How to find Real Part of PolyLog[3,(1-i)] in closed form

Question

$ \Re \bigg(\text{Li}_3(1-i)\bigg)=\frac{\pi^a}{b}\ln(2)+\frac{c}{d}\zeta(e)$ has an approximate value of .8711588834109380
if $a=1 , b=-3415 , c=34 , d=39 , e=19$ are substituted into the closed form expression
I get .871158882044199 but that's as far as my numerical efforts take me. Having 8 decimal digits in agreement is not enough. Has anyone come across the closed form expression for the real part of this polylogarithm or knows how to derive it? Since the |(1-i)|>1 it is harder still.

Answer 1

Take the functional identity

$$ \text{Li}_3(z)+\text{Li}_3(1-z)+\text{Li}_3(1-z^{-1})=\zeta(3)+\frac{1}{6}\log^3(z)+\frac{\pi^2}{6}\log(z)-\frac{1}{2}\log^2(z)\log(1-z) $$

Putting $z=i$ we get

$$ 2\Re[\text{Li}_3(1-i)]=-\text{Li}_3(i)+\zeta(3)+\frac{i\pi^3}{32}+\frac{\pi^2}{16}\log(2) \quad \color{green}{(*)} $$

But on the other hand

$$ \text{Li}_3(z)+\text{Li}_3(-z)=\frac{1}{4}\text{Li}_3(z^2)\quad\color{red}{(1)}\\ \text{Li}_3(-z)-\text{Li}_3(-z^{-1})=-\frac{1}{6}\log^3(z)-\frac{\pi^2}{6}\log(z)\quad\color{blue}{(2)} $$

setting $z=i$ and calculating $\color{red}{(1)}-\color{blue}{(2)}$ we obtain

$$ \text{Li}_3(i)=\frac{i \pi^3}{32}+\frac{1}{8}\text{Li}_3(-1) \quad \color{orange}{(**)} $$

Putting $\color{orange}{(**)}$ into $\color{green}{(*)}$ we get

$$ 2\Re[\text{Li}_3(1-i)]=-\frac{1}{8}\text{Li}_3(-1)+\zeta(3)+\frac{\pi^2}{16}\log(2) $$

Togehter with the special value $\text{Li}_3(-1)=\frac{-3}{4}\zeta(3)$ this yields the final result

$$ \Re[\text{Li}_3(1-i)]=\frac{35}{64}\zeta(3)+\frac{\pi^2}{32}\log(2) $$

which seems to fit numerical results but disproves the conjecture