Is there a function that is not absolutely integrable in $[-\pi,\pi]$ so that its Fourier Series Exists?

Question

For existence of Fourier coefficients of a function $f$ is sufficient that $f$ is absolutely integrable in $[-\pi,\pi]$ but, is this condition necessary? that is, is there a function that is not absolutely integrable in $[-\pi,\pi]$ so that its Fourier series Exists?
Consider the usual trigonometric system.

Answer 1

Fourier coefficients (and thus Fourier series) are defined for integrable functions. There are however trigonometric series that are not integrable (elements of $L_1(\mathbb{S}^1)$. Is this what you mean?

Here is an example: let $\{c_n:n\in\mathbb{Z}\}\subset\mathbb{R}_+$ be such that both $c_n\searrow0$ and $c_{-n}\searrow0$ as $n\rightarrow\infty$. Dirichlet's convergence test shows that $f(x)=\sum_n\operatorname{sign}(n)c_ne^{inx}$ is a pointwise convergent series. In particular, for $c_n=\frac{1}{\log |n|}\mathbb{1}(|n|\geq2)$ we have that \begin{align*} f(x)=\sum_{|n|\geq2}\frac{e^{inx}}{\operatorname{sign}(n)\log |n|}=2i\sum^\infty_{n=2}\frac{\sin nx}{\log n} \end{align*} is pointwise convergent. However, as $\sum^\infty_{n=2}\frac{1}{n\log n}$ diverges, $f\notin\mathcal{L}_1(\mathbb{S}^1)$.

Here I have used a well know fact that states that if $f\in L_1(\mathbb{S}^1)$, then $\sum^\infty_{n=1}\frac{\hat{f}(n)-\hat{f}(-n)}{n}$ converges, where $\hat{f}(n)=\frac{1}{2\pi}\int^{\pi}_{-\pi}f(t)e^{-int}\,dt$.