はじめに

　私は無限級数と実関数積分をささやかに嗜む浅学素人です。どちらかといえば級数のほうが直感的で好ましく、積分は級数を考えるための道具という認識です。これから自分で得た級数の結果を記事にしていきたいと思っています。結果自体は他のサービスで公開していますが、ここではその結果に至るまでの過程も公開します。
　現在までに得た結果の数は少なくないので、それぞれの結果ごとに記事を書きます。
　それでは、よろしくお願いします。

今回の等式

　今回証明する等式はこちら

$$\begin{aligned} &\frac{\pi}{8}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2}\\ =&\sum_{0\le n}\frac{2^{6n}}{(2n+1)^3\binom{2n}{n}\binom{4n+2}{2n+1}}-\frac{\pi}{32}\sum_{0<n}\frac{2^{2n}\binom{2n}{n}}{n^2\binom{4n}{2n}}\\ =&\frac{\pi^3}{16}-\sqrt{2}\sum_{0\le m\le n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{1}{2^n(2n+1)^2}\\ =&\frac{1}{4}\sum_{0\le m<n}\frac{(-1)^m}{2m+1}\frac{1}{n^2}+\frac{1}{2}\sum_{0<m\le n}\frac{1}{m}\frac{(-1)^{n-1}}{(2n+1)^2}\\ =&\frac{\pi^3}{32}+2\sum_{0\le m<n}\frac{1}{2m+1}\frac{(-1)^n}{(2n+1)^2}\\ =&2\sum_{0\le m<n}\frac{(-1)^m}{2m+1}\frac{1}{(2n+1)^2} \end{aligned}$$

証明

${\bf Proof~1.}$

まず，$\sin^{-1}e^{ix}$の実部と虚部を求めます。$\sin^{-1}e^{ix}=A+iB$とし，

$$\begin{aligned} \sin(\sin^{-1}e^{ix})&=\sin(A+iB)\\ e^{ix}&=\sin A\cosh B+i \cos A\sinh B \end{aligned}$$

実部と虚部を比較し，$A,B$について解くと，

$$\begin{aligned} A&=\cos^{-1}\sqrt{\sin x}\\ B&=\sinh^{-1}\sqrt{\sin x} \end{aligned}$$

すなわち，

$$\begin{aligned} \cos^{-1}\sqrt{\sin x}=\sum_{0\le n}\frac{\binom{2n}{n}\cos(2n+1)x}{2^{2n}(2n+1)}\\ \sinh^{-1}\sqrt{\sin x}=\sum_{0\le n}\frac{\binom{2n}{n}\sin(2n+1)x}{2^{2n}(2n+1)} \end{aligned}$$

となります。
${\rm Parceval's~Identity}$により

$$\begin{aligned} \int_0^\frac{\pi}{2}\left(\cos^{-1}\sqrt{\sin x} \right)^2dx =\int_0^\frac{\pi}{2}\left(\sinh^{-1}\sqrt{\sin x} \right)^2dx =\frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2} \end{aligned}$$

が成り立ちます。真ん中の式を変形していきます。

$$\begin{aligned} &\int_0^\frac{\pi}{2}\left(\sinh^{-1}\sqrt{\sin x} \right)^2dx\\ =&\frac{1}{2}\sum_{0<n}\frac{(-1)^{n-1}2^{2n}}{n^2\binom{2n}{n}}\int_0^\frac{\pi}{2}\sin^n x\,dx\\ =&\frac{1}{2}\sum_{0\le n}\frac{2^{4n+2}}{(2n+1)^2\binom{4n+2}{2n+1}}\int_0^\frac{\pi}{2}\sin^{2n+1}x\,dx -\frac{1}{2}\sum_{0<n}\frac{2^{4n}}{(2n)^2\binom{4n}{2n}}\int_0^\frac{\pi}{2}\sin^{2n}x\,dx\\ =&2\sum_{0\le n}\frac{2^{6n}}{(2n+1)^3\binom{2n}{n}\binom{4n+2}{2n+1}} -\frac{\pi}{16}\sum_{0<n}\frac{2^{2n}\binom{2n}{n}}{n^2\binom{4n}{2n}} \end{aligned}$$

よって

$$\frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2} =2\sum_{0\le n}\frac{2^{6n}}{(2n+1)^3\binom{2n}{n}\binom{4n+2}{2n+1}} -\frac{\pi}{16}\sum_{0<n}\frac{2^{2n}\binom{2n}{n}}{n^2\binom{4n}{2n}}$$

