I asked the question below on Math Stack Exchange, https://math.stackexchange.com/questions/2592555/convergence-of-integral-formula-for-fourier-inversion-and-hilbert-transform-fo, but [despite it having 6 upvotes and two favourites] received no answer even after a 300-point bounty. Therefore, I am re-posting it here.
Let $f \colon \mathbb{R} \to \mathbb{R}$ be a bounded $L^1$ function that is piecewise-smooth (with the boundaries of the pieces having no accumulation points), but not necessarily continuous. Define the Fourier transform $\hat{f} \colon \mathbb{R} \to \mathbb{C}$ of $f$ by $$ \hat{f}(\xi) \ = \ \int_\mathbb{R} f(t) e^{-2\pi i \xi t} \, dt $$ and write $\hat{f}(\xi)=A_\xi e^{i\phi_\xi}\,$ where $A_\xi \geq 0$ and $\phi_\xi \in (-\pi,\pi]$.
Is it the case that for almost all $t \in \mathbb{R}$, the limit \begin{align*} \mathring{f}(t) :=& \lim_{R \to \infty} \int_0^R \mathrm{Im} \Big(\hat{f}(\xi) e^{2\pi i \xi t} \Big) \, d\xi \\ =& \lim_{R \to \infty} \int_0^R A_\xi\sin(2\pi\xi t + \phi_\xi) \, d\xi \end{align*} exists and is finite? (And if so, then is the Hilbert transform of $f$ equal to $2\mathring{f}$?)
Now I have read in a few places that for every $t \in \mathbb{R}$, $$ \lim_{R \to \infty} \int_{-R}^R \hat{f}(\xi) e^{2\pi i \xi t} \, d\xi \ = \ \tfrac{1}{2}\Big(\lim_{s \to t+} f(s) + \lim_{s \to t-}f(s)\Big). $$
Hence my question is equivalent to the following: For almost all $t$, can one let the lower and upper boundaries of the above integral tend to $\infty$ independently?
To see the equivalence: Note that $A_{-\xi}=A_\xi\,$ and if $A_\xi \neq 0$ then $\,\phi_{-\xi}=-\phi_\xi$ (mod $2\pi$). Hence, for all $R_1,R_2>0$, \begin{align*} \int_{-R_1}^{R_2} \hat{f}(\xi) & e^{2\pi i \xi t} \, d\xi \\ &= \ \left( \int_0^{R_1} + \int_0^{R_2} \right) A_\xi\cos(2\pi\xi t + \phi_\xi) \, d\xi \ + \ i\int_{R_1}^{R_2} A_\xi\sin(2\pi\xi t + \phi_\xi) \, d\xi. \end{align*}
