The Fourier expansion for the floor funcion The Fourier expansion for the floor funcion is given by:
$$\lfloor x \rfloor = x- \frac{1}{2} + \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\sin 2 \pi k x}{k}$$
for all $x$ not an integer.
My question is where I can find the proof of this formula.
 A: Here:
First, the function $\lfloor x \rfloor$ is not periodic, thus it does not have a Fourier expansion. (It has a Fourier-transform, but it is another topic.)
This formula solves the problem by that it unfolds $\lfloor x \rfloor$ as the difference of $x$ and its fractional part: $\lfloor x \rfloor = x - \{ x \}$. The first, $x$ is not fourier expanded, this is what you see as the first literals of the formula.
The second part, the fractional part, is a well known signal in the signal processing, the Sawtooth wave.
From practical reasons, also this is unfolded into a constant part (which is $\frac{1}{2}$) and to a zero-constant periodic one. This already has a fourier expansion, which can be already calculated by the formula:
$$\mathcal{F}_n = \int\limits_0^1 x\cdot e^{2\pi i n x} dx$$
Here we also use the identity $e^{i\omega} = \cos \omega + i \sin \omega$, to be able to calculate both the $\cos$ and $\sin$ members of the expansion in the same formula. However, knowing that the Sawtooth wave is anti-periodic, the $\cos$ - members will fall out (which is also an easy way to convince ourself from the correctness of the final result :-) ).
This we can already calculate easily, because it is easy to integrate any function like $\int x\cdot e^{cx} dx$ with the integration by parts:
$$\int\limits_a^b u(x)\cdot v'(x) = \left[ u(x)v(x)\right]_a^b - \int\limits_a^b u'(x)v(x)dx$$
The result will be what you got.
