We often define Fourier transform $\mathcal F$ on $L^2$ by continuously extending it from $L^1$ or the Swartz space $\mathcal S$. There are of course other ways to do it.
Note: by "integral" I am referring to the Lebesgue integral.
But now, I am interested in the convergence of the integral $$ \int_{\mathbb R^d} f(x) e^{-i\langle x, \xi\rangle} dx $$ itself. Even if the function $f$ is not in $L^1$, this integral could converge, sometimes for almost every $\xi$. For example, if we pick a real-valued function $g(x)$ that is not $L^1$ but $L^2$, then the Fourier transform integral of $g(x)e^{i\langle x,\xi_0 \rangle}$ is undefined when $\xi=\xi_0$, but is defined for other values of $\xi$.
Does there exists a function $f\in L^2 (\mathbb R)$ such that the integral $$ \int_{\mathbb R} f(x) e^{-ix \xi} dx $$ diverges for $\xi$ on a set of non-zero measure?
The inversion integral for $f\in L^2$ converges in $L^2(\mathbb{R})$. That is, $$ \lim_{R\rightarrow\infty} \left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\hat{f}(s)e^{-isx}ds-f\right\|_{L^2(\mathbb{R})}=0. $$ You can choose a subsequence that converges pointwise a.e. to $f$.