Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 30A (Asymptotic Methods)

If $g^{\prime }$ never vanishes on $[a,b]$, integration by parts gives $I(\lambda )=[f(t)e^{i\lambda g(t)}/(i\lambda g^{\prime }(t)){]}_a^b-(i\lambda {)}^{-1}{\int }_a^b(f/g^{\prime }{)}^{\prime }{e}^{i\lambda g(t)}dt$, hence $\boxed{I(\lambda )=O({\lambda }^{-1})}$ for smooth $f,g$. Endpoint terms usually attain this order; special cancellation can give faster decay.

For the final integral, take the imaginary part of ${\int }_0^1{e}^{i\lambda (t^3-t)}dt$. Its sole stationary point is $t_0=1/\sqrt{3}$, with $g(t_0)=-2/(3\sqrt{3})$ and $g^{\prime \prime }(t_0)=2\sqrt{3}$. The stationary phase formula in part (a) therefore gives

Near the nondegenerate stationary point, write $g(t)=g(t_0)+g^{\prime \prime }(t_0)(t-t_0{)}^2/2+O((t-t_0{)}^3)$ and replace $f(t)$ by $f(t_0)$ to leading order. Scaling $t-t_0$ by ${\lambda }^{-1/2}$ reduces the local integral to the Fresnel integral, while the remaining nonstationary region contributes $O({\lambda }^{-1})$. The stationary phase result is

Use disjoint small neighborhoods of the finitely many nondegenerate stationary points $t_j$, apply the stationary phase formula in each, and integration by parts elsewhere. The leading contributions add:Interference between these oscillatory contributions can cancel the leading coefficient at particular values of $\lambda$.

For a simple stationary point at the endpoint $b$, only one half of the local quadratic neighborhood lies in the integration interval. The even Gaussian integrand therefore gives half the interior stationary phase contribution: