Choice of deformed contour in steepest descent method

Question

I'm trying to learn about the method of steepest descent (from https://people.math.osu.edu/tanveer.1/m805/supplement.steepest.pdf ).

An example there is $$\int_{-\infty}^{+\infty}e^{i\lambda(t+t^3/3)}dt$$ where $\lambda\rightarrow \infty$. We have two saddle points, $t_0=\pm i$ at which the first derivative of $p=-i(t+t^3/3)$ is zero. Now, I'm trying to deform the contour to pass from either or both these points. Due to arguments that can be found in the link, the two deformations are (see p.7):

1. $C_1$: connects $\infty e^{5i\pi/6}$ to $\infty e^{i\pi/6}$, passing through $+i$. At the latter, it's parallel with the real axis.
2. $C_2$: connects $\infty e^{-5i\pi/6}$ to $\infty e^{-i\pi/6}$, passing through $-i$. At the latter, it's parallel with the real axis.

The deformed contours cannot cross the real axis, so we can't deform either in order to get to both points with one contour. What I don't understand is how we choose one over the other. The author chooses $C_1$ and simply states:

"It is to be noted that even though $Re (p)$ is smaller at the other saddle $t = − i$, there is no contribution from this since the steepest descent path equivalent to the original contour integral does not pass through $t = − i$".

So, not only I do not understand why $C_2$ is equivalent to the original contour while $C_1$ is not, I don't get why the author even studied the $C_2$ case. Any help will be appreciated.

EDIT: Since physicists tend to use a lot of saddle point approximations, maybe the question should transition to physics stackexchange?

Answer 1

Let $\phi(t) = \mathrm i(t + t^3/3)$, and let $t$ = $|t|e^\mathrm{ i\theta}$. Then $$ \Re(\phi(t)) = -\left(\sin\theta|t| + \frac{1}{3}\sin(3\theta)|t|^3\right) \sim \frac{1}{3}\sin(3\theta)|t|^3, \quad |t|\to\infty. $$ Thus $e^{\lambda\phi(t)}$ will decay exponentially as $|t|\to\infty$ in the sectors where $\Re(\phi(t)) < 0$, which is when $$ 0 < \theta < \pi/3, \qquad 2\pi/3 < \theta < \pi, \qquad 4\pi/3 < \theta < 5\pi/3. $$ This allows you to connect (loosely speaking) $-\infty$ to the start of $C_1$ and the end of $C_1$ to $+\infty$, without changing the value of the integral, which is a necessary part of deforming the contour onto $C_1$. The same is not true of $C_2$, since then you would be trying to connect the end points to the real line through sectors where the integrand is growing exponentially.

Answer 2

Here are the main points:

1. The argument of the exponential function (divided by $\lambda$) is $-p(t)=i(t+t^3/3)$ with stationary points $t_{\pm}=\pm i$. The direction of steepest descent from $t_{\pm}$ in the complex $t$-plane is horizontal (vertical), respectively.

2. We can define anti-Stokes curves ${\rm Re}p(t)={\rm Re}p(t_{\pm})=\pm\frac{2}{3}$ emanating from the critical points $t_{\pm}$. It is clear that asymptotically for $|t|\to\infty$, the anti-Stokes curves approach lines with slopes $\arg t\in\frac{\pi}{3}\mathbb{Z}$.

3. The anti-Stokes curves divide the complex $t$-plane into sectors. It turns out that the steepest descent contour should only cross anti-Stokes curves at stationary points.

4. The undeformed integration contour (= the real $t$-axis) can be deformed to the steepest descent contour $C_1$ for $t_+$ within the (closure of the) same sector.

5. The same is not possible for the steepest descent contour for $t_-$.

6. $C_2$ is the steepest ascent contour for $t_-$.