It is trivial that for $f \in L^1$ and $y \in \mathbb{R}$ the integral $$\int_{\mathbb{R}} f(x)e^{-2\pi i xy} dx$$ converges. I was just wondering whether the converse is true: Given a (let's say continuous) function $f:\mathbb{R} \to \mathbb{C}$ such that the integral $$\int_{\mathbb{R}} f(x)e^{-2\pi i xy} dx$$ is convergent for all $y$, does it imply $f \in L^1$? I'm really not so sure what I should believe here since I neither see a counterexample nor a way to prove this. This question must have been considered before, but I couldn't find any references online.
Edit: As mentioned in the comments, the integrals can of course not be absolutely convergent, otherwise the result would be trivial. By saying that $\int_{\mathbb{R}} \dots$ exists I really mean that $\lim_{T \to \infty} \int_0^T \dots$ and $\lim_{T \to \infty} \int_{-T}^0 \dots$ both exist.
The Fourier transform can exist even if we have conditional convergence of $\int_{-\infty}^{\infty} f(x) \, dx$.
If $f \in L^1$ then by the Riemann-Lebesgue lemma, the Fourier transform $$\hat{f}(y) = \int_{-\infty}^\infty f(x) e ^{iyx} \, dy$$
must decay to $0$ as $|y| \to \infty$. Any function with a convergent Fourier transform without this property is not in $L^1$.
An example is
$$f(x) = \begin{cases}\sin(x^2) & x \geqslant 0 \\ 0 & x < 0 \end{cases}$$
where the Fourier (cosine) transform is
$$\hat{f}(y) = \int_0^\infty \sin (x^2)\, \cos(yx) \, dx = \sqrt{\frac{\pi}{8}}\left[\cos \frac{y^2}{4} - \sin \frac{y^2}{4}\right]$$