一个含有超几何函数的积分

\[\Large\int_0^{\large \frac{\pi}{2}} \ln^2\left(\tan\frac{x}{2}\right){ _3F_2}\left(\frac12,1,1;\frac32,\frac32;\sin^2 x \right)\mathrm{d}x = \frac{5 \pi^5}{384} \]

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\(\Large\mathbf{Proof:}\)
The \({\;}_3F_2\) is just a deception.

\[{\;}_3F_2\left(\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2};\sin^2(x) \right) =\sum_{n=0}^\infty\frac{(2\sin x)^{2n}}{(2n+1)^2\dbinom{2n}{n}}= \frac{1}{\sin x}\int_0^x\frac{\theta}{\sin\theta}\mathrm{d}\theta \]

This is obtained by the simple observation

\[\frac{\left(\dfrac{1}{2}\right)_n (1)_n^2}{\left(\dfrac{3}{2}\right)_n^2 n!}=\frac{4^n}{\dbinom{2n}{n} (2n+1)^2} \]

Also, note that \(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\ln\left(\tan \frac{x}{2} \right)=\frac{1}{\sin x}\). Therefore, by repeated integration by parts:

\[\begin{align*} &\int_0^{\large \frac{\pi}{2}} \ln^2\left(\tan\frac{x}{2}\right){ _3F_2}\left(\frac12,1,1;\frac32,\frac32;\sin^2 x \right)\mathrm{d}x=\frac{1}{12}\int_0^{\frac{\pi}{2}}\ln^4\left(\tan\frac{x}{2} \right)\mathrm{d}x\\ &=\frac{1}{6}\int_0^1\frac{\ln^4 x}{1+x^2}\mathrm{d}x=\frac{1}{6}\sum_{n=0}^{\infty }\left ( -1 \right )^{n}\int_{0}^{1}x^{2n}\ln^{4}x\mathrm{d}x=\frac{1}{6} \cdot 24\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n+1 \right )^{5}}\\&=\frac{1}{6}\cdot 24\cdot \beta \left ( 5 \right )=\frac{1}{6}\cdot 24\cdot \frac{5\pi ^{5}}{1536}=\Large\boxed{\color{blue}{\dfrac{5\pi ^{5}}{384}}} \end{align*}\]