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Consider the following integral: $$ I(t)=\int_{\mathbb{R}}e^{itp(z)}dz $$

where $p(z)$ is a real-valued polynomial. And suppose it has both real and non-real critical points, how to find the asymptotics when $t$ goes to positive infinity. Do we only need to consider the real critical points(i.e. stationary point)? If not, then when we choose the steepest descent contour, do we need path all the critical points, or only need to find a contour (homotopic to the real line) passing some of the saddle points?

A specifical case: take $p=(z+1)(z-1)(z+i)(z-i)$.

Correction: Specifical case take $p'(z)=(z+1)(z-1)(z+i)(z-i)$, so the critical points are $\pm 1,\pm i$

DuFong
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  • the asymptotic of a polynomial at infinity is defined by the order of the polynomial and the coefficient of it maximum power. Im not sure if this is hat you are asking – Masacroso Apr 12 '19 at 03:18
  • @Masacroso, I need the asymptotic expansion of that integral, or probably just the leading term – DuFong Apr 13 '19 at 13:16

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In your example, the only critical point of $p$ is $z = 0$. $\operatorname{Re} i p$ is constant on the lines $u v = 0$ and $v = \pm u$, dividing the plane into eight octants, and $\operatorname{Im} i p$ is constant on the bisectors of the octants. The goal is to prove that the integral is asymptotically equivalent to the integral over a small part of a bisector around $z = 0$, then any other critical points will be irrelevant.

For that, you need to choose the bisectors of those angles inside which the contour can be deformed to the real line while $\operatorname{Re} i p$ remains less than its value at $z = 0$. There is only one pair of angles that fits the requirement.

Maxim
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  • Sorry about the example. What I really mean is the critical points are $\pm i, \pm 1$, which are not all real. – DuFong Apr 13 '19 at 13:13
  • This is determined by the configuration of the problem. You have to analyze where the valleys and mountains corresponding to the critical points are located. Take this integral: $$\int_{\mathbb R} \exp \left( t \left( -\frac {z^4} 4 - \frac {1 - 2 i} 3 z^3 + \frac {1 + 2 i} 2 z^2 + z \right) \right) dz.$$ The steepest descent contour has to enter the critical point $z = i$ in the direction $e^{i \pi/4}$ and to leave the critical point in the direction $e^{-i \pi/12}$. The critical point $z = -1$ can be ignored. – Maxim Apr 13 '19 at 15:02
  • I cannot figure out the direction $e^{i\pi/4}$. Due to my calculation, $\phi'''(z)|_{z=i}=-2-2i$, hence the steepest descent angel $\theta$ need to max/min the $\phi'''(i)/3(z-i)^3$, that is $-2\sqrt{2}\frac{r^3e^{3i\theta+i5\pi/4}}{3}$, and we need $3\theta+5\pi/4=0,\pi$, hence $\theta=-5\pi/12$ and $\theta=-\pi/12$. However, if we choose $arg(\phi'''(i))=-3\pi/4$, then the we have $\theta=\pi/4$. What's the reason ? – DuFong Apr 14 '19 at 00:29
  • $3 \theta + 5 \pi/4 = 2 \pi$ is just as good as $3 \theta + 5 \pi/4 = 0$ :). You still need to determine which of the six sectors cover the real axis: https://i.imgur.com/Xp4NCK1.png (also notice the fourth mountain region). – Maxim Apr 14 '19 at 05:34