= Solution
With the stated <Fourier transform> convention, <integration by parts> gives
$$
\widetilde{f'}(\omega)=\int_{-\infty}^\infty f'(t)e^{-i\omega t}\,dt
=\left[f(t)e^{-i\omega t}\right]_{-\infty}^{\infty}
+i\omega\int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt.
$$
The boundary term vanishes, so
$$
\boxed{\widetilde{f'}(\omega)=i\omega\widetilde f(\omega).}
$$
For an <absolutely continuous function>, this calculation requires enough integrability to justify the displayed operations. Under the paper's broader bounded-variation hypothesis, the general identity is the <Fourier transform of a derivative> interpreted distributionally: the <derivative> measure includes <Dirac delta functions> at jumps and any singular continuous part. If one instead differentiates only the smooth pieces and discards jumps, the formula needs the corresponding jump corrections. Absolute integrability and <bounded variation> ensure the limits of $f$ at both infinities are zero, but do not by themselves make its ordinary <derivative> capture those singular parts.
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