= Solution
Take the nondegenerate <stationary point> in the interior of the integration interval. Choose a smooth cutoff equal to one near $k_\star$. Away from its support, $\psi'$ is bounded away from zero. <Integration by parts> using $e^{it\psi}=(it\psi')^{-1}\partial_ke^{it\psi}$ makes that nonstationary contribution $O(t^{-1})$, under the usual smooth-phase assumptions.
Near the <stationary point> use the signed coordinate
$$
s=\operatorname{sgn}(k-k_\star)\sqrt{2(\psi(k)-\psi(k_\star))}.
$$
Since $\psi''(k_\star)>0$, this is locally invertible, with $ds/dk=\sqrt{\psi''(k_\star)}$ at the point. The local <integral> becomes
$$
e^{it\psi(k_\star)}\int a(s)e^{its^2/2}\,ds,\qquad a(0)=\frac{f(k_\star)}{\sqrt{\psi''(k_\star)}}.
$$
Scale $s=\sqrt{2/t}\,r$. The supplied <Fresnel integral> gives $\int_{-\infty}^{\infty}e^{ir^2}dr=e^{i\pi/4}\sqrt\pi$. Approximating the smooth <amplitude> by its value at zero, with a cutoff and a nonstationary estimate for the tails, therefore gives
$$
\boxed{I(t)=f(k_\star)e^{it\psi(k_\star)+i\pi/4}\sqrt{\frac{2\pi}{t\psi''(k_\star)}}+o(t^{-1/2}).}
$$
If $f(k_\star)\ne0$, this is the requested asymptotic equivalence. The phase shift $\pi/4$ comes from the positive quadratic curvature. A <stationary point> at an integration endpoint instead needs a one-sided formula.
The printed auxiliary estimate asserting an $O(\gamma^{-1})$ Fourier <integral> for every continuous <amplitude> is too strong without extra regularity. The argument here uses <integration by parts> only for the actual smooth, cutoff nonstationary amplitudes, where that estimate is justified; continuity alone generally guarantees convergence to zero, not this rate.
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