Suppose I want to compute $$\sum\_{n=1}^\infty \cos(n x)/n^{2k}.$$ Now, For any fixed $k$ this is, at least in principle, doable (see the excellent answer to [my math.SE question](https://math.stackexchange.com/questions/985778/weighted-sum-of-cosines) a while back), but the question is whether the sequence of polynomials one gets has some sort of a closed form.



Summing cosines
This can be written as
$$ F\_{2k}(x) = \dfrac{1}{2} \left(\text{polylog}(2k, e^{ix}) + \text{polylog}(2k, e^{-ix})\right)$$
Note that
 $F\_{k}''(x) = -F\_{k-2}(x)$, with $F\_k(0) = \zeta(k)$ and $F\_k'(0) = 0$, so
that $F\_k(x) = \zeta(k) + \int\_0^x dt\; (x-t) F\_{k-2}(t)$
It seems we have
$$F\_{2k}(x) = (-1)^{k+1} \dfrac{x^{2k}}{2(2k)!} + (-1)^k \dfrac{\pi |x^{2k-1}|}{2(2k-1)!}
+ \sum\_{j=1}^{k} (-1)^{k-j} \zeta(2j) \dfrac{x^{2k-2j}}{(2k-2j)!}$$


