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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.