今回の等式

今回証明する等式はコチラ

$$\frac{\pi \alpha}{2}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}(n+\alpha)} =\sum_{n=0}^\infty \frac{1}{2n+1}\frac{{\left(\frac{1}{2}+\alpha \right)}_n}{{(1+\alpha)}_n}$$

$(z)_n$は${\rm Pochhammer~Symbol}$で，

$${(z)}_n=\frac{\Gamma(z+n)}{\Gamma(z)}$$

という定義です。
$\\$

証明

まず

$$\mathcal{I}=\int_0^1 \frac{(tx)^{2\alpha}}{(1-t^2x^2)\sqrt{1-x^2}}\,dx$$

と定義します。

$$\begin{aligned} \mathcal{I} &=\sum_{m=0}^\infty \int_0^1 \frac{(tx)^{2m+2\alpha}}{\sqrt{1-x^2}}\,dx\\ &=\frac{\sqrt\pi}{2}\sum_{m=0}^\infty \frac{\Gamma\left(\frac{1}{2}+m+\alpha \right)}{\Gamma(1+m+\alpha)}t^{2m+2\alpha} \end{aligned}$$

つぎに

$$\mathcal{J}_n=\int_0^t \frac{x^{2n+2\alpha-1}}{\sqrt{1-x^2}}\,dx$$

と定義します。

$$\begin{aligned} \mathcal{J}_{n-1}-\mathcal{J}_n &=\int_0^t x^{2n+2\alpha-3}\sqrt{1-x^2}\,dx\\ &=\left[\frac{x^{2n+2\alpha-2}\sqrt{1-x^2}}{2n+2\alpha-2} \right]_0^t-\frac{1}{2n+2\alpha-2}\int_0^t x^{2n+2\alpha-2}\cdot\frac{-2x}{2\sqrt{1-x^2}}\,dx\\ &=\frac{t^{2n+2\alpha-2}\sqrt{1-t^2}}{2n+2\alpha-2}+\frac{\mathcal{J}_n}{2n+2\alpha-2}\\ \frac{n+\alpha-\frac{1}{2}}{n+\alpha-1}\,\mathcal{J}_n &=-\frac{t^{2n+2\alpha-1}\sqrt{1-t^2}}{2(n+\alpha-1)}+\mathcal{J}_{n-1}\\ \frac{\Gamma(n+\alpha+\frac{1}{2})}{\Gamma\left(n+\alpha\right)}\,\mathcal{J}_n &=-\frac{\Gamma\left(n+\alpha-\frac{1}{2}\right)t^{2n+2\alpha-2}\sqrt{1-t^2}}{2\Gamma\left(n+\alpha\right)} +\frac{\Gamma\left(n+\alpha-\frac{1}{2}\right)}{\Gamma\left(n+\alpha-1\right)}\,\mathcal{J}_{n-1}\\ &=\frac{\Gamma\left(\frac{1}{2}+\alpha\right)}{\Gamma\left(\alpha\right)}\,\mathcal{J}_0 -\sum_{m=1}^n\frac{\Gamma\left(m+\alpha-\frac{1}{2}\right)t^{2m+2\alpha-2}\sqrt{1-t^2}}{2\Gamma\left(m+\alpha\right)}\\ &=\sum_{m=n+1}^\infty\frac{\Gamma\left(m+\alpha-\frac{1}{2}\right)t^{2m+2\alpha-2}\sqrt{1-t^2}}{2\Gamma\left(m+\alpha\right)}\\ &=\sum_{m=0}^\infty \frac{\Gamma\left(m+n+\alpha+\frac{1}{2}\right)t^{2m+2n+2\alpha}\sqrt{1-t^2}}{2\Gamma(1+m+n+\alpha)}\\ \end{aligned}$$

いま

$$\begin{aligned} \mathcal{J}_0&=\frac{\Gamma(\alpha)\sqrt{1-t^2}}{2\Gamma\left(\frac{1}{2}+\alpha \right)} \sum_{m=0}^\infty \frac{\Gamma\left(\frac{1}{2}+m+\alpha\right)}{\Gamma(1+m+\alpha)}t^{2m+2\alpha}\\ \mathcal{I}&=\frac{\sqrt{\pi}}{2}\sum_{m=0}^\infty \frac{\Gamma\left(\frac{1}{2}+m+\alpha \right)}{\Gamma(1+m+\alpha)}t^{2m+2\alpha} \end{aligned}$$

なので

$$\begin{aligned} \mathcal{I} &=\int_0^1 \frac{(tx)^{2\alpha}}{(1-t^2x^2)\sqrt{1-x^2}}\,dx\\ &=\frac{\sqrt{\pi}}{2}\sum_{m=0}^\infty \frac{\Gamma\left(\frac{1}{2}+m+\alpha \right)}{\Gamma(1+m+\alpha)}t^{2m+2\alpha}\\ &=\frac{\sqrt\pi}{2}\frac{2\Gamma\left(\frac{1}{2}+\alpha \right)}{\Gamma(\alpha)\sqrt{1-t^2}}\,\mathcal{J}_0\\ &=\frac{\sqrt\pi\,\Gamma\left(\frac{1}{2}+\alpha \right)}{\Gamma(\alpha)\sqrt{1-t^2}} \int_0^t \frac{x^{2\alpha-1}}{\sqrt{1-x^2}}\,dx\\ &=\frac{\sqrt\pi\,\Gamma\left(\frac{1}{2}+\alpha \right)}{\Gamma(\alpha)\sqrt{1-t^2}} \sum_{m=0}^\infty\frac{\binom{2m}{m}}{2^{2m}}\int_0^t x^{2m+2\alpha-1}\,dx\\ &=\frac{\sqrt\pi\,\Gamma\left(\frac{1}{2}+\alpha \right)}{\Gamma(\alpha)\sqrt{1-t^2}} \sum_{m=0}^\infty\frac{\binom{2m}{m}t^{2m+2\alpha}}{2^{2m}(2m+2\alpha)} \end{aligned}$$

