= Solution
Split $I$ into the two tails and the interval $|k-k_*|<\delta$. On each tail $\psi'$ 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 change
$$
z=\operatorname{sgn}(k-k_*)\sqrt{2[\psi(k)-\psi(k_*)]/\psi''(k_*)}.
$$
It has $dk/dz=1$ at zero and makes the phase exactly $\psi(k_*)+\psi''(k_*)z^2/2$. The transformed amplitude tends to $f(k_*)$. Scale $z$ by $t^{-1/2}$ and use the <Fresnel integral> to obtain
$$
\boxed{I(t)=f(k_*)e^{it\psi(k_*)+i\pi/4}\sqrt{\frac{2\pi}{t\psi''(k_*)}}+o(t^{-1/2}).}
$$
The $O(t^{-1})$ tails are smaller than this leading stationary contribution. If $\psi''(k_*)<0$, complex conjugation of the Gaussian phase changes the phase correction to $-\pi/4$ and replaces the second derivative by its absolute value:
$$
\boxed{I(t)=f(k_*)e^{it\psi(k_*)-i\pi/4}\sqrt{\frac{2\pi}{t|\psi''(k_*)|}}+o(t^{-1/2}).}
$$
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.
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