 Solution

Interpret the Fourier transform as a distributional Fourier transform. Exponentially regularize the Heaviside step function:As $a\downarrow 0$, the first term tends to $\pi \delta (k)$ in the tempered distribution sense, since its integral is $\pi$ and it concentrates at zero. The second tends to $-i\mathrm{PV}(1/k)$. ThereforeThe principal-value convention is essential: the printed $1/(ik)$ is not an ordinary integrable function at zero. This normalization also satisfies $ik\hat{H}=1$, the Fourier transform of a derivative version of $H^{\prime }=\delta$.