を得ます。
　
${\bf Proof~2.}$

$$\begin{aligned} &\int_0^\frac{\pi}{2}\left(\cos^{-1}\sqrt{\sin x}\right)^2dx\\ =&\int_0^\frac{\pi}{2}\left(\frac{\pi}{2}-\sin^{-1}\sqrt{\sin x} \right)^2dx\\ =&\frac{\pi^3}{8}-\pi\int_0^\frac{\pi}{2}\sin^{-1}\sqrt{\sin x}\,dx+\int_0^\frac{\pi}{2}\left(\sin^{-1}\sqrt{\sin x}\right)^2dx\\ \end{aligned}$$

ここで

$$\begin{aligned} &\int_0^\frac{\pi}{2}\sin^{-1}\sqrt{\sin x}\,dx\\ =&\int_0^\frac{\pi}{2}\sin^{-1}\sin y\,\frac{2\sin y\cos y}{\cos x}\,dy　　　(\sin x=\sin^2 y)\\ =&2\int_0^\frac{\pi}{2}\frac{y\sin y}{\sqrt{1+\sin^2 y}}\,dy \\ =&\sqrt{2}\int_0^\frac{\pi}{2}\frac{y\sin y}{\sqrt{1-\frac{\cos^2 y}{2}}}\,dy \\ =&-2\left[y\sin^{-1}\frac{\sin y}{\sqrt{2}} \right]_0^\frac{\pi}{2}+2\int_0^\frac{\pi}{2}\sin^{-1}\frac{\sin y}{\sqrt2}\,dy \\ =&2\int_0^\frac{\pi}{2}\sin^{-1}\frac{\sin y}{\sqrt2}\,dy\\ =&\sqrt{2}\sum_{0\le n}\frac{\binom{2n}{n}}{2^{3n}(2n+1)}\int_0^\frac{\pi}{2}\sin^{2n+1}y\,dy\\ =&\sqrt2\sum_{0\le n}\frac{1}{2^n(2n+1)^2}\\ \\ &\int_0^\frac{\pi}{2}\left(\sin^{-1}\sqrt{\sin x} \right)^2dx\\ =&4\int_0^\frac{\pi}{2} y\sin^{-1}\frac{\sin y}{\sqrt2}\,dy　　　　\left(\sin x=\sin^2 y,~IBP\right)\\ =&2\sqrt{2}\sum_{0\le n}\frac{\binom{2n}{n}}{2^{3n}(2n+1)}\int_0^\frac{\pi}{2}y\sin^{2n+1}y\,dy\\ =&2\sqrt{2}\sum_{0\le n}\frac{\binom{2n}{n}}{2^{3n}(2n+1)}\frac{2^{2n}}{(2n+1)\binom{2n}{n}}\sum_{n<m}\frac{\binom{2m}{m}}{2^{2m}(2n+1)}　　　　\left(\small IBPにより得る漸化式を解く\right)\\ =&2\sqrt2\sum_{0\le n<m}\frac{1}{2^n(2n+1)^2}\frac{\binom{2m}{m}}{2^{2m}(2m+1)} \end{aligned}$$

よって

$$\begin{aligned} &\frac{\pi^3}{8}-\pi\int_0^\frac{\pi}{2}\sin^{-1}\sqrt{\sin x}\,dx+\int_0^\frac{\pi}{2}\left(\sin^{-1}\sqrt{\sin x}\right)^2dx\\ =&\frac{\pi^3}{8}-\sqrt2\pi\sum_{0\le n}\frac{1}{2^n(2n+1)^2}+2\sqrt2\sum_{0\le n<m}\frac{1}{2^n(2n+1)^2}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\\ =&\frac{\pi^3}{8}-2\sqrt{2}\sum_{0\le n}\frac{1}{2^n(2n+1)^2}\left(\frac{\pi}{2}-\sum_{n<m}\frac{\binom{2m}{m}}{2^{2m}(2m+1)} \right)\\ =&\frac{\pi^3}{8}-2\sqrt{2}\sum_{0\le m\le n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{1}{2^n(2n+1)^2} \end{aligned}$$

すなわち

$$\begin{aligned} \frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2} =\frac{\pi^3}{8}-2\sqrt{2}\sum_{0\le m\le n}\frac{\binom{2m}{m}}{2^{2m}(2m+1)}\frac{1}{2^n(2n+1)^2} \end{aligned}$$

