Can anyone prove this formula for the harmonic progression easily?
For $a$ and $b$ integers: \begin{multline}\sum _{k=1}^n \frac{1}{a k+b}=-\frac{1}{2b}+\frac{1}{2(a n+b)}\\+\int_0^1 2\pi (1-u)\sin[a n\pi u]\sin[(a n+2b)\pi u]\cot[a\pi u]\,du\end{multline}
PS This formula was created by me, am just looking for an indirect proof to show its validity (not sure if it's a good idea to divulge it before publishing the result but anyway, perhaps being here already shows I am the creator).
Now, even more surprising are the patterns of the higher orders:
$$\sum_{k=1}^n \frac{1}{(a k+b)^3}=-\frac{1}{2 b^3}+\frac{1}{2(a n+b)^3}\\-\frac{4\pi ^3}{3}\int _0^1 \left(-u+u^3\right) \sin{[a n\pi(1-u)]}\sin{[(a n+2b)\pi(1-u)]}\cot[a\pi (1-u)]du$$
The general formula for odd powers is:
$$\sum_{j=1}^{n}\frac{1}{(a j+b)^{2k+1}}=-\frac{1}{2b^{2k+1}}+\frac{1}{2(a n+b)^{2k+1}}\\+(-1)^{k}(2\pi)^{2k+1}\int_{0}^{1}\sum_{j=0}^{k}\frac{B_{2k-2j}\left(2-2^{2k-2j}\right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n)\,du$$
where $f(u,n)=\sin{[a n\pi(1-u)]}\sin{[(a n+2b)\pi(1-u)]}\cot[a\pi (1-u)]$,
and the general formula for even powers is:
$$\sum_{j=1}^{n}\frac{1}{(a j+b)^{2k}}=-\frac{1}{2b^{2k}}+\frac{1}{2(a n+b)^{2k}}\\-(-1)^{k}(2\pi)^{2k}\int_{0}^{1}\sum_{j=0}^{k}\frac{B_{2k-2j}\left(2-2^{2k-2j}\right)}{(2k-2j)!(2j)!}u^{2j}g(u,n)\,du$$
where $g(u,n)=\sin{[a n\pi(1-u)]}\cos{[(a n+2b)\pi(1-u)]}\cot[a\pi (1-u)]$
These formulae work for the harmonic numbers too ($a=1,b=0$), if term $-1/(2b^k)$ is discarded.
Looks like my post was vandalized. I'm able to edit this answer though, so let me include a link to the paper on arXiv, which has now been accepted.
https://arxiv.org/abs/1811.11305