Localization condition for Laplace's method

A unique nondegenerate interior minimum does not by itself ensure that a integral treated by Laplace's method is localized there. A sufficient condition is a positive gap above the minimum outside every sufficiently small fixed neighbourhood, together with finiteness of the integral for some positive exponent. The exterior contribution is then exponentially negligible. Without localization, $g(y)=y^2(1-y{)}^4$ on $(-1,1)$ has its unique interior minimum at zero with $g^{\prime \prime }(0)=2$, but its boundary contribution is at least $e^{-1}{n}^{-1/4}$ and dominates the nominal $n^{-1/2}$ interior Gaussian contribution.