Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

In astrophysics, specifically in partially degenerated matter, are used what is called Fermi-Dirac Integrals, which are written in terms of polylogaritms, and the z-value is a degeneracy parameter. In some papers I found that in fact they need the inverse of the Fermi-Dirac Integrals, that is, the inverse of the Polylogarithm.

share|cite|improve this answer
Thanks Michael, can you give some specifics? – Simon Nov 17 '11 at 4:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.