Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 3 / 5A (Methods)

Integration by parts gives ${\tilde{g}}_{\sigma }\prime (k)=-{\sigma }^{2}k{\tilde{g}}_{\sigma }(k)$ and ${\tilde{g}}_{\sigma }(0)=1$, so ${\tilde{g}}_{\sigma }=e^{-{\sigma }^2{k}^2/2}$. The convolution theorem says $\widetilde{f\ast g}=\tilde{f}\tilde{g}$; multiplying the two Gaussian transforms and inverting yields $g_{\sqrt{{\sigma }_1^2+{\sigma }_2^2}}$.