A hard log definite integral: $\int_0^{\pi/4}\ln^3\sin x\,\mathrm dx$ Show that:
$$\int_0^{\pi/4}\ln ^3\sin x\text{d}x=\frac34\text{Im}\left(\text{Polylog}\left(4,i\right)\right)-3\text{Im}\left(\text{Polylog}\left(4,\frac12+\frac{i}{2}\right)\right)-\frac{23}{128}\pi^3\ln 2+\frac32\text{Im}\left(\text{Polylog}\left(3,\frac12+\frac{i}{2}\right)\right)\ln 2-\frac38G\ln^22-\frac5{16}\pi\ln^32-\frac38\pi\zeta\left(3\right)$$

A related problem. Here is a different form for the solution in terms of the hypergeometric function
$$\int_0^{\frac{\pi}{4}}\ln ^3\sin x\text{d}x = -\frac{\pi\,\ln^3(2)}{32}-\frac{3}{8}\sqrt {2} \left(\ln^2( 2 ) + 4\,\ln 
\left( 2 \right) +8 \right)\times$$
$${_7F_6\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\bar{\alpha}, \alpha\, ;\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\bar{\beta} , \beta;\,\frac{1}{2}\right )}\sim -6.037581109, $$
where $i=\sqrt{-1},$ 
$$ \alpha = \frac{3\ln(2)+2+2i }{2\ln(2)}, \quad \beta = \frac{\ln(2)+2+2i }{2\ln(2)}, $$
and $ \bar{\alpha},\bar{\beta} $ are the complex conjugates of $\alpha,\beta$. 
