arcsin絡みの級数

259

はじめに

どうも、色数です。
今回は$\arcsin x$や反復ベータ積分などを用いて級数を解いていきます。
1番大変な積分は数楽さんに解いていただきました。ありがとうございます。
証明などを共有する意図で記事を書いているわけではないのでおおまかな流れ以外は割愛します。

内容

$\arcsin^k x$の級数展開

$\displaystyle \arcsin^{2r-1} x=\frac{(2r-1)!}{2^{2r-1}}\sum_{0< n_1<\cdots< n_{r}}\frac{2^{2n_{r}}}{n_1^2\cdots n_{r-1}^2n_{r}\binom{2n_{r}}{n_{r}}}x^{2n_{r}-1}\sqrt{1-x^2}$

$\displaystyle \arcsin^{2r} x=\frac{(2r)!}{2^{2r}}\sum_{0< n_1<\cdots< n_r}\frac{2^{2n_{r}}}{n_1^2\cdots n_{r}^2\binom{2n_{r}}{n_{r}}}x^{2n_{r}}$

証明のスケッチ

\begin{align} \arcsin^3x&=6\int_0^x\frac{dt_3}{\sqrt{1-t_3^2}}\int_0^{t_3}\frac{dt_2}{\sqrt{1-t_2^2}}\int_0^{t_2}\frac{dt_1}{\sqrt{1-t_1}}\\&=3\int_0^x\frac{\arcsin^2 t_3}{\sqrt{1-t_3^2}}dt_3\\&=\frac{3}{2}\sum_{n=1}^\infty\frac{2^{2n}}{n^2\binom{2n}{n}}\int_0^x\frac{t_3^{2n}}{\sqrt{1-t_3^2}}dt_3\\&=\frac{3}{2}\sum_{n=1}^\infty\frac{2^{2n}}{n^2\binom{2n}{n}}\frac{\binom{2n}{n}}{2^{2n}}\sum_{n< m}\frac{2^{2m}}{2m\binom{2m}{m}}x^{2m-1}\sqrt{1-x^2}\\&=\frac{3}{4}\sum_{0< n_1< n_2}\frac{2^{2n_2}}{n_1^2n_2\binom{2n_2}{n_2}}x^{2n_2-1}\sqrt{1-x^2} \end{align}

\begin{align} \arcsin^4x&=4\int_0^x\frac{\arcsin^3t}{\sqrt{1-t^2}}dt\\&=3\sum_{0< n_1< n_2}\frac{2^{2n_2}}{n_1^2n_2\binom{2n_2}{n_2}}\int_0^xt^{2n_2-1}dt_2\\&=3\sum_{0< n_1< n_2}\frac{2^{2n_2}}{2n_1^2n_2^2\binom{2n_2}{n_2}}x^{2n_2} \end{align}
\begin{align} \arcsin^5x&=5\int_0^x\frac{\arcsin^4t}{\sqrt{1-t^2}}dt\\&=\frac{15}{2}\sum_{0< n_1< n_2}\frac{2^{2n_2}}{n_1^2n_2^2\binom{2n_2}{n_2}}\int_0^x\frac{t^{2n_2}}{\sqrt{1-t^2}}dt\\&=\frac{15}{4}\sum_{0< n_1< n_2< n_3}\frac{2^{2n_3}}{n_1^2n_2^2n_3\binom{2n_3}{n_3}}x^{2n_3-1}\sqrt{1-x^2} \end{align}

$\displaystyle \sum_{0< n_1< n_2}\frac{1}{n_1^2n_2\binom{2n_2}{n_2}}=\frac{\pi^3}{162\sqrt{3}}$
$\displaystyle \sum_{0< n_1< n_2< n_3}\frac{1}{n_1^2n_2^2n_3\binom{2n_3}{n_3}}=\frac{\pi^5}{29160\sqrt{3}}$
$\vdots$

