For any smooth amplitude $v(x,\lambda)$,
$$
D_x^\alpha(e^{i\lambda\cdot x}v)=e^{i\lambda\cdot x}(\lambda+D_x)^\alpha v
$$
because $D_x=-i\partial_x$. Applying each term of the differential operator under the <oscillatory integral> gives
$$
\boxed{Q(x,D)A_1(x)=\frac1{(2\pi)^n}\int e^{i\lambda\cdot x}Q(x,\lambda+D_x)a_1(x,\lambda)\,d\lambda}.
$$
The identity is justified distributionally by regularizing the frequency integral and integrating by parts.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 / 3 / c / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-327/3/c/iii/solution