Show that $\int_{0}^{\infty} e^{ix(\frac{t^3}{3}+t)}dt \sim \frac{i}{x}$ Show that $\int_{0}^{\infty} e^{ix(\frac{t^3}{3}+t)}dt \sim \frac{i}{x}$
I thought that you would use the method of stationary phase, but the maximum of $\frac{t^3}{3}+t$ occurs at $+/- i$. So how do I progress?
 A: The idea is not to use stationary phase here, but note that $t+\frac13t^3$ is monotonically increasing to make a change of variables
$$
u=t+\tfrac13t^3
$$
Inverting the function, we get near $u=0$,
$$
t=u-\tfrac13u^3+\tfrac13u^5-\tfrac49u^7+O(u^9)
$$
As $u\to\infty$, $t\sim(3u)^{1/3}$. In any case, when $t$ stays within the cone between $1$ and $e^{i\pi/12}$, $u$ stays within the cone between $1$ and $e^{i\pi/4}$. There is a similar bound for $t$ in terms of $u$. The point is that we can modify the contour in both $t$ and $u$ to be at a small angle from the real axis. That is, $u([0,e^{i\pi/12}\infty])$ is a curved path that goes from $0$ to $e^{i\pi/4}\infty$. Thus, we get
$$
\int_0^{e^{i\pi/12}\infty} e^{ix(t+\frac13t^3)}\,\mathrm{d}t=\int_0^{e^{i\pi/4}\infty}e^{ixu}\left(1-u^2+\tfrac53u^4-\tfrac{28}9u^6+O(u^8)\right)\,\mathrm{d}u
$$
Now we can evaluate the individual terms by changing the contour from $u([0,e^{i\pi/12}\infty])$ to $[0,i\infty]$. Thus, we get
$$
\begin{align}
\int_0^\infty e^{ix(t+\frac13t^3)}\,\mathrm{d}t
&\sim i\int_0^\infty e^{-xu}\left(1+u^2+\tfrac53u^4+\tfrac{28}9u^6+O(u^8)\right)\,\mathrm{d}u\\[6pt]
&=i\left(\frac1x+\frac2{x^3}+\frac{40}{x^5}+\frac{2240}{x^7}+O\left(\frac1{x^9}\right)\right)
\end{align}
$$

Numerical Verification
Using Mathematica to numerically evaluate the integral at $x=1000$, I got
$$
i\,0.\underbrace{001}_{\large\frac1{x}}\underbrace{000002}_{\large\frac2{x^3}}\underbrace{000040}_{\large\frac{40}{x^5}}\underbrace{002240}_{\large\frac{2240}{x^7}}\dots
$$
