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 $g(z)$.


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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