= Solution
{c}
For $x>0$, the real and imaginary parts of the exponent divided by $x$ are
$$
\operatorname{Re}(it^3)=v^3-3u^2v=-r^3\sin3\theta,\qquad
\operatorname{Im}(it^3)=u^3-3uv^2=r^3\cos3\theta.
$$
Thus <constant-phase contours> satisfy $u^3-3uv^2=c$. Through zero, $c=0$ gives the six rays $\theta=\pi/6+j\pi/3$. The three rays $\theta=\pi/6,5\pi/6,3\pi/2$ have decreasing real part of the exponent and are <steepest descent contours>; the alternating three are steepest ascent rays.
Through one, the phase is $c=1$. Its branch passing through that point can be written
$$
t=(1+is)^{1/3},\qquad s\in\mathbb R,
$$
with the cube root continuous from $t=1$. The exponent is $ix-xs$, so increasing $s$ from zero gives descent and decreasing $s$ gives ascent.
Deform the finite interval into the ray $t=e^{i\pi/6}r$, $0\leq r<\infty$, followed by the reverse of the descent branch from one. The integrand is entire, so <Cauchy's integral theorem> permits the deformation.
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2009-cubic-descent.png]
{title=Cubic descent deformation with the endpoint branch traversed back toward one}
For a precise connector at infinity, use $t=(a+iR)^{1/3}$, $0\leq a\leq1$. Its length is $O(R^{-2/3})$ and the exponential has modulus $e^{-xR}$. Its contribution is therefore $O(R^{-2/3}e^{-xR})\to0$. This avoids relying on an undifferentiated claim that every large circular arc decays.
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