Question 1: How can we prove $$I_3=\int_{-\pi/2}^{\pi/2}\operatorname{Li}_3(\sin x)dx=\frac{\pi}{4}\zeta(3)+\frac{1}{6}\pi\ln^32-\frac{1}{24}\pi^3\ln2?$$
(where $\displaystyle\operatorname{Li}_s(x):=\sum_{n=1}^\infty\frac{x^n}{n^s}$)
Question 2: Moreover, can we find a general method (not necessary to be an explicit formula) to find $$I_n=\int_{-\pi/2}^{\pi/2}\operatorname{Li}_n(\sin x)dx$$ where $n\in\mathbb{Z}^+$?
My attempt
Recall the definition of polylogarithm, $$I_3=\int_{-\pi/2}^{\pi/2}\sum_{n=1}^\infty\frac{\sin^nx}{n^3}dx\\
=\sum_{m=1}^\infty\frac1{8m^3}\int_0^{\pi/2}2\sin^{2m}xdx\\
=\sum_{m=1}^\infty\frac{\pi(1/2)_m}{8m^3m!}$$
where $(1/2)_n$ denotes the Pochhammer symbol)
Therefore, we are able to deduce that $$I_3=\frac{1}{16} \pi \, _5F_4\left(1,1,1,1,\frac{3}{2};2,2,2,2;1\right).$$
Moreover, $I_n$ can be represented to similar $_pF_q$ terms. But, I'm not familiar with $_5F_4$. I have no idea how to turn the hypergeometric term into the answer.
Some equivalent forms of $I_n$:$$I_n=\int_{-1}^1\operatorname{Li}_n(x)\frac{dx}{\sqrt{1-x^2}}$$
$$I_n=\int_0^12^{1-n}\operatorname{Li}_n(x^2)\frac{dx}{\sqrt{1-x^2}}$$
Any answer without using hypergeometric techniques will be highly appreciated.
Best Answer
If $n$ is odd the integral $\int_{-\pi/2}^{\pi/2}(\sin\theta)^{n}\,d\theta$ equals zero, so the computation of $\int_{-\pi/2}^{\pi/2}\text{Li}_3(\sin\theta)\,d\theta$ boils down to the computation of $$ \sum_{n\geq 1}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{1}{n^3}=\frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}\left[\frac{1}{4^n}\binom{2n}{n}\right]x^{n-1}\log^2(x)\,dx $$ or (by integration by parts) $$\int_{0}^{1}\frac{\log^3(x)}{(1-x)^{3/2}}\,dx.$$ This can be easily tackled through the partial derivatives of the Beta function.