Prove $\text{Li}_2(e^{-2 i x})+\text{Li}_2(e^{2 i x})=\frac{1}{3} (6 x^2-6 \pi x+\pi ^2)$ when $0<x<\pi$

Question

This is an identity I deduced when playing with the initial-boundary value problem of heat conduction equation asked here. It's easy to verify numerically with Mathematica:

Plot[{PolyLog[2, E^(-2 I x)] + PolyLog[2, E^(2 I x)], 
  1/3 (π^2 - 6 π x + 6 x^2)}, {x, -1, 4}, 
 PlotStyle -> {Automatic, {Thick, Dashed}}]

However, Mathematica doesn't know how to simplify $\text{Li}_2(e^{-2 i x})+\text{Li}_2(e^{2 i x})$ to $\frac{1}{3} (6 x^2-6 \pi x+\pi ^2)$ symbolically, so I'm wondering, how can I prove it by hand?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\left.\vphantom{\LARGE A}% \mrm{Li}_{2}\pars{\expo{-2\ic x}} + \mrm{Li}_{2}\pars{\expo{2\ic x}}\,\right\vert_{\ 0\ <\ x\ <\ \pi}} = \mrm{Li}_{2}\pars{\exp\pars{2\pi\ic\,{x \over \pi}}} + \mrm{Li}_{2}\pars{\exp\pars{-2\pi\ic\,{x \over \pi}}} \\[5mm] = &\ -\,{\pars{2\pi\ic}^{2} \over 2!}\,\mathrm{B}_{2}\pars{x \over \pi} \end{align}

which is Jonqui$\grave{\mrm{e}}$re Inversion Formula. $\ds{\mrm{B}_{n}}$ is a Bernoulli Polynomial.

Note that $\ds{\mrm{B}_{2}\pars{z} = z^{2} - z + {1 \over 6}}$ such that

\begin{align} &\bbox[10px,#ffd]{\left.\vphantom{\LARGE A}% \mrm{Li}_{2}\pars{\expo{-2\ic x}} + \mrm{Li}_{2}\pars{\expo{2\ic x}}\,\right\vert_{\ 0\ <\ x\ <\ \pi}} = 2\pi^{2}\bracks{\pars{x \over \pi}^{2} - {x \over \pi} + {1 \over 6}} \\[5mm] = &\ \bbx{2x^{2} - 2\pi x + {\pi^{2} \over 3}} \end{align}

Answer 2

$$\operatorname{Li}_2(e^{-2 i x})+\operatorname{Li}_2(e^{2 i x})\\ =2\Re\operatorname{Li}_2(e^{2i x})\\ =2\operatorname{Sl}_2(2x)\\ =2(\frac{\pi^2}{6}+\pi x+x^2)$$ Where $\operatorname{Sl}$ is the SL-type Clausen function.

Answer 3

Try a series expansion around $x=0$ and get $$\text{Li}_2\left(e^{-2 i x}\right)+\text{Li}_2\left(e^{2 i x}\right)=\frac{\pi ^2}{3}-2 \pi x+2 x^2+O\left(x^{99}\right)$$