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Laplace's method estimates an integral whose real exponential weight concentrates near a phase maximum. A useful precise case is a smooth real phase $\varphi$ on a compact interval, with a unique interior maximum $t_0$, ${\varphi }^{\prime \prime }(t_0)<0$, and a continuous amplitude $a$ satisfying $a(t_0)\ne 0$. As $x\to +\infty$,The quadratic Taylor expansion of the phase and the rescaling $s=\sqrt{x}(t-t_0)$ turn the local contribution into a Gaussian integral. Outside any fixed neighborhood the compact-domain strict maximum supplies a positive phase gap, making the remainder exponentially smaller. On an unbounded interval one must separately check the localization condition for Laplace's method.^[ref]

For example,The peak width is $x^{-1/2}$. A nonstationary endpoint maximum has width $x^{-1}$ and a different prefactor, given by Laplace endpoint estimate. If the amplitude vanishes at the maximum, the displayed equivalence cannot be used with a zero right side: higher amplitude terms or cancellation, as in endpoint asymptotics after odd Gaussian cancellation, decide the leading contribution.

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.