Is it possible to prove that $\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^a}\geq 0$ for $a\geq 1$ and $x\in [0,\pi]$?

Question

I was trying to prove that $$\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^a}\geq 0,$$

for $0\leq x\leq \pi$, and $a\geq1$.

For $a$ being an odd integer, this is not really a problem, as the sums may then be expressed using Bernoulli polynomials. However, it seems that the assumption is true for all (real) values of $a$, not just integers, so I tried to split the sums and estimate the remainders, but unfortunately to no avail.

Does anyone have an idea on how to tackle this problem?

Thank you very much in advance.

Answer 1

We can assume that $x\neq 0$. Then your sum is $$ \Im \sum\limits_{k = 1}^\infty {\frac{{e^{ixk} }}{{k^a }}} = \Im \operatorname{Li}_a (e^{ix} ) = \frac{1}{{\Gamma (a)}}\Im \int_0^{ + \infty } \!\!\frac{{t^{a - 1} }}{{e^{t - ix} - 1}}dt = \frac{\sin x}{{\Gamma (a)}}\int_0^{ + \infty } \!\!\frac{{t^{a - 1} e^t }}{{e^{2t} - 2e^t \cos x + 1}}dt , $$ which has the desired property since $ e^{2t} - 2e^t \cos x + 1 > e^{2t} - 2e^t + 1 = (e^t - 1)^2 > 0$ for $t>0$. Here $\operatorname{Li}_a$ denotes the polylogarithm, see http://dlmf.nist.gov/25.12.ii.