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*}