 Solution

With the stated Fourier transform convention, integration by parts givesThe boundary term vanishes, soFor 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.