I am looking to evaluate the following asymptotic integral:

Find the leading term of asymptotics as $\lambda\to\infty$

$$I(\lambda)=\int_0^1\cos(\lambda x^3)dx$$

Using method of steepest descent along a certain contour. I am having trouble approaching this problem as I don't understand it well. Any help would be appreciated.

  • $\begingroup$ Writing $\int_0^1dx=\int_0^{\infty}dx-\int_1^{\infty}dx$ simple integration by parts will give everything you need $\endgroup$
    – tired
    Oct 19 '16 at 12:39

To start, recognize that

$$ I(\lambda) = \operatorname{Re} \int_0^1 e^{i\lambda x^3}\,dx. $$

Now there are several questions that you can ask to get yourself going:

  1. Where is the saddle point?

  2. What are the paths of steepest descent away from the saddle point?

  3. How can I deform my contour so that it follows this path of steepest descent?

The last one is a bit tricky since the endpoints of the contour are finite. You only need to follow a portion of the path of steepest descent though; you can have the contour return to its start/endpoint afterwards.

Let me know if you get stuck on any of these.


A general blueprint to solve this type of problems is explained in my Math.SE answer here. Applied to OP's example, one derives that

$$ \int_0^1 \! \mathrm{d}z~e^{i\lambda z^3}~=~J(0)-e^{i\lambda}J(1), \tag{1}$$


$$ J(a)~=~\frac{i}{3} \int_0^{\infty} \! \mathrm{d}u~ \frac{e^{-\lambda u}}{(iu+a^3)^{\frac{2}{3}}} .\tag{2} $$

The integral associated with the upper endpoint $a=1$ yields

$$ J(a\!=\!1)~\stackrel{(2)}{=}~\frac{i}{3} \int_0^{\infty} \! \mathrm{d}u~ e^{-\lambda u}\left( 1 - \frac{2i}{3}u + O(u^2) \right) ~=~ \frac{i}{3\lambda} + \frac{2}{9\lambda^2} + O(\lambda^{-3}) .\tag{3} $$

The integral associated with the lower endpoint $a=0$ yields

$$ J(a\!=\!0)~\stackrel{(2)}{=}~ \frac{e^{\frac{i\pi}{6}}}{3} \int_0^{\infty} \! \mathrm{d}u~ \frac{e^{-\lambda u}}{u^{\frac{2}{3}}}~=~ \frac{e^{\frac{i\pi}{6}}\Gamma\left(\frac{1}{3}\right)}{3\lambda^{\frac{1}{3}}}.\tag{4} $$

This leads to the OP's sought-for expansion

$${\rm Re} \int_0^1 \! \mathrm{d}z~e^{i\lambda z^3}~\stackrel{(1)+(3)+(4)}{=}~ \frac{\Gamma\left(\frac{1}{3}\right)}{2\sqrt{3}\lambda^{\frac{1}{3}}} + \frac{\sin(\lambda)}{3\lambda} + O(\lambda^{-2}).\tag{5} $$

  • $\begingroup$ I am wondering if the trick on P284 of B&O still works here, there the authors employed a vertical segment path instead of the true path, it worked well for their problem $\exp(i\lambda t^2)$. But here it's $z^3$ instead of $z^2$. Meaning with the same trick of substituting $z\to 1+i v$, the exponent contains $- 3\lambda v + \lambda v^3$, the second term diverges at infinity. $\endgroup$
    – Taozi
    Oct 26 '20 at 20:45

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