Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 5 / a / iii / Solution

The reciprocal in this transform must mean the principal-value reciprocal distribution. Its inverse transform isusing the given Dirichlet integral and the sign of $t$. Also ${\mathcal{F}}^{-1}[\delta ](t)=1/(2\pi )$. The inverse of the proposed expression is consequently $(iA/2)\mathrm{sgn}t+B/(2\pi )$. Comparison with $H(t)=[1+\mathrm{sgn}t]/2$ for $t\ne 0$ givesThe value assigned at the single point $t=0$ does not affect the distribution. This is the Fourier transform of the Heaviside step function; treating $1/\omega$ as an ordinary integral would miss the required prescription.