Suppose $f\in L^1(\mathbf{R})$. Recall the Fourier transform of $f$ is given by $$ \hat f(x) = \int_{-\infty}^\infty f(t) e^{-2\pi i x t}\, \mathrm{d} t . $$ The Fourier transform extends from $L^1\cap L^2$ to a unitary operator $L^2 \to L^2$. I am wondering: suppose $f\in L^2\cap C^\infty$. Can we conclude that the integral above converges conditionally? Or, at least, can we conclude that $$ \limsup_{T\to \infty} \left| \int_{-T}^T f(t) e^{-2\pi i x t}\, \mathrm{d} t\right| < \infty ? $$ In general, I am wondering under what conditions the conditional convergence or boundedness of $\int_{-\infty}^\infty f(t) e^{-2\pi i x t}\, \mathrm{d} t$ can be shown.
If you could point me to a reference which considers these types of questions, I'd be very grateful!
Edit: if $f(t) = e^{2\pi i x t} / \phi(t)$, where $\phi(t) \sim t$ as $t\to \infty$, but $\phi$ is nonvanishing, then at least the integral for $\widehat f(x)$ diverges, and in fact $\left|\int_{-T}^T f(t) e^{-2\pi i x t}\, \mathrm{d} t\right| \sim \log T$. So the best I can hope for is almost everywhere conditional convergence, or even almost-everywhere boundedness. Is this known to hold?
Carleson's theorem says that $\int_{-\infty}^\infty f(t) e^{-2\pi i t x}\, \mathrm{d} t = \widehat f(x)$ almost everywhere. I should have realized this earlier, and I'm happy to delete the whole question if that's more appropriate.