TITLE: Integrable function whose Fourier transform is not integrable
QUESTION [9 upvotes]: I am looking for an example of a function $f: \mathbb R \rightarrow \mathbb R$ such that $f \in L^1$ in the sense that $\int_{\mathbb R} |f| < \infty$ but its Fourier transform $\hat f$ is not in $L^1$. Does anyone have one?
Thanks.

REPLY [12 votes]: Note that any function whose fourier transform is in $L^1$ must be equal to a continuous function almost everywhere, since $\mathcal F(\mathcal F(f)) = f$ a.e. in this case. This follows from the inversion formula and because the Fourier transform of a function is continuous.
This gives us many examples of functions you are looking for. For example $f(x) = \chi_{[-1,1]}(x)$ must necessarily be such a function.