Converting polylogarithms to Dirichlet L functions

Question

When trying to simplify polylogarithms evaluated at some root of unity, namely $\text{Li}_s(\omega)$ for $\omega=e^{2\pi i ~r/n}$, it is reasonable to convert it to Hurwitz zeta functions or Dirichlet L functions, since further manipulation (e.g. taking derivatives, numerical computations or transforming) will be easier.

Anyway, this can be done by a straightforward manipulation

$$ \text{Li}_s(\omega)=\sum_{k\ge1}\frac{\omega^k}{k^s}=\frac{1}{n^s}\sum _{j=1}^n \omega^j~ \zeta \left(s,\frac{j}{n}\right) $$

and orthogonality of characters to get

$$ \zeta \left(s,\frac{j}{n}\right)=\frac{n^s}{\varphi (n)} \sum_\chi L(s,\chi) \chi(j){}^*\qquad \gcd(j,n)=1 $$

However, since $j,n$ may be not coprime when $j$ ranges over $[1,n]$ the second formula cannot be substituted directly, resulting in a bad-looking double sum. Moreover each $\chi$ could be not primitive so further simplification can be applied.

In fact, actual computation shows that the result can be rather nice-looking, especially for $n$ composite. For example (in Mathematica notation)

$$ \text{Li}_s(e^{7\pi i/6})=-\frac{1}{2} i 3^{-s} \left(3^s+3\right) L_{4,2}(s)-\frac{1}{2} \sqrt{3} L_{12,4}(s)+i \sqrt{3} 2^{-2 s-1} \left(2^s+2\right) L_{3,2}(s)+2^{-2 s-1} 3^{-s} \left(2^s-2\right) \left(3^s-3\right) \zeta (s) $$

Question : Can we bypass the lengthy procedure described above and directly obtain the neat expression (with only primitive characters and simplified coefficients?

Any help would be appreciated.

Answer 1

I tried to do this once. The form is (roughly!):

$$ \text{Li}_s(e^{2 \pi i a /q}) = \frac{1}{\phi(q)} \sum_{\chi} G(a, \overline{\chi}) L^{*}(s, \chi) $$

where $G$ is a gauss sum and $L^{*}$ is the L function adjusted for missing euler factors due to imprimitive characters.

A neat expression in some sense, although in the general the gauss sums will not be as straightforward in your example and will lie in the $\phi(q)$th cyclotomic field. I guess it depends on what you want to do (?).