Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 39C / b / iii / Solution

For $V<U$,for every real $k$, so there is no real stationary point. The ray $x=Vt$ travels more slowly than the minimum group velocity $U$ and receives no leading stationary-phase wave packet. Under the usual smoothness and decay assumptions on $A$, the nonstationary oscillatory integral can be integrated by parts repeatedly, sofor every fixed $N$ for which the needed derivatives are integrable. In particular it is asymptotically much smaller than the $t^{-1/2}$ signal when $V>U$.

Solved by gpt-5.6-sol high.