I'm trying to find the value of $\mathbf S$ where
$$\mathbf S = \sum_{k=1}^\infty \frac{\sin(2\pi k x)}{k}; k \in \mathbb N, x \in \mathbb R^*, x \not \in \mathbb N$$
I had a look to WA which lead me to this result.
I think this result can be simplified. So I did some research and some hours later, I found this result:
$$\mathbf S = \pi (x+\frac{1}{2}-(2n+m)) ; n,m \in \mathbb N$$
I'm not satisfied with this result either. I think I miss something on the way.
I'm stuck. What is the result of this infinite sum?
Indeed, WA missed some simplification.
$$\log(1-e^{i2\pi x})-\log(1-e^{-i2\pi x})=\log\left(\frac{1-\cos(2\pi x)+i\sin(2\pi x)}{1-\cos(2\pi x)-i\sin(2\pi x)}\right)\\ =\log\left(\frac{2\sin^2(\pi x)-i2\cos(\pi x)\sin(\pi x)}{2\sin^2(\pi x)+i2\cos(\pi x)\sin(\pi x)}\right)=\log\left(-\frac{ie^{i\pi x}}{ie^{-i\pi x}}\right)=i2\pi x+i(2k+1)\pi.$$
So with the factor $\frac i2$ the sum should be
$$\pi\left(k+\frac12\right)-\pi x$$ where $k$ is indeterminate.
As the sine is periodic so is the Fourier series, and the indeterminate $k$ must be some integer offset from the integer part of $x$,
$$\pi\left(\lfloor x\rfloor+n+\frac12\right)-\pi x.$$
Then we can evaluate the series at $x=\frac12$, which is an ordinary point ($x=0$ wouldn't work), get $S(\frac12)=0$, and conclude
$$S(x)=\pi\left(\frac12-\{x\}\right).$$