The Polylogarithm

Define the k-logarithm, L_k by,

L_k (z) = sum_[n=1 to +infinity] z^n / n^k , for |z|<1

We will derive an interesting identity for this function which illustrates the discrete Fourier transform.

Let ζ be an N th root of unity with ζ not =1 (i.e- ζ is primitive). Consider the sum,

N^(k-1) sum_[ζ^N = 1] L_k (ζz)

Using the definition of L_k (z) we can write this as,

N^(k-1) sum_[n=1 to +infinity] (z^n / n^k) sum_[ζ^N = 1] ζ^n

Note that, sum_[i=0 to N-1] ζ^i = (1 - ζ^N) / (1 - ζ) = 0

by summing the geometric progression. Now note that this is satisfied also by ζ^2, ... , ζ^(N-1) so these are the other roots of unity apart from 1 and ζ (at ζ=N we again get 1 and cycle through). But this is again the same geometric series and so,

sum_[ζ^N = 1] ζ^n = 0 , for all n>1 with n not =N.

Then in our sum,

N^(k-1) sum_[ζ^N = 1] L_k (ζz)

N^(k-1) sum_[lN

(N^(k-1)).z^(nN) / [(n^k).(N^(k-1))] = z^(nN) / n^k for n=1,...

Then,

N^(k-1) sum_[ζ^N = 1] L_k (ζz) = L_k (z^N)