= Solution
{c}
The endpoint branch has $t=(1+is)^{1/3}$ and $dt=(i/3)(1+is)^{-2/3}ds$. Its reversed orientation in the deformed contour gives
$$
I_1(x)=-\frac{ie^{ix}}3\int_0^\infty e^{-xs}(1+is)^{-2/3}\,ds.
$$
At $s=0$,
$$
(1+is)^{-2/3}=1-\frac{2i}{3}s+O(s^2).
$$
<Watson's lemma>, or elementary <Laplace's method> with a Taylor remainder, now yields
$$
I_1(x)=-\frac{ie^{ix}}{3x}-\frac{2e^{ix}}{9x^2}+O(x^{-3}).
$$
Adding the exact cubic-saddle contribution gives the two leading terms
$$
\boxed{I(x)=e^{i\pi/6}\Gamma(4/3)x^{-1/3}-\frac{ie^{ix}}{3x}+O(x^{-2}).}
$$
If the next term is retained, it is $-2e^{ix}/(9x^2)$. The different powers reflect a degenerate saddle at zero and an ordinary endpoint at one.
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