Explicit value of $\operatorname{Li}_2(1/2-{\rm i}/2)$

Question

When you ask Wolfram Alpha about the value of $\operatorname{Li}_{2}\left(1/2-{\rm i}/2\right)$ it gives you $$ \frac{5\pi^{2}}{96} - \frac{\ln^{2}\left(2\right)}{8} + {\rm i}\left[\frac{\pi\ln\left(2\right)}{8} - G\right], $$ where $G$ is Catalan's constant defined as $\displaystyle\sum_{n = 0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n + 1\right)^{2}}$.

• I would like to know how one could get such value. I tried to put together the real an imaginary parts of $\operatorname{Li}_{2}\left(1/2 - {\rm i}/2\right)$ but I was unable to get to the result.
• Could you, in the same way, calculate $\operatorname{Li}_{3}\left(1/2 - {\rm i}/2\right)$?

It's pure curiosity and I would appreciate any thoughts about it. Thanks.

Answer 1

The value of $\newcommand{\li}{\operatorname{Li}}\li_2(1/2-i/2)$ can be deduced from the identity $$\li_2\left(\frac1{1-z}\right)=\li_2(z)+\frac{\pi^2}6+\ln(-z)\ln(1-z)-\frac12\ln^2(1-z),$$ valid for $z\in\mathbb{C}\setminus\mathbb{R}_{\geqslant 0}$ with the principal branch of $\color{blue}{\ln}$. In turn, this can be deduced from other known identities, or verified directly (like these), by taking derivatives and checking $z\to 0$.

In our case, we put $z=-i$ and use $$\li_2(-i)=\sum_{n=1}^\infty\frac{(-i)^n}{n^2}=\sum_{n=1}^\infty\frac{(-1)^n}{(2n)^2}-i\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}=-\frac{\pi^2}{48}-iG.$$

Regarding $\li_3(1/2\pm i/2)$: see this question.