EllipticKを含む積分　【解けるだけ解いてみた】

\begin{eqnarray} κ(1-x)=π\sum_{0≤n}(-1)^n\left(2n+\frac{1}{2}\right)β_n^3ρ_n(x) \end{eqnarray}
を用いる.
これにより
\begin{eqnarray} \int_0^1 κ(1-x)^2x^{-1/2}dx=\frac{π^2}{2}\sum_{0≤n}(4n+1)β_n^6 \end{eqnarray}
がわかり,
\begin{eqnarray} \int_0^1 κ(1-x)^2x^{-1/2}dx=\frac{8}{π^2}\int_0^1 Κ'(x)^2dx \end{eqnarray}
であるため,命題の右側が成り立つ.
また,Landen Transformを用いれば
\begin{eqnarray} \int_0^1 Κ'(x)^2dx&=&\int_0^1 \frac{x}{\sqrt{1-x^2}}Κ(x)^2dx\\ &=&\int_0^1 \left(\frac{2x}{1+x^2}Κ\left(\frac{2x}{1+x^2}\right)\right)^2\frac{dx}{x}　\left(x→\frac{2x}{1+x^2}\right)\\ &=&4\int_0^1 xΚ(x^2)^2dx \end{eqnarray}
より命題が示された

\begin{eqnarray} \int_0^1 Κ(x)\frac{\ln\frac{1}{1-x^2}}{\sqrt{1-x^2}}dx&=&\frac{π}{2}\sum_{0≤n}β_n^2\int_0^1\frac{\ln\frac{1}{1-x^2}}{\sqrt{1-x^2}}dx\\ &=&\frac{π^2A}{2}\ln2+\frac{π^2}{4}\sum_{0< k≤n}\frac{β_n^3}{k}\\ &=&\frac{π^3}{6}A \end{eqnarray}

\begin{eqnarray} \int_0^1\frac{Κ(x)\ln\frac{1}{x}}{1-x^2}dx&=&\frac{π}{2}\int_0^1\ln\frac{1}{x}\sum_{0≤k≤n}β_k^2x^{2n}dx\\ &=&\frac{π}{2}\sum_{0≤k≤n}\frac{β_k^2}{(2n+1)^2}\\ &=&\frac{π}{2}\int_0^1\sum_{0≤n}β_n^2x^{2n+1}Κ(x)dx\\ &=&\int_0^1xΚ(x)^2dx \end{eqnarray}

\begin{eqnarray} \int_0^1\frac{Κ'(x)\ln\frac{1}{x}}{1-x^2}dx&=&\frac{1}{2}\int_0^1 Κ(x)\frac{\ln\frac{1}{1-x^2}}{x\sqrt{1-x^2}}dx\\ &=&\frac{π}{4}\left(\int_0^1\frac{\ln\frac{1}{1-x^2}}{x\sqrt{1-x^2}}dx+\sum_{0≤k< n}\frac{β_n}{(2k+1)n}\right)\\ &=&\frac{π}{2}\int_0^1\frac{\ln\frac{1}{1-x^2}}{x\sqrt{1-x^2}}dx\\ &=&\frac{π}{4}\sum_{0< n}\frac{1}{n^2β_n} \end{eqnarray}

先ほどの公式に,$f(x)=κ(x)$,$f(x)=κ(x)^2$
を順次いれる.
\begin{eqnarray} \int_0^1x^{n-1/2}κ(1-x)dx=π[x^n]κ(x) \end{eqnarray}
\begin{eqnarray} ⇔\int_0^1\frac{κ(1-x)}{1-tx}\frac{dx}{\sqrt{x}}=πκ(t) \end{eqnarray}
\begin{eqnarray} \int_0^1x^n\Im(κ(x)^2+2iκ(x)κ(1-x)-κ(1-x)^2)dx=π[x^n]κ(x)^2 \end{eqnarray}
\begin{eqnarray} ⇔\int_0^1\frac{κ(x)κ(1-x)}{1-xt}dx=\frac{π}{2}κ(t)^2 \end{eqnarray}
より2つの命題がわかる.

\begin{eqnarray} \int_0^1\frac{Κ(x)}{1+x}dx&=&\int_0^1\frac{2x}{1+x^2}Κ(x^2)dx\\ &=&\int_0^1\frac{2x}{(1+x^2)^2}Κ\left(\frac{2x}{1+x^2}\right)dx\\ &=&\frac{1}{2}\int_0^1\frac{xΚ(x)}{\sqrt{1-x^2}}dx　\left(\frac{2x}{1+x^2}→x\right)\\ &=&\frac{π}{4}\sum_{0≤n}\frac{β_n}{2n+1} \end{eqnarray}