Vrftsjtryk

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 2pi (1-u)sin[a npi u]sin[(a n+2b)pi u]cot[api u],duend{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:

$$sum_{k=1}^n frac{1}{(a k+b)^3}=-frac{1}{2 b^3}+frac{1}{2(a n+b)^3}\-frac{4pi ^3}{3}int _0^1 left(-u+u^3right) sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]du$$

The general formula for odd powers is:

$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k+1}}=-frac{1}{2b^{2k+1}}+frac{1}{2(a n+b)^{2k+1}}\+(-1)^{k}(2pi)^{2k+1}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j+1)!}u^{2j+1}f(u,n),du$$

where
$f(u,n)=sin{[a npi(1-u)]}sin{[(a n+2b)pi(1-u)]}cot[api (1-u)]$,

and the general formula for even powers is:

$$sum_{j=1}^{n}frac{1}{(a j+b)^{2k}}=-frac{1}{2b^{2k}}+frac{1}{2(a n+b)^{2k}}\-(-1)^{k}(2pi)^{2k}int_{0}^{1}sum_{j=0}^{k}frac{B_{2k-2j}left(2-2^{2k-2j}right)}{(2k-2j)!(2j)!}u^{2j}g(u,n),du$$

where
$g(u,n)=sin{[a npi(1-u)]}cos{[(a n+2b)pi(1-u)]}cot[api (1-u)]$

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