今回の等式

今回証明する等式はコチラ

$$\sum_{n=1}^\infty \frac{2^{2n}}{n^2\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}}{2^{2m}}x^{2m} =4\sum_{n=0}^\infty \frac{\left(1-x^2\right)^n}{(2n+1)^2}+\frac{2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,\ln\frac{1+\sqrt{1-x^2}}{x}$$

準備

$$\begin{aligned} \int_0^\frac{\pi}{2}\frac{t}{1-x\sin t}\,dt &=\int_0^1 \frac{2\tan^{-1}u}{1-x\cdot\frac{2u}{1+u^2}}\frac{2du}{1+u^2}\\ &=4\int_0^1 \frac{\tan^{-1}u}{1-2ux+u^2}\,du\\ &=4\left(\left[\frac{\tan^{-1}u}{\sqrt{1-x^2}}\tan^{-1}\frac{u-x}{\sqrt{1-x^2}} \right]_0^1-\frac{1}{\sqrt{1-x^2}}\int_0^1 \frac{1}{1+u^2}\tan^{-1}\frac{u-x}{\sqrt{1-x^2}}du \right)\\ &=\frac{4}{\sqrt{1-x^2}}\left(\frac{\pi}{4}\tan^{-1}\frac{1-x}{\sqrt{1-x^2}}-\int_0^\frac{\pi}{4}\tan^{-1}\frac{\tan\theta-x}{\sqrt{1-x^2}}\,d\theta \right) \end{aligned}$$

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$$\begin{aligned} \int_0^\frac{\pi}{2} \tan^{-1}(x\tan \theta)\,d\theta &=\int_0^\frac{\pi}{2}\int_0^x \frac{\tan\theta}{1+t^2\tan^2\theta}\,dt\,d\theta\\ &=\int_0^x\int_0^\frac{\pi}{2} \frac{\tan\theta}{1+t^2\tan^2\theta}\,d\theta\,dt\\ &=\int_0^x\int_0^\infty \frac{1}{(1+t^2s)(1+s)}\,ds\,dt\\ &=\int_0^x \frac{1}{1-t^2}\ln\frac{1}{t}\,dt\\ &=\sum_{n=0}^\infty \int_0^x t^{2n}\ln\frac{1}{t}\,dt\\ &=\sum_{n=0}^\infty \left(\left[\frac{t^{2n+1}}{2n+1}\ln\frac{1}{t} \right]_0^x+\frac{1}{2n+1}\int_0^x t^{2n}dt \right)\\ &=\ln\frac{1}{x}\,\tanh^{-1}x+\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2} \end{aligned}$$

証明

前回から着想を得ました。

$$\int_0^1 \frac{\sin^{-1}t}{(1-x^2t^2)\sqrt{1-t^2}}\,dt$$

を２通りに変形していきます。

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$$\begin{aligned} \int_0^1 \frac{\sin^{-1}t}{(1-x^2t^2)\sqrt{1-t^2}}\,dt &=\sum_{n=0}^\infty x^{2n}\int_0^1\frac{t^{2n}\sin^{-1}t}{\sqrt{1-t^2}}\,dt\\ &=\sum_{n=0}^\infty x^{2n}\int_0^\frac{\pi}{2} \theta\sin^{2n}\theta\,d\theta\\ &=\sum_{n=0}^\infty x^{2n}\frac{\binom{2n}{n}}{2^{2n+2}}\left(\frac{\pi^2}{2}+\sum_{m=1}^n\frac{2^{2m}}{m^2\binom{2m}{m}} \right)\\ &=\frac{\pi^2}{8\sqrt{1-x^2}}+\frac{1}{4}\sum_{n=1}^\infty \frac{\binom{2n}{n}}{2^{2n}}x^{2n}\sum_{m=1}^n\frac{2^{2m}}{m^2\binom{2m}{m}} \end{aligned}$$

