Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 31C (Asymptotic Methods)

Suppose first that the unique interior minimum is nondegenerate, ${\varphi }^{\prime \prime }(c)>0$, and $f(c)\ne 0$. Outside a small neighbourhood of $c$ the phase exceeds $\varphi (c)$ by a fixed positive amount, so that region is exponentially smaller. Inside, expandThe rescaling $s=\sqrt{\lambda {\varphi }^{\prime \prime }(c)}(t-c)$ and a dominated local Gaussian calculation give the Laplace methodIf the minimum is at $a$ and ${\varphi }^{\prime }(a)>0$, the scale is $t-a=O({\lambda }^{-1})$, giving $f(a)e^{-\lambda \varphi (a)}/[\lambda {\varphi }^{\prime }(a)]$. At $b$ with ${\varphi }^{\prime }(b)<0$, replace the denominator by $\lambda |{\varphi }^{\prime }(b)|$. A stationary quadratic endpoint gives half the interior Gaussian factor.

The printed assumptions do not guarantee nondegeneracy or $f(c)\ne 0$. More generally, if $\varphi (t)-\varphi (c)\sim A|t-c{|}^m$ with $A>0$ and even $m$ at an interior minimum, the leading phase scale is ${\lambda }^{-1/m}$ and, for $f(c)\ne 0$, the factor isAt a one-sided endpoint the factor two is absent and the first positive one-sided phase power need not be even. If the amplitude vanishes, expand it to its first contributing term; parity can cancel interior terms. A phase flat to all orders requires its actual local behavior rather than a nonexistent quadratic Taylor coefficient.

The unique minimum is at zero with phase $t^2$ and amplitude $\cos 0=1$. The interior Laplace method givesFor an additional check, extending to the real line gives the exact Gaussian Fourier integral $\sqrt{\pi /\lambda }{e}^{-1/(4\lambda )}$. The omitted tails are exponentially small, so the displayed leading term follows directly.

This is endpoint asymptotics after odd Gaussian cancellation. All interior algebraic terms cancel because $\sin t{e}^{-\lambda t^2}$ is odd. Split the interval into $[-1,1]$ and the remaining tail:The right-hand integral has its minimum phase at its left endpoint, with ${\varphi }^{\prime }(1)=2$ and nonzero amplitude $\sin 1$. Endpoint Laplace method therefore giveThe answer is exponentially small, rather than zero; applying only the interior Gaussian approximation misses the asymmetric boundary tail.