Past exam of the theoretical mechanics course of the University of Oxford / 2011 / B / 5 / a / ii / Solution

For a negative second derivative, the corresponding local normal form is $\psi =\psi (k_{\star })-s^2/2$. The oscillatory Gaussian integral is the complex conjugate of the positive-curvature one, so the phase shift changes sign. ThusWhen the leading amplitude is nonzero, the equality may again be written as an asymptotic equivalence. Both signs are summarized by the factor $e^{i\pi \mathrm{sgn}({\psi }^{\prime \prime })/4}$.