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

  1. 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?

  2. Is there something similar possible for $\mathfrak{Im}(\textrm{Li}_2(e^{ix}))$?

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

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    $\begingroup$ Hi, please put your commented link in the question so it's easier for people to see :) $\endgroup$
    – Benjamin Wang
    Commented Oct 5, 2023 at 0:00
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    $\begingroup$ The imaginary component of the dilogarithm is equivalent to the Clausen function. $\endgroup$
    – David H
    Commented Oct 5, 2023 at 0:05
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    $\begingroup$ 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. $\endgroup$
    – garondal
    Commented Oct 5, 2023 at 0:25
  • $\begingroup$ @garondal You can also think about it as a Fourier series. $\endgroup$
    – Gary
    Commented Oct 5, 2023 at 3:22
  • $\begingroup$ Does this answer your question? Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$, combined with $$ \Re(\text{Li}_2(e^{ix}))=\Re\left(\sum_{k=1}^{\infty}\frac{e^{ikx}}{k^2}\right)= \sum_{k=1}^{\infty}\frac{\cos(kx)}{k^2}. $$ $\endgroup$
    – Gonçalo
    Commented Jan 22 at 20:43

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