Recently I've been playing around with some trigonometric series like $$\sum\limits_{n=1}^\infty \frac{\sin{nx}}{n^k}, \ \sum\limits_{n=1}^\infty \frac{\cos nx}{n^k}$$ and quickly found out that for certain values of $k$ there are well-known results: $$\sum\limits_{n=1}^\infty \frac{\sin nx}{n^{2k+1}} = \frac{(-1)^{k+1}(2\pi)^{2k+1} {B}_{2k+1}\left(\frac{x}{2\pi}\right)}{2(2k+1)!}$$ $$\sum\limits_{n=1}^\infty \frac{\cos nx}{n^{2k}} = \frac{(-1)^{k+1}(2\pi)^{2k}{B}_{2k}\left(\frac{x}{2\pi}\right)}{2(2k)!}$$ They look really similar to formula for $\zeta(2k)$: $$\zeta(2k) = \frac{(-1)^{k+1}(2\pi)^{2k}{B_{2k}}}{2(2k)!}$$ Actually you can just plug in $x=2\pi$ into the second formula to get this. But here we are witnessing the same problem as for $\zeta(2k+1)$, because we don't have any exact formulas for this two sums: $$\sum\limits_{n=1}^\infty \frac{\sin nx}{n^{2k}}, \ \sum\limits_{n=1}^\infty \frac{\cos nx}{n^{2k+1}}$$ Except for $k=0$ for which we know $$\sum\limits_{n=1}^\infty \frac{\cos{nx}}{n} = \Re {\sum\limits_{n=1}^\infty \frac{e^{inx}}{n}} = -\Re \ln(1-e^{ix}) = -\frac{\ln{(2-2\cos{x})}}{2}$$ But for $\zeta(2k+1)$ there is a Ramanujan formula, which for $\zeta(4k-1)$ looks like this: $$\zeta(4k-1) = (2\pi)^{4k-1}\sum\limits_{m=0}^{2k}\frac{(-1)^{m+1}B_{2m}B_{4k-2m}}{2(2m)!(4k-2m)!} - 2\sum\limits_{n=1}^\infty\frac{1}{n^{4k-1}(e^{2\pi n}-1)}$$ And so I thought that if there is a similarity between sums that can be calculated, there should be some for sums that can't. And then that's what I came up with: $$\sum\limits_{n=1}^\infty \frac{\sin nx}{n^{2k}} = (2\pi)^{2k}\sum\limits_{m=0}^k \frac{(-1)^{m+1}{B}_{2m+1}\left(\frac{x}{2\pi}\right){B}_{2k-2m}}{(2m+1)!(2k-2m)!} + \sum\limits_{n=1}^\infty \frac{(-1)^k \sinh((\pi-x)n)}{n^{2k}\sinh(\pi n)} - 2\sum\limits_{n=1}^\infty \frac{\sin nx}{n^{2k}(e^{2\pi n}-1)}$$ $$\sum\limits_{n=1}^\infty \frac{\cos nx}{n^{2k-1}} = (2\pi)^{2k-1}\sum\limits_{m=0}^k \frac{(-1)^{m+1}{B}_{2m}\left(\frac{x}{2\pi}\right){B}_{2k-2m}}{(2m)!(2k-2m)!} - \sum\limits_{n=1}^\infty \frac{(-1)^k \cosh((\pi-x)n)}{n^{2k-1}\sinh(\pi n)} - 2\sum\limits_{n=1}^\infty \frac{\cos nx}{n^{2k-1}(e^{2\pi n}-1)}$$ Looks quite natural. This formulas can be used to calculate polylogarithm values at the unit circle (see my question). Actually they can be optimized to accelerate their convergence so that general term in all series is $O(e^{-\pi n})$. There is also a generalization that can be found in my another question with interesting case of alternating series.
I can't believe that these formulas are not well-known. They look too natural for me. Do you know something about this?