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I want to compute the asymptotics of \begin{align*} \mathrm{AI}(x) = \frac{1}{2 \pi i} \int_C e^{i z x + i z^3/3} \frac{ dz}{z}, \quad x \in \mathbb R, \end{align*} as $x \to \pm \infty$. Here $C$ is a contour in the upper-half of the complex $z$-plane described by \begin{align*} C = \{ u(s) : s \in \mathbb R\}, \end{align*} for a sufficiently nice function $u$, oriented in the direction of increasing $s$. Suppose that \begin{align*} \lim_{s \to \pm \infty} |u(s)| = \infty, \quad \lim_{s \to + \infty} \frac{u(s)}{|u(s)|} = e^{i \pi/6}, \quad \lim_{s \to - \infty} \frac{u(s)}{|u(s)|} = e^{i 5\pi/6}. \end{align*} I need to compute the first term as $x \to +\infty$ and the first two terms as $x \to -\infty$ and determine the order of the first term neglected in both cases.

I set $x = |x| \sigma$ for $\sigma = \pm 1$, $x \neq 0$. Then use $|x|$ to rescale the integration variable $z$ so that the exponential in the integral can be written as $e^{|x|^\alpha h(z)}$ for a function $h$ depending only on $\sigma$ and $\alpha > 0$. The method of steepest descent applies to this integral.

I found four $z$ values for which $h'(z) = 0$. What should I do next to get the contour?

Gary
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hanamontana
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  • Look at $C$ after the change of integration variables. It should look like one a steepest descent contour passing though one of the saddles. Deform the contour so that it is exactly that steepest path (by an appeal to Cauchy's theorem). – Gary Feb 02 '24 at 01:11
  • I'm sorry I don't understand what you meant. The change of integration variables did not change C at all. – hanamontana Feb 02 '24 at 05:18
  • Happy to hear that. Now looking at your $C$ which path of steepest descent through which of the saddles looks very similar to your $C$? – Gary Feb 02 '24 at 05:22

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