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$$　　　\begin{aligned} \int_0^1 \frac{\sin^{-1}t}{(1-x^2t^2)\sqrt{1-t^2}}\,dt &=\int_0^\frac{\pi}{2} \frac{\theta}{1-x^2\sin^2\theta}\,d\theta\\ &=\frac{1}{2}\int_0^\frac{\pi}{2} \left(\frac{\theta}{1-x\sin\theta}+\frac{\theta}{1+x\sin\theta} \right)d\theta\\ &=\frac{2}{\sqrt{1-x^2}}\left(\frac{\pi}{4}\tan^{-1}\frac{1-x}{\sqrt{1-x^2}}-\int_0^\frac{\pi}{4}\tan^{-1}\frac{\tan\theta-x}{\sqrt{1-x^2}}\,d\theta \right) +\frac{2}{\sqrt{1-x^2}}\left(\frac{\pi}{4}\tan^{-1}\frac{1+x}{\sqrt{1-x^2}}-\int_0^\frac{\pi}{4}\tan^{-1}\frac{\tan\theta+x}{\sqrt{1-x^2}}\,d\theta \right)\\ &=\frac{\pi}{2\sqrt{1-x^2}}\left(\tan^{-1}\sqrt{\frac{1-x}{1+x}}+\tan^{-1}\sqrt{\frac{1+x}{1-x}} \right) -\frac{2}{\sqrt{1-x^2}}\int_0^\frac{\pi}{4}\left(\tan^{-1}\frac{\tan\theta-x}{\sqrt{1-x^2}}+\frac{\tan\theta+x}{\sqrt{1-x^2}} \right)d\theta\\ &=\frac{\pi^2}{4\sqrt{1-x^2}}-\frac{2}{\sqrt{1-x^2}}\int_0^\frac{\pi}{4}\tan^{-1}\frac{\frac{\tan\theta-x}{\sqrt{1-x^2}}+\frac{\tan\theta+x}{\sqrt{1-x^2}}}{1-\frac{\tan\theta-x}{\sqrt{1-x^2}}\frac{\tan\theta+x}{\sqrt{1-x^2}}}\,d\theta\\ &=\frac{\pi^2}{4\sqrt{1-x^2}} -\frac{2}{\sqrt{1-x^2}}\int_0^\frac{\pi}{4}\tan^{-1}\frac{2\sqrt{1-x^2}\tan\theta}{1-x^2-(\tan^2\theta-x^2)}\,d\theta\\ &=\frac{\pi^2}{4\sqrt{1-x^2}} -\frac{2}{\sqrt{1-x^2}}\int_0^\frac{\pi}{4}\tan^{-1}\left(\sqrt{1-x^2}\tan2\theta\right)d\theta\\ &=\frac{\pi^2}{4\sqrt{1-x^2}} -\frac{1}{\sqrt{1-x^2}}\int_0^\frac{\pi}{2}\tan^{-1}\left(\sqrt{1-x^2}\tan\theta\right)d\theta\\ &=\frac{\pi^2}{4\sqrt{1-x^2}} -\frac{1}{\sqrt{1-x^2}}\left(\ln\frac{1}{\sqrt{1-x^2}}\,\tanh^{-1}\sqrt{1-x^2}+\sqrt{1-x^2}\sum_{n=0}^\infty\frac{(1-x^2)^n}{(2n+1)^2} \right)\\ &=\frac{\pi^2}{4\sqrt{1-x^2}} -\frac{1}{4\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,\ln\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}-\sum_{n=0}^\infty\frac{(1-x^2)^n}{(2n+1)^2}\\ &=\frac{\pi^2}{4\sqrt{1-x^2}} -\frac{1}{2\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,\ln\frac{1+\sqrt{1-x^2}}{x}-\sum_{n=0}^\infty\frac{(1-x^2)^n}{(2n+1)^2}\\ \end{aligned}$$

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よって

$$\frac{\pi^2}{8\sqrt{1-x^2}}+\frac{1}{4}\sum_{n=1}^\infty \frac{\binom{2n}{n}}{2^{2n}}x^{2n}\sum_{m=1}^n\frac{2^{2m}}{m^2\binom{2m}{m}} =\frac{\pi^2}{4\sqrt{1-x^2}} -\frac{1}{2\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,\ln\frac{1+\sqrt{1-x^2}}{x}-\sum_{n=0}^\infty\frac{(1-x^2)^n}{(2n+1)^2}$$

が得られます。
整理すると

$$\frac{\pi^2}{2\sqrt{1-x^2}}-\sum_{n=1}^\infty \frac{\binom{2n}{n}}{2^{2n}}x^{2n}\sum_{m=1}^n\frac{2^{2m}}{m^2\binom{2m}{m}} =4\sum_{n=0}^\infty\frac{(1-x^2)^n}{(2n+1)^2}+\frac{2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,\ln\frac{1+\sqrt{1-x^2}}{x}$$

いま

$$\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}x^{2n}=\frac{1}{\sqrt{1-x^2}}，\sum_{m=1}^\infty \frac{2^{2m}}{m^2\binom{2m}{m}}=\frac{\pi^2}{2}$$

であることに注意して，

$$\sum_{n=1}^\infty \frac{2^{2n}}{n^2\binom{2n}{n}}\sum_{m=0}^{n-1}\frac{\binom{2m}{m}}{2^{2m}}x^{2m} =4\sum_{n=0}^\infty\frac{{(1-x^2)}^n}{(2n+1)^2}+\frac{2}{\sqrt{1-x^2}}\ln\frac{1}{1-x^2}\,\ln\frac{1+\sqrt{1-x^2}}{x}$$

となります。

ありがとうございました。