# Inverse of the polylogarithm

The polylogarithm can be defined using the power series $$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s}.$$ Contiguous polylogs have the ladder operators $$\operatorname{Li}_{s+1}(z) = \int_0^z \frac {\operatorname{Li}_s(t)}{t}\,\mathrm{d}t\,, \qquad \operatorname{Li}_{s-1}(z) = z \,{\partial \operatorname{Li}_s(z) \over \partial z}\ ,$$ and the sequence can be started with either $$\operatorname{Li}_{1}(z) = -\ln(1-z)\,,\qquad \operatorname{Li}_{0}(z) = {z \over 1-z} \ .$$

Both $\operatorname{Li}_0$ and $\operatorname{Li}_1$ have inverse functions (up to a choice of branchcut) $$\operatorname{Li}_0^{-1}(z)=\frac{z}{z+1}\,,\quad \operatorname{Li}_1^{-1}(z)=1-e^{-z}\,,$$ $$\operatorname{Li}_0\left(\frac{z}{z+1}\right) =z= \operatorname{Li}_1\left(1-e^{-z}\right) + 2 n \pi i\,,\quad n\in\mathbb{Z}$$

Is there a nice/useful inverse function for the dilog ($\operatorname{Li}_2(z)$) and higher polylogs?

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As $Li_s'(0) \neq 0,$ an inverse exists as a formal powerseries. I would start there. – jspecter Oct 18 '11 at 1:25
@jspecter: $\text{Li}_n^{-1}(z) \approx z-2^{-n} z^2+(2^{1-2 n}-3^{-n}) z^3 + \dots$... but is the general coefficient known in closed form? Can it be summed as, e.g., a hypergeometric? – Simon Oct 18 '11 at 1:27
Certainly, one can use Lagrangian inversion to derive series for the inverse polylogarithms. I haven't encountered any situation where the inverses are needed,though. – J. M. Oct 18 '11 at 1:29
@Simon If the inverse polylogarithm is indeed hypergeometric function, it should satisfy a differential equation, and this is unlikely for non-integer $n$. – Sasha Oct 18 '11 at 4:34
@Sasha: You're probably (almost certainly) right about non-integer $n$, but more often than not, the integer case is the one that occurs. In particular, my question asked about the inverse dilog separately from the general polylog. – Simon Jan 16 '12 at 22:01