi^2A}\beta_k^3\right)\\
\sum_{n=0}^{\infty}\frac{\beta_n^3}{m-n-\frac{1}{2}}&=&\frac{1}{m^3\beta_m^3}\sum_{k=0}^{m-1}\left(\frac{A}{2}\left(k+\frac14\right)\beta_k^3+\frac{1}{2\pi^2A}\beta_k^3\right)
\end{eqnarray}
と,
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n\beta_n}\frac{1}{n-m-\frac12}\sum_{k=0}^{n-1}\beta_k^2=\pi^2\beta_m
\end{align}
より,
\begin{align}
\int_0^1K\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)^2\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx
\end{align}
\begin{eqnarray}
&=&\frac{\pi^2}{8}\int_0^1\sum_{n=0}^{\infty}\beta_n^3x^{2n}\sum_{m=1}^{\infty}\frac{x^{2m-1}}{m\beta_m}\sum_{k=0}^{m-1}\beta_k^2dx\\
&=&\frac{\pi^2}{16}\sum_{n\geq0,m\gt0}\frac{\beta_n^3}{n+m}\frac{1}{m\beta_m}\sum_{k=0}^{m-1}\beta_k^2\\
&=&\frac{\pi^2}{16}\left(\sum_{n\geq0,m\gt0}\frac{\beta_n^3}{m-n-\frac12}
\frac{1}{m\beta_m}\sum_{k=0}^{m-1}\beta_k^2-\frac{1}{\pi^2A}\sum_{m=1}^{\infty}\frac1{m^4\beta_m^4}\left(\sum_{k=0}^{m-1}\beta_k^2\right)\sum_{k=0}^{m-1}\beta_k^3\right)\\
&=&\frac{\pi^4}{16}\sum_{n=0}^{\infty}\beta_n^4-\frac{1}{16A}\sum_{n=1}^{\infty}\frac{1}{n^4\beta_n^4}\left(\sum_{k=0}^{n-1}\beta_k^2\right)\sum_{k=0}^{n-1}\beta_k^3
\end{eqnarray}
ここで,
\begin{align}
\sum_{n=0}^{\infty}\frac{\beta_n^2}{n+m}=\sum_{n=0}^{\infty}\frac{\beta_n^2}{m-n-\frac12}=\frac{1}{\pi m^2\beta_m^2}\sum_{k=0}^{m-1}\beta_k^2
\end{align}
と,
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n^2\beta_n^2}\left(\frac{1}{n+m}+\frac{1}{n-m-\frac12}\right)\sum_{k=0}^{n-1}\beta_k^3=\beta_m^2\sum_{n\gt m}\frac{1}{n^3\beta_n^3}
\end{align}
より,
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac1{n^4\beta_n^4}\left(\sum_{k=0}^{n-1}\beta_k^2\right)\sum_{k=0}^{n-1}\beta_k^3&=&\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{1}{n^2\beta_n^2}\sum_{k=0}^{n-1}\beta_n^3\sum_{m=0}^{\infty}\beta_m^2\left(\frac{1}{n+m}+\frac{1}{n-m-\frac12}\right)\\
&=&\frac{\pi}{2}\sum_{m=0}^{\infty}\beta_m^2\sum_{n=1}^{\infty}\frac{1}{n^2\beta_n^2}\left(\frac{1}{n+m}+\frac{1}{n-m-\frac12}\right)\sum_{k=0}^{n-1}\beta_k^3\\
&=&\frac{\pi}{2}\sum_{m=0}^{\infty}\beta_m^4\sum_{n\gt m}\frac{1}{n^3\beta_n^3}\\
&=&\frac{\pi}{2}\sum_{n=1}^{\infty}\frac{1}{n^3\beta_n^3}\sum_{k=0}^{n-1}\beta_k^4
\end{eqnarray}
以上から,
\begin{align}
\int_0^1K\left(\sqrt{\frac{1-\sqrt{1-x^2}}{2}}\right)^2\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx=\frac{\pi^4}{16}\sum_{n=0}^{\infty}\beta_n^4-\frac{\pi}{32A}\sum_{n=1}^{\infty}\frac1{n^3\beta_n^3}\sum_{k=0}^{n-1}\beta_k^4
\end{align}
二つ目の式について,
便利
さんが次の変形を出してくれました.
