Past exam of the theoretical mechanics course of the University of Oxford / 2011 / B / 5 / a / i / Solution

Take the nondegenerate stationary point in the interior of the integration interval. Choose a smooth cutoff equal to one near $k_{\star }$. Away from its support, ${\psi }^{\prime }$ is bounded away from zero. Integration by parts using $e^{it\psi }=(it{\psi }^{\prime }{)}^{-1}{\partial }_k{e}^{it\psi }$ makes that nonstationary contribution $O(t^{-1})$, under the usual smooth-phase assumptions.

Near the stationary point use the signed coordinateSince ${\psi }^{\prime \prime }(k_{\star })>0$, this is locally invertible, with $ds/dk=\sqrt{{\psi }^{\prime \prime }(k_{\star })}$ at the point. The local integral becomesScale $s=\sqrt{2/t}r$. The supplied Fresnel integral gives ${\int }_{-\infty }^{\infty }{e}^{i{r}^2}dr=e^{i\pi /4}\sqrt{\pi }$. Approximating the smooth amplitude by its value at zero, with a cutoff and a nonstationary estimate for the tails, therefore givesIf $f(k_{\star })\ne 0$, this is the requested asymptotic equivalence. The phase shift $\pi /4$ comes from the positive quadratic curvature. A stationary point at an integration endpoint instead needs a one-sided formula.

The printed auxiliary estimate asserting an $O({\gamma }^{-1})$ Fourier integral for every continuous amplitude is too strong without extra regularity. The argument here uses integration by parts only for the actual smooth, cutoff nonstationary amplitudes, where that estimate is justified; continuity alone generally guarantees convergence to zero, not this rate.