 Solution

Split $I$ into the two tails and the interval $|k-k_{\ast }|<\delta$. On each tail ${\psi }^{\prime }$ has a fixed nonzero sign, so changing variable to $s=\psi (k)$ produces a nonstationary Fourier integral. Under the usual decay and regularity needed for integration by parts, these pieces are $O(t^{-1})$.

Near the nondegenerate stationary point, make the signed local changeIt has $dk/dz=1$ at zero and makes the phase exactly $\psi (k_{\ast })+{\psi }^{\prime \prime }(k_{\ast })z^2/2$. The transformed amplitude tends to $f(k_{\ast })$. Scale $z$ by $t^{-1/2}$ and use the Fresnel integral to obtainThe $O(t^{-1})$ tails are smaller than this leading stationary contribution. If ${\psi }^{\prime \prime }(k_{\ast })<0$, complex conjugation of the Gaussian phase changes the phase correction to $-\pi /4$ and replaces the second derivative by its absolute value:When the leading coefficient is nonzero this may be written as an asymptotic equivalence. For full rigor, continuity of an arbitrary amplitude alone is not sufficient for the stated nonstationary $O(t^{-1})$ rate, nor does it ensure convergence of an infinite integral. Piecewise differentiability with integrable derivative, bounded variation, or an equivalent localization/decay hypothesis supplies the usual missing assumption. The exponentially decaying beam amplitude below has the required piecewise regularity.