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

The Heaviside step function restricts the integral defining the Fourier transform to $t\ge 0$, soThe convergence condition is essential. For real $\alpha$ it means $\alpha >0$, which is the decay assumption used in the subsequent parts. At $\alpha =0$ a distributional Fourier transform is needed; for negative real $\alpha$ the defining improper integral diverges.