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I have the following expression, $$\mathrm{Li}_3(-e^{+iy})-\mathrm{Li}_3(-e^{-iy})+\left[\mathrm{Li}_4(-e^{-iy})+\mathrm{Li}_4(-e^{+iy})\right],$$ where $0<y<\pi$. I know that this expression can be simplified to polynom $P(y)$. Which manipulations can help me to see polynomial structure of this expression?

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For ${\rm Li}_3$ and ${\rm Li}_4$ we have \begin{align*} {\rm Li}_3(-z)-{\rm Li}_3(-z^{-1})&=-\frac16\log^3(z)-\frac{\pi^2}6\log(z)\tag1\\ {\rm Li}_4(-z)+{\rm Li}_4(-z^{-1})&=-\frac1{24}\log^4(z)-\frac{\pi^2}{12}\log^2(z)-\frac{7\pi^4}{360}\tag2 \end{align*} These are special cases of a more general Inversion Formula for Polylogarithms, given for example in L. Lewin's "Polylogarithms and Associated Functions" as

$${\rm Li}_n(-z)+(-1)^n{\rm Li}_n(-z^{-1})~=~-\frac1{n!}\log^n(z)+2\sum_{k=1}^{\lfloor n/2\rfloor}\frac{\log^{n-2k}(z)}{(n-2k)!}{\rm Li}_{2k}(-1)$$

This formula is straightforward to prove. Given the usual inversion formula for ${\rm Li}_2$ $${\rm Li}_2(-z)+{\rm Li}_2(-z^{-1})=-\frac{\pi^2}6-\frac12\log^2(z)\tag{$\star$}$$ Divide by $z$, integrate both sides and determine the constant of integration by evaluating at $z=1$. Then, repeat. $(\star)$ on the other hand follows immediately from differentiating ${\rm Li}_2(-z^{-1})$ and integrating back.

However, apply $(1)$ and $(2)$ in your case for $z=e^{iy}$ to obtain \begin{align*} &[{\rm Li}_3(e^{iy})-{\rm Li}_3(e^{-iy})]+[{\rm Li}_4(e^{iy})+{\rm Li}_4(e^{iy})]\\ =&\left[-\frac16\log^3(e^{iy})-\frac{\pi^2}6\log(e^{iy})\right]+\left[-\frac1{24}\log^4(e^{iy})-\frac{\pi^2}{12}\log^2(e^{iy})-\frac{7\pi^4}{360}\right]\\ =&\left[-\frac16(iy)^3-\frac{\pi^2}6(iy)\right]+\left[-\frac1{24}(iy)^4-\frac{\pi^2}{12}(iy)^2-\frac{7\pi^4}{360}\right]\\ =&-\frac1{24}y^4+\frac i6y^3+\frac{\pi^2}{12}y^2-\frac{i\pi^2}6y-\frac{7\pi^4}{360} \end{align*}

$$\therefore~P(y)~=~-\frac1{24}y^4+\frac i6y^3+\frac{\pi^2}{12}y^2-\frac{i\pi^2}6y-\frac{7\pi^4}{360}$$

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