\begin{align}
\int_0^1x^{2n}K(x)dx=\beta_n\int_0^1x^{2n}\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx=\beta_n^2\left(2\beta(2)+\sum_{m=1}^{n}\frac{1}{(2k)^2\beta_k^2}\right)
\end{align}
より,
\begin{eqnarray}
(2つ目の式)&=&\frac{\pi^2}{4}\sum_{n=0}^{\infty}\beta_n^3\int_0^1x^{2n}\frac{\tanh^{-1}x}{\sqrt{1-x^2}}dx\\
&=&\frac{\pi^2}{4}\sum_{n=0}^{\infty}\beta_n^2\int_0^1x^{2n}K(x)dx\\
&=&\frac{\pi}{2}\int_0^1K(x)^2dx
\end{eqnarray}
これについて
黑イト
さんが題意の式を示してくれました.
\begin{align}
\int_0^1x^{n-1/2}\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=\beta_n^3\left(\pi-\frac{1}{\pi^2A}\sum_{n\gt k}\frac{1}{k^3\beta_k^3}\right)
\end{align}
より,
\begin{align}
\int_0^1\frac{dx}{\sqrt{x(1-x)}}\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)=\pi\sum_{n=0}^{\infty}\beta_n^4-\frac{1}{\pi^2A}\sum_{n=0}^{\infty}\beta_n^4\sum_{n\gt k}\frac{1}{k^3\beta_k^3}
\end{align}
\begin{eqnarray}
(LHS)&=&2\int_0^1\kappa\left(\frac{1+x}{2}\right)\kappa\left(\frac{1-x}{2}\right)\frac{dx}{\sqrt{1-x^2}}-\int_0^1\frac{dx}{\sqrt{x(1-x)}}\sum_{n=0}^{\infty}\beta_n^3x^n\\
&=&\int_{-1}^{1}\kappa\left(\frac{1+x}{2}\right)\kappa\left(\frac{1-x}{2}\right)\frac{dx}{\sqrt{1-x^2}}-\pi\sum_{n=0}^{\infty}\beta_n^4\\
&=&\int_0^1\kappa(x)\kappa(1-x)\frac{dx}{\sqrt{x(1-x)}}-\pi\sum_{n=0}^{\infty}\beta_n^4
\end{eqnarray}
ここで,
\begin{eqnarray}
\kappa(x)\kappa(1-x)&=&\frac{\pi}{2}\sum_{n\geq0}(4n+1)\beta_n^4P_{2n}(2x-1)\\
\frac1{\sqrt{x(1-x)}}&=&\pi\sum_{n\geq0}(4n+1)\beta_n^2P_{2n}(2x-1)
\end{eqnarray}
より,
\begin{align}
\int_0^1\frac{\kappa(x)\kappa(1-x)}{\sqrt{x(1-x)}}dx=\frac{\pi^2}{2}\sum_{n\geq0}(4n+1)\beta_n^6
\end{align}
となる.よって,
\begin{eqnarray}
\frac{\pi^2}{2}\sum_{n\geq0}(4n+1)\beta_n^6=2\pi\sum_{n\geq0}\beta_n^4-\frac{1}{\pi^2A}\sum_{n\gt k\geq0}\frac{1}{n^3\beta_n^3}\beta_k^4
\end{eqnarray}
であり,
\begin{align}
\int_0^1K(x)^2dx=\frac{\pi^4}{32}\sum_{n\geq0}(4n+1)\beta_n^6
\end{align}
なので題意が示される.
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