= Solution
The endpoint $t=0$ is treated by rotating onto its <method of steepest descent> ray. It contributes
$$
\Gamma(3/2)(-ix)^{-3/2}=\frac{\sqrt\pi}{2}(-ix)^{-3/2}.
$$
At $t=1$, put $s=1-t$ and expand
$$
(1-s)^{1/2}=\sum_{n=0}^\infty(-1)^n\binom{1/2}{n}s^n.
$$
Termwise endpoint integration, using $\int_0^\infty s^ne^{-ixs}ds=n!(ix)^{-n-1}$ on the rotated ray, gives
$$
I(x)\sim\frac{\sqrt\pi}{2}(-ix)^{-3/2}
-e^{ix}\sum_{n=0}^\infty a_n(-ix)^{-n-1},
$$
where
$$
a_n=n!\binom{1/2}{n}.
$$
The two displayed contributions come from the two endpoints; the remaining deformed contour is exponentially or algebraically smaller after any fixed truncation.
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