Is there a function that is not absolutely integrable in [−π,π] so that its Fourier Series Exists? For existence of Fourier coefficients of a function f is sufficient that f is absolutely integrable in [−π,π] but, is this condition necessary? that is, is there a function that is not absolutely integrable in [−π,π] so that its Fourier series Exists?
Consider the usual trigonometric system.
 A: All $2π$ periodic distributions are temperate and since all temperate distributions have a Fourier transform, you have plenty of examples. Note that the previous statements prove the power of abstract nonsense in Mathematics: you define for $u\in \mathscr S(\mathbb R)$ the Fourier transform
$$
\hat u(\xi)=\int_{\mathbb R} e^{-2iπ x \xi } u(x) dx,
$$
and you can prove that it is an isomorphism of $\mathscr S(\mathbb R)$, with the inverse given by
$
u(x)=\int_{\mathbb R} e^{2iπ x \xi } \hat u(\xi) d\xi.
$
Well, not trivial but very standard with direct proofs. Then you dramatically increase the generality by defining the Fourier transform of a temperate distribution $T$, as
$$
\langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R)}
=\langle T,\hat \phi\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R)},
$$
and with $\mathcal F$ standing for the Fourier transform and $\mathcal C$ for the mapping $T(x)\mapsto T(-x)$ (clearly defined for a function and also by duality for distributions). Then (and this is the power referred to above), you get trivially that for any tempered distribution $T$
$$
\mathcal C\mathcal F \mathcal F T=T.
$$
For instance you get with $H=\mathbf 1_{\mathbb R_+}$,
$
\hat H=\frac12\delta_0+\frac{1}{2π i}\textrm{pv}\frac1x.
$
A: How about $1/x$, for distributional Fourier transform