を得ます。

${\bf Proof~3.}$

$$\begin{aligned} &\frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2}\\ =&\frac{1}{2}\sum_{0\le n}\frac{\binom{2n}{n}}{2^{2n}(2n+1)^2}\int_0^\frac{\pi}{2}\sin^{2n}x\,dx\\ =&\frac{1}{2}\int_0^\frac{\pi}{2}\frac{1}{\sin x}\int_0^x \frac{t}{\tan t}\,dt\,dx \\ =&\sum_{0<n}\int_0^\frac{\pi}{2}\frac{1}{\sin x}\int_0^x t\sin2nt \,dt\,dx \\ =&\sum_{0<n}\int_0^\frac{\pi}{2}\frac{1}{\sin x}\frac{\sin 2nx-2nx\cos 2nx}{4n^2}\,dx\\ =&\sum_{0<n}\frac{1}{4n^2}\int_0^\frac{\pi}{2}\left(2\sum_{m=0}^{n-1}\cos(2m+1)x-2nx\cdot 2\sum_{n\le m}\sin(2m+1)x \right)dx\\ =&\sum_{0<n}\frac{1}{2n^2}\left(\sum_{m=0}^{n-1}\frac{(-1)^m}{2m+1}-2n\sum_{n\le m}\frac{(-1)^m}{(2m+1)^2} \right)\\ =&\frac{1}{2}\sum_{0\le m<n}\frac{(-1)^m}{2m+1}\frac{1}{n^2}+\sum_{0<m\le n}\frac{1}{m}\frac{(-1)^{n-1}}{(2n+1)^2} \end{aligned}$$

よって

$$\begin{aligned} \frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2} =\frac{1}{2}\sum_{0\le m<n}\frac{(-1)^m}{2m+1}\frac{1}{n^2}+\sum_{0<m\le n}\frac{1}{m}\frac{(-1)^{n-1}}{(2n+1)^2} \end{aligned}$$

を得ます。

${\bf Proof~4.}$

$$\begin{aligned} &\frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2}\\ =&\frac{1}{2}\int_0^\frac{\pi}{2}\frac{1}{\sin x}\int_0^x \frac{t}{\tan t}\,dt\,dx \\ =&\frac{1}{2}\int_0^\frac{\pi}{2}\frac{x}{\tan x}\int_x^\frac{\pi}{2}\frac{1}{\sin t}\,dt\,dx\\ =&\sum_{0\le n}\int_0^\frac{\pi}{2}\frac{x}{\tan x}\int_x^\frac{\pi}{2}\sin(2n+1)t\,dt\,dx\\ =&\sum_{0\le n}\int_0^\frac{\pi}{2}\frac{x}{\tan x}\frac{\cos(2n+1)x}{2n+1}\,dx\\ =&\sum_{0\le n}\frac{1}{2n+1}\int_0^\frac{\pi}{2}x\left(\sin(2n+1)x+2\sum_{n<m}\sin(2m+1)x \right)dx\\ =&\sum_{0\le n}\frac{1}{2n+1}\left(\frac{(-1)^n}{(2n+1)^2}+2\sum_{n<m}\frac{(-1)^m}{(2m+1)^2} \right)\\ =&\sum_{0\le n}\frac{(-1)^n}{(2n+1)^3}+2\sum_{0\le m<n}\frac{1}{2m+1}\frac{(-1)^n}{(2n+1)^2}\\ =&\frac{\pi^3}{32}+2\sum_{0\le m<n}\frac{1}{2m+1}\frac{(-1)^n}{(2n+1)^2} \end{aligned}$$

よって

$$\begin{aligned} \frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2} =\frac{\pi^3}{32}+2\sum_{0\le m<n}\frac{1}{2m+1}\frac{(-1)^n}{(2n+1)^2} \end{aligned}$$

を得ます。

${\bf Proof~5.}$

$$\begin{aligned} &\int_0^\frac{\pi}{2}\left(\cos^{-1}\sqrt{\sin x} \right)^2dx\\ =&2\int_0^\frac{\pi}{2}\frac{y^2\cos y}{\sqrt{1+\cos^2 y}}\,dy　　　　(\sin x=\cos^2 y)\\ =&2\sum_{0\le n}(-1)^n\frac{\binom{2n}{n}}{2^{2n}}\int_0^\frac{\pi}{2}y^2\cos^{2n+1}y\,dy\\ =&2\sum_{0\le n}(-1)^n\frac{\binom{2n}{n}}{2^{2n}}\frac{2^{2n+1}}{(2n+1)\binom{2n}{n}}\sum_{n<m}\frac{1}{(2m+1)^2}　　　　\left(\small IBPにより得る漸化式を解く\right)\\ =&4\sum_{0\le m<n}\frac{(-1)^m}{2m+1}\frac{1}{(2n+1)^2} \end{aligned}$$

よって

$$\begin{aligned} \frac{\pi}{4}\sum_{0\le n}\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2} =4\sum_{0\le m<n}\frac{(-1)^m}{2m+1}\frac{1}{(2n+1)^2} \end{aligned}$$

を得ます。

ありがとうございました。