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We have the complex valued function: $f(z)=\sum_{n=0}^{\infty}a_{n}\text{Li}_{-n}(z)\;\;\;\;\;\;\;(\left | z\right |<1)$ We wish to recover the coefficients $a_{n}$. The only thing I though would work is to try and come up with a function $\phi(n,x)$, such that: $\int f(z)\phi(n,z)dz=a_{n}$ or: $\int\text{Li}_{-n}(z)\phi(m,z)dz=\delta_{nm}$ but that's about as far as I've gotten. Any help is appreciated. The question is motivated by the following:

Suppose that for some analytic function $g(x)$ we have the values of the function at positive integers, so we can write a Taylor development : $g(m)=\sum_{n=0}^{\infty}a_{n}m^{n}$ Now suppose that the following summation is convergent in the open unit disk: $\sum_{m=1}^{\infty}g(m)z^{m}$ Using the above Taylor expansion, and the definition of the Polylogarithm function, we have: $\sum_{m=1}^{\infty}g(m)z^{m}=:f(z)=\sum_{n=0}^{\infty}a_{n}\text{Li}_{-n}(z)\;\;\left | z\right |<1$ The plan is to recover the coefficients $a_{n}$, and the thus the Taylor expansion of $g(z)$.

EDIT:

By Ramanujan's master theorem, $g(z)$ is given by: $\frac{\pi}{\sin(\pi s)}g(-s)=\int_{0}^{\infty}\left(f(-x)+g(0)\right)x^{s-1}dx\;\;\;\;(0<\Re(s)<1)$ However, the function $f(x)$ is not always convergent along the real line, hence the quest for an alternative.

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For $n\ge0$, the polylogarithm becomes a rational function, with $(1-z)^{n+1}$ as its denominator. With partial fraction decomposition, it's possible to split the polylogarithms into descending powers of $1-z$. I don't know the pattern of their coefficients off the top of my head, but you could collect like terms to create a Laurent series.