一般に$R(\{2\}^r,1)$が求まります。

$1/\beta,\beta$の級数間の関係式

$\displaystyle \sum_{0< n_1<\cdots< n_{r}}\frac{2^{2n_{r}}}{n_1^2\cdots n_{r-1}^2n_{r}\binom{2n_{r}}{n_{r}}}x^{2n_{r}-1}\sqrt{1-x^2}=\sum_{0\le n_1<\cdots< n_r}\frac{\binom{2n_r}{n_r}}{2^{2n_r}(n_1+\frac{1}{2})^2\cdots(n_{r-1}+\frac{1}{2})^2(n_r+\frac{1}{2})}x^{2n_r+1}$

メモ

\begin{align} \int_0^1t^{2r-1}\arcsin^2 tdt&=\left[\frac{t^{2r}}{2r}\arcsin^2t\right]_0^1-\int_0^1\frac{t^{2r}\arcsin t}{r\sqrt{1-t^2}}dt\\&=\frac{\pi^2}{8r}-\left[\frac{\binom{2r}{r}}{r2^{2r}}\sum_{r< m}\frac{2^{2m}t^{2m-1}\sqrt{1-t^2}}{2m\binom{2m}{m}}\arcsin t\right]_0^1+\frac{\binom{2r}{r}}{r2^{2r}}\sum_{r< m}\frac{2^{2m}}{2m\binom{2m}{m}}\int_0^1t^{2m-1}dt\\&=\frac{\pi^2}{8r}-\frac{\binom{2r}{r}}{r2^{2r}}\sum_{r< m}\frac{2^{2m}}{4m^2\binom{2m}{m}} \end{align}
\begin{align} \int_0^1\arcsin^2t\textup{arctanh}\;tdt&=\sum_{n=1}^\infty\frac{1}{2n-1}\int_0^1t^{2n-1}\arcsin^2tdt\\&=\frac{\pi^2}{8}\sum_{n=1}^\infty\frac{1}{(2n-1)n}+\frac{1}{4}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}n_1^2n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}}-\frac{1}{2}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}(2n_1-1)n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}}\\&=\frac{\pi^2}{4}\ln2+\frac{1}{4}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}n_1^2n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}}-\frac{1}{2}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}(2n_1-1)n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}} \end{align}
$\displaystyle \frac{1}{4}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}n_1^2n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}}-\frac{1}{2}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}(2n_1-1)n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}}=\frac{7}{8}\zeta(3)-2\ln2+\frac{1}{4}\pi^2\ln2-\frac{\pi^2}{4}$
\begin{align} \int_0^1\arcsin^2t\tanh^{-1}tdt&=\sum_{n=1}^\infty\frac{1}{2n-1}\int_0^1t^{2n-1}\arcsin^2tdt\\&=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2n-1}\sum_{m=1}^\infty\frac{2^{2m}}{m^2\binom{2m}{m}}\int_0^1t^{2m+2n-1}dt\\&=\frac{1}{2}\sum_{0< n,m}\frac{2^{2m}}{(2n-1)(2n+2m)m^2\binom{2m}{m}} \end{align}

\begin{align} \frac{7}{8}\zeta(3)-2\ln2+\frac{1}{2}\pi^2\ln2-\frac{\pi^2}{4}&=\frac{1}{2}\sum_{0< n,m}\frac{2^{2m}}{(2n-1)(2n+2m)m^2\binom{2m}{m}}\\&=\frac{\pi^2}{4}\ln2+\frac{1}{4}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}n_1^2n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}}-\frac{1}{2}\sum_{0< n_1< n_2}\frac{1}{2^{2n_1-2n_2}(2n_1-1)n_2^2}\frac{\binom{2n_1}{n_1}}{\binom{2n_2}{n_2}} \end{align}