よって

$$\int_0^1 \frac{x^{2\alpha}}{(1-t^2x^2)\sqrt{1-x^2}}\,dx =\frac{\sqrt\pi\,\Gamma\left(\frac{1}{2}+\alpha \right)}{\Gamma(\alpha)\sqrt{1-t^2}} \sum_{m=0}^\infty\frac{\binom{2m}{m}t^{2m}}{2^{2m}(2m+2\alpha)}$$

両辺を$0<t<1$で定積分すると
左辺は

$$\int_0^1 \int_0^1 \frac{x^{2\alpha}}{(1-t^2x^2)\sqrt{1-x^2}}\,dx\,dt=\int_0^1 \frac{x^{2\alpha-1}\tanh^{-1}x}{\sqrt{1-x^2}}\,dx$$

右辺は

$$\frac{\sqrt\pi\,\Gamma\left(\frac{1}{2}+\alpha \right)}{2\Gamma(\alpha)} \sum_{m=0}^\infty\frac{\binom{2m}{m}}{2^{2m}(m+\alpha)}\int_0^1 \frac{t^{2m}}{\sqrt{1-t^2}}\,dt =\frac{\pi\sqrt\pi\,\Gamma\left(\frac{1}{2}+\alpha \right)}{4\Gamma(\alpha)} \sum_{m=0}^\infty\frac{\binom{2m}{m}^2}{2^{4m}(m+\alpha)}$$

となります。左辺について

$$\begin{aligned} \int_0^1 \frac{x^{2\alpha-1}\tanh^{-1}x}{\sqrt{1-x^2}}\,dx &=\sum_{m=0}^\infty\frac{1}{2m+1}\int_0^1 \frac{x^{2m+2\alpha}}{\sqrt{1-x^2}}\,dx\\ &=\sum_{m=0}^\infty\frac{1}{2m+1}\frac{\sqrt\pi}{2}\frac{\Gamma\left(\frac{1}{2}+m+\alpha \right)}{\Gamma(1+m+\alpha)}\\ &=\frac{\sqrt\pi}{2}\frac{\Gamma\left(\frac{1}{2}+\alpha \right)}{\Gamma(1+\alpha)} \sum_{m=0}^\infty\frac{1}{2m+1}\frac{{\left(\frac{1}{2}+\alpha \right)}_m}{{(1+\alpha)}_m}\\ \end{aligned}$$

となっていますので，

$$\frac{\pi\sqrt\pi\,\Gamma\left(\frac{1}{2}+\alpha \right)}{4\Gamma(\alpha)}\sum_{m=0}^\infty\frac{\binom{2m}{m}^2}{2^{4m}(m+\alpha)} =\frac{\sqrt\pi}{2}\frac{\Gamma\left(\frac{1}{2}+\alpha \right)}{\Gamma(1+\alpha)} \sum_{m=0}^\infty\frac{1}{2m+1}\frac{{\left(\frac{1}{2}+\alpha \right)}_m}{{(1+\alpha)}_m}$$

すなわち

$$\frac{\pi \alpha}{2}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}(n+\alpha)} =\sum_{n=0}^\infty \frac{1}{2n+1}\frac{{\left(\frac{1}{2}+\alpha \right)}_n}{{(1+\alpha)}_n}$$

を得ます。
$\\$

おまけ

$\alpha=\frac{1}{2}$を代入すると

$$\frac{\pi}{2}\sum_{n=0}^\infty \frac{\binom{2n}{n}^2}{2^{4n}(2n+1)} =\sum_{n=0}^\infty \frac{1}{2n+1}\frac{(1)_n}{\left(\frac{3}{2} \right)_n} =\sum_{n=0}^\infty \frac{2^{2n}}{(2n+1)^2\binom{2n}{n}}$$

となります。
また，両辺を$\alpha$で微分し，$\alpha=\frac{1}{2}$を代入すると

$$\begin{aligned} \pi\sum_{n=0}^\infty\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)}-\pi\sum_{n=0}^\infty\frac{\binom{2n}{n}^2}{2^{4n}(2n+1)^2} &=\sum_{n=0}^\infty \frac{2^{2n}}{(2n+1)^2\binom{2n}{n}}\sum_{m=0}^{n-1}\left(\frac{1}{m+1}-\frac{1}{m+\frac{3}{2}} \right)\\ &=2\sum_{n=0}^\infty \frac{2^{2n}}{(2n+1)^2\binom{2n}{n}}\left(1-\sum_{m=1}^{2n+1}\frac{(-1)^{m-1}}{m} \right) \end{aligned}$$

となるので

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

がわかります。

ありがとうございました。