= 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. Thus
$$
\boxed{I(t)=f(k_\star)e^{it\psi(k_\star)-i\pi/4}\sqrt{\frac{2\pi}{t|\psi''(k_\star)|}}+o(t^{-1/2}).}
$$
When the leading <amplitude> is nonzero, the equality may again be written as an asymptotic equivalence. Both signs are summarized by the factor $e^{i\pi\operatorname{sgn}(\psi'')/4}$.
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