Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 4 / f / Solution

For $T>0$ the difference of Heaviside step functions is an interval indicator, and $v=1/2$ on $(-T,T)$ and zero elsewhere. Its Fourier transform isThe apparent singularity at $\omega =0$ is removable. The formula remains valid for real negative $T$, because the original step-function difference and the oriented integral both reverse sign; for $T=0$ both vanish. This is the Fourier transform of an interval indicator and produces a scaled sinc function.