How can we calculate $$L_n=\operatorname{Li}_n(1-i)+(-1)^n\operatorname{Li}_n\left(\frac{1+i}2\right)?$$
I have found the following result manually: $$L_1=-\frac12\pi i+\ln\frac{1-i}2=-\frac34\pi i-\frac12\ln2$$ $$L_2=\frac1{16}\pi^2-i\left(\frac14\pi\ln2+G\right)+\frac5{96}\pi^2-\frac18\ln^22+i\left(G-\frac18\pi\ln2\right)\\ =\frac{11}{96}\pi^2-\frac18\ln^22-i\frac38\pi\ln2$$ Something I have found by using Mathematica: $$L_3=-\frac{1}{48} \ln^32+\frac{11}{192} \pi ^2 \ln2+i \left(-\frac{7 \pi ^3}{128}-\frac{1}{32} 3 \pi \ln^22\right)$$ $$L_4=\frac{1313 \pi ^4}{92160}-\frac{1}{384} \ln ^42+\frac{11}{768} \pi ^2 \ln ^22+i \left(-\frac{1}{64} \pi \ln ^32-\frac{7}{256} \pi ^3 \ln 2\right)$$ It seems like there is a method that Mathematica knows to simplify $L_n$. What is the method?
Note that $(1-i)^{-1}=(1+i)/2$ and from here, $${\rm{Li}}_n(1-i)=(-1)^{n-1}{\rm{Li}}_n\left(\frac{1+i}2\right)-\frac{(2\pi i)^n}{n!}B_n\left(\frac{\ln(i-1)}{2\pi i}+\frac12\right)$$ and since $(-1)^{n-1}=-(-1)^n$, $$L_n=-\frac{(2\pi i)^n}{n!}B_n\left(\frac{\frac{\ln2}2+\frac{3i\pi}4}{2\pi i}+\frac12\right)=-\frac{(2\pi i)^n}{n!}B_n\left(\frac78-\frac{\ln2}{4\pi}i\right)$$ where $B_n(\cdot)$ is the Bernoulli polynomial from which you get an $n$th degree complex polynomial.