Fourier series of $\frac{1}{x}$ 
What is the Fourier series expansion of $\frac{1}{x}$ ?

The best method I could come up with was shifting the function by 'k' (shifting the function to $\frac{1}{x - k}$), so that while calculating the coefficients you don't run into the discontinuity of 1/x. 
Is there a different method to calculate the Fourier series of $\frac{1}{x}$.
 A: The Fourier series only exists for periodic functions which are integrable over a period. You can choose an interval and consider the periodic extension of $\frac{1}{x}$ over that interval, but if that interval contains $0$ (even as an endpoint), it will not be integrable.
A: Let $\hat{f}$ be the Fourier transform of some function $f$.
Consider the image of the integer numbers under this Fourier transform
$$\{\hat{f}(n) | \ n\in\mathbb{Z}\}$$
Observe that the inverse Fourier series of this set,
$$\sum\limits_{n\in\mathbb{Z}} \hat{f}(n) e^{i n}, $$
equals (a.e) some periodic function. If you calculate the Fourier Transform of $1/x$, as it is done here and here, you get the Fourier series coefficients $$c_n = \hat{f}(n).$$
In this case, if the domain of $\frac{1}{x}$ is $[-\pi, \pi)$ then  $$c_n = −\frac{i}{2}sgn(n).$$
The same way it must be possible to make the same analysis for the Fourier Series.
A: It is true that $f(x)=1/x$ is not locally integrable at $0$, so there is a problem.
As @reuns points out, once choice is to interpret "$1/x$" as being the Cauchy principal value integral $\lim_{\epsilon\to 0}\int_{\epsilon\le|x|\le\pi}{e^{2\pi inx}\,dx\over x}$. A similar "principal value integral" can be arranged on $[0,2\pi]$, instead, if desired.
But the non-local-integrability cannot be evaded. The "principal value integral" is not a literal integral, since the literal (improper) integral would require that the limits below $0$ and above $0$ be independent... which is not possible.
So, truly, one is probably asking about the Fourier series of a distribution, given by the principal-value-integral (as in @reuns' answer).
