Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 5 / b / Solution

The inverse Fourier transform isFor $t>0$, close in the upper half-plane, where $e^{i\omega t}$ decays. The single enclosed pole has residue $-i{e}^{-t}$, so counterclockwise integration gives $f=e^{-t}$. For $t<0$, close clockwise in the lower half-plane; there is no enclosed pole and the result is zero. ThusHere $H$ is the Heaviside step function. At the jump, symmetric Fourier inversion gives $f(0)=1/2$, the average of the one-sided limits. The transform itself cannot fix a function’s value at a single point, so this is the conventional representative rather than extra frequency information.

A direct check is ${\int }_0^{\infty }{e}^{-(1+i\omega )t}dt=(1+i\omega {)}^{-1}=-i/(\omega -i)$. Distributionally, $f^{\prime }=\delta (t)-e^{-t}H(t)$.