= Solution
Use the entire function $e^{ikz^3}$. The <steepest descent contour> starting at $1$ is
$$
\boxed{z(u)=(1+iu)^{1/3},\quad0\leq u<\infty,}
$$
with the cube-root branch starting at $1$. Along it $ikz^3=ik-ku$, so the real part decreases and the imaginary part stays constant. It approaches the ray $\arg z=\pi/6$, on which $z=e^{i\pi/6}u^{1/3}$ and $ikz^3=-ku$.
Close a contour using the interval $[0,1]$, these two outgoing paths truncated at parameter $R$, and a short connecting arc. The two endpoints differ by $O(R^{-2/3})$, and the integrand on the connector decays exponentially in $R$. <Cauchy integral theorem> therefore gives $\int_0^1e^{ikt^3}\,dt=I_1-I_2$.
The specified substitutions give
$$
I_1=\frac{e^{i\pi/6}}3\int_0^\infty e^{-ku}u^{-2/3}\,du=\frac{e^{i\pi/6}\Gamma(1/3)}{3k^{1/3}},
$$
and
$$
I_2=\frac{ie^{ik}}3\int_0^\infty e^{-ku}(1+iu)^{-2/3}\,du.
$$
Expand $(1+iu)^{-2/3}=1-2iu/3+O(u^2)$ near zero, and apply <Watson lemma>, or split the integral at a fixed small $u$ and integrate the Taylor remainder. This gives \b[the first two asymptotic terms]
$$
\boxed{\int_0^1e^{ikt^3}\,dt=\frac{e^{i\pi/6}\Gamma(1/3)}{3k^{1/3}}-\frac{ie^{ik}}{3k}+O(k^{-2}).}
$$
Keeping the next term gives $-2e^{ik}/(9k^2)+O(k^{-3})$. The fractional power comes from the degenerate stationary endpoint at zero; the oscillatory integer-power terms come from the ordinary endpoint at one.
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