級数•積分botの級数を示してみる①

$$\begin{array}{rl}{\displaystyle {\int }_0^1{x}^{2n}(\log x{)}^{2}dx}& ={\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)t}{t}^{2}dt(x\to e^{-t})}\\ & ={\displaystyle \frac{1}{(2n+1{)}^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-u}{u}^{2}du((2n+1)t\to u)}\\ & ={\displaystyle \frac{\mathrm{\Gamma }(3)}{(2n+1{)}^3}}\\ & ={\displaystyle \frac{2}{(2n+1{)}^3}}\end{array}$$

まめけびさんの記事で証明されています。
リンク

$$\begin{array}{r}S={\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(\genfrac{}{}{0ex}{}{2n}{n})}{2^{2n}(2n+1{)}^3}}\end{array}$$
とします。
補題1,2より
$$\begin{array}{rl}S& ={\displaystyle \frac{1}{2}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(\genfrac{}{}{0ex}{}{2n}{n})}{4^n}{\displaystyle {\int }_0^1{x}^{2n}(\log x{)}^{2}dx}}\\ & ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1(\log x{)}^2{\displaystyle \sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2n}{n}){\left(\frac{x^2}{4}\right)}^{n}dx}}\\ & ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1\frac{(\log x{)}^2}{\sqrt{1-x^2}}dx※}\\ & ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\frac{\pi }{2}}(\log \sin \theta {)}^{2}d\theta (x\to \sin \theta )}\\ & ={\displaystyle \frac{{\pi }^3}{48}}+{\displaystyle \frac{\pi \log {\left(2\right)}^2}{4}}\end{array}$$

※の変形では中央二項係数の母関数を用いています。
$$\begin{array}{rl}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2n}{n})x^n}& ={\displaystyle \frac{1}{\sqrt{1-4x}}}\end{array}$$