= Solution
{c}
On the selected <steepest descent contour> put $t=e^{i\pi/6}r$, $r\geq0$. Then $it^3=-r^3$ and $dt=e^{i\pi/6}\,dr$. Its exact contribution is
$$
I_0(x)=e^{i\pi/6}\int_0^\infty e^{-xr^3}\,dr
=\frac{e^{i\pi/6}}{3x^{1/3}}\int_0^\infty e^{-s}s^{-2/3}\,ds.
$$
Use the <Gamma function> integral and its recurrence $\Gamma(4/3)=\Gamma(1/3)/3$ to obtain
$$
\boxed{I_0(x)=e^{i\pi/6}\Gamma(4/3)x^{-1/3}.}
$$
There is no remainder here: the complete infinite descent ray was integrated exactly.
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