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In these lecture notes (http://www.math.ucla.edu/~tao/247b.1.07w/notes8.pdf, section 3), Terence Tao approximates the following integral for large $\lambda$ using the stationary phase method:

$\begin{align} I_{a,\phi}(\lambda) &:= \int_{\mathbf{R}^d}\ a(x) e^{i \lambda \phi(x) } dx\\ &= \sum_{n=0}^N c_n \lambda^{-n - \frac{d}{2}} e^{i\lambda \phi(x_0)} + O_{N,a,d,\phi}(\lambda^{-N -\frac{d}{2}-1}) \end{align}$

Here, $\phi$ has a single non-degenerate critical point. That much is fine. My question is about the following comment (pg 11):

The situation gets significantly more complicated when the det $\nabla^2 \phi$ vanishes; for instance, factors of log $\lambda$ begin to appear in the asymptotic expansion.

-What's a simple example of an integral that has such log $\lambda$ terms?

-Is there a heuristic way of finding the leading log $\lambda$ term (the way the leading term of the usual stationary phase approximation comes from a Gaussian integral)?

marlow
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1 Answers1

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Here is an example:

$I(\lambda) = \int_0^1 dx \int_0^x dy\ e^{i \lambda(x^2 - y^2) }$

Note that $\phi(x,y) = x^2 - y^2$ has a single stationary point, $(x,y)= (0,0)$, and that $\text{det}\ \nabla^2 \phi =0 $.

Direct evaluation

With some help from Mathematica, we can evaluate this integral in terms of a generalized hypergeometric function and then take $\lambda$ to be large:

$\begin{align} I(\lambda) &=\frac{1}{2}\ _2 F_2(1,1;\frac{3}{2}, 2; i \lambda)\\ &\to \frac{i}{4}\frac{\ln \lambda}{\lambda} +O\left( \frac{1}{\lambda} \right) \qquad \left(\lambda \to \infty\right) \