Known exact values of the $\operatorname{Li}_3$ function We know some exact values of the trilogarithm $\operatorname{Li}_3$ function.
Known real analytic values for $\operatorname{Li}_3$:


*

*$\operatorname{Li}_3(-1)=-\frac{3}{4} \zeta(3)$

*$\operatorname{Li}_3(0)=0$

*$\operatorname{Li}_3\left( \frac{1}{2} \right) = \frac{7}{8} \zeta(3) - \frac{1}{12} \pi^2 \ln 2 + \frac{1}{6} \ln^3 2$

*$\operatorname{Li}_3(1) = \zeta(3),$ where $\zeta(3)$ is the Apéry's constant

*$\operatorname{Li}_3\left(\phi^{-2}\right)=\frac{4}{5} \zeta(3) + \frac{2}{3} (\ln \phi)^3 - \frac{2}{15} \pi^2 \ln \phi,$ where $\phi$ is the golden ratio.


Using identities for the list above we could also get:


*

*$\operatorname{Li}_3(2) = \frac{\pi^2}{4} \ln 2 + \frac{7}{8} \zeta(3) - \frac{\pi}{2} \ln^2(2) \cdot i,$ or we could write into this alternate form.

*$\operatorname{Li}_3\left(\phi^2\right) = \frac{4}{5} \zeta(3) - \frac{2}{3} \ln^3 \phi + \frac{8\pi^2}{15}\ln \phi - 2\pi\ln^2(\phi) \cdot i,$ or there is an alternate form here.


We know even less about complex argumented values:


*

*$\operatorname{Li}_3(i)=-\frac{3}{32}\zeta(3) +\frac{\pi^3}{32} i$

*$\operatorname{Li}_3(-i)=-\frac{3}{32}\zeta(3) -\frac{\pi^3}{32} i.$


There are some partial result for complex cases:


*

*$\Im\left[ \operatorname{Li}_3 \left( \frac{1 \pm i}{\sqrt{2}} \right) \right] = \pm \frac{7\pi^3}{256}$, and there is an expression for the real part in term of derivatives of digamma function,

*other related specific values around the unit circle like this or this, etc.


Lucian said the following in this question:


*

*$\Re\left[\text{Li}_3\left(\dfrac{1+i}2\right)\right]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).$


While working on Lucian's problem I was able to specify this one


*

*$\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$


Than I got the idea to write a question, maybe someone could give us some more specific values of the trilogarithm function.

It appears that
$$
\Im\left[\mathrm{Li}_3\left((-1)^{\frac{1}{2^n}}\right) \right]=\frac{(2^n-1)(2^{n+1}-1)}{3\cdot2^{3n+2}}\pi^3
$$
but that's by experimentation.
