Show that $\mathfrak{Re}(\textrm{Li}_2(e^{ix}))=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$ (polylogarithm)

Asked Oct 04 '23 at 22:39

Active Jan 24 '24 at 20:38

Viewed 113 times

I am working with the polylogarithm function and want to find closed expressions for $\textrm{Li}_2(e^{ix})$.

If I plot the function $\mathfrak{Re}(\textrm{Li}_2(e^{ix}))$ I get $y=\dfrac{x^2}{4}-\dfrac{\pi x}{2}+\dfrac{\pi^2}{6}$ in the interval from 0 to $2\pi$. How can I see this connection?
Is there something similar possible for $\mathfrak{Im}(\textrm{Li}_2(e^{ix}))$?

Edit: I saw this but was not able to use it.

edited Jan 24 '24 at 20:38

Robin

asked Oct 04 '23 at 22:39

garondal

1

Hi, please put your commented link in the question so it's easier for people to see :) – Benjamin Wang Oct 05 '23 at 00:00
3

The imaginary component of the dilogarithm is equivalent to the Clausen function. – David H Oct 05 '23 at 00:05
Ok, and the real part of $\text{Li}_2(e^{ix})$ is just the SL-type Clausen function $\text{Sl}_2$, which has a closed form. – garondal Oct 05 '23 at 00:25
@garondal You can also think about it as a Fourier series. – Gary Oct 05 '23 at 03:22
Does this answer your question? Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$, combined with $$ \Re(\text{Li}2(e^{ix}))=\Re\left(\sum{k=1}^{\infty}\frac{e^{ikx}}{k^2}\right)= \sum_{k=1}^{\infty}\frac{\cos(kx)}{k^2}. $$ – Gonçalo Jan 22 '24 at 20:43

0 Answers0