\begin{align} \int_0^1t^{2r-1}\arcsin^3 tdt&=\left[\frac{t^{2r}}{2r}\arcsin^3 t\right]_0^1-3\int_0^1\frac{t^{2r}}{2r}\frac{\arcsin^2t}{\sqrt{1-t^2}}\\&=\frac{\pi^3}{16r}-\frac{3}{2r}\left[\sum_{r< m}\frac{2^{2m}t^{2m-1}\sqrt{1-t^2}}{2m\binom{2m}{m}}\arcsin^2t\right]_0^1+\frac{3}{4r}\sum_{r< m}\int_0^1\frac{2^{2m}t^{2m-1}\sqrt{1-t^2}}{m\binom{2m}{m}}\arcsin^2tdt\\&=\frac{\pi^3}{16r}+\frac{3}{4r}\sum_{r< m}\frac{2^{2m}}{m\binom{2m}{m}}\left(\left[\left(\frac{t^{2m}}{2m}\sqrt{1-t^2}+\frac{1}{2m}\frac{2^{2m+2}}{2(m+1)\binom{2m+2}{m+1}}\sum_{m+1\le n}\frac{2^{2n}t^{2n-1}\sqrt{1-t^2}}{2n\binom{2n}{n}}\right)\arcsin^2t\right]_0^1-2\int_0^1\left(\frac{t^{2m}}{2m}\sqrt{1-t^2}+\frac{1}{2m}\frac{2^{2m+2}}{2(m+1)\binom{2m+2}{m+1}}\sum_{m+1\le n}\frac{2^{2n}t^{2n-1}\sqrt{1-t^2}}{2n\binom{2n}{n}}\right)\frac{\arcsin t}{\sqrt{1-t^2}}dt\right)\\&= \frac{\pi^3}{16r}-\frac{3}{4r}\sum_{r< m}\frac{2^{2m}}{m^2\binom{2m}{m}}\left(\int_0^1t^{2m}\arcsin tdt+\frac{2^{2m+2}}{2(m+1)\binom{2m+2}{m+1}}\sum_{m+1\le n}\frac{2^{2n}}{2n\binom{2n}{n}}\int_0^1t^{2n-1}\arcsin tdt\right)\\&=\frac{\pi^3}{16r}-\frac{3}{4r}\sum_{r< m}\frac{2^{2m}}{m^2\binom{2m}{m}}\left(\frac{\pi}{4m+2}-\frac{1}{2m+1}\frac{2^{2m+2}}{2(m+1)\binom{2m+2}{m+1}}+\frac{2^{2m+2}}{2(m+1)\binom{2m+2}{m+1}}\sum_{m< n}\frac{2^{2n}}{2n\binom{2n}{n}}\frac{\pi}{4}\left(\frac{1}{n}-\frac{\binom{2n}{n}}{n2^{2n}}\right)\right) \end{align}

\begin{align} \frac{3\pi}{8}\sum_{m=1}^\infty\frac{1}{2m-1}\sum_{0< n_1< n_2}\frac{2^{2n_2}}{n_1^2n_2\binom{2n_2}{n_2}}\frac{\binom{2n_2+m}{n_2+m}}{2^{2n_2+2m}}&= \sum_{0< n_1< n_2< n_3}\frac{1}{2n_1-1}\left(\frac{\pi^3}{16n_1}-\frac{3}{4n_1}\frac{2^{2n_2}}{n_2^2\binom{2n_2}{n_2}}\left(\frac{\pi}{4n_2+2}-\frac{1}{2n_2+1}\frac{2^{2n_2+2}}{2(n_2+1)\binom{2n_2+2}{n_2+1}}+\frac{2^{2n_2+2}}{2(n_2+1)\binom{2n_2+2}{n_2+1}}\frac{2^{2n_3}}{2n_3\binom{2n_3}{n_3}}\frac{\pi}{4}\left(\frac{1}{n_3}-\frac{\binom{2n_3}{n_3}}{n_32^{2n_3}}\right)\right)\right) \end{align}