 Solution

Apply the spherical identity of part (b) with $x=v-v_{\ast }$, $y=\xi$ and $\varphi (s)=e^{-is/2}$. The angular exponential in part (c) can be replaced, after angular integration, by $e^{-i|\xi |(v-v_{\ast })\cdot \sigma /2}$. The full phase is thenThe Fubini theorem factors the two velocity integrals into Fourier transforms. With ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$, the gain integral is exactly ${\int }_{{\mathbb{S}}^2}\hat{f}({\xi }^{+})\hat{f}({\xi }^{-})d\sigma$.

The loss Fourier transform is $\hat{f}(\xi )\hat{f}(0)$. Dividing the gain by the sphere area gives the Bobylev identity for the Maxwell molecule collision operator:The initial Fourier datum is ${\hat{f}}_0$. The surface measure on a sphere here is two-dimensional surface area, not the restriction of ambient three-dimensional Lebesgue measure, which would give the sphere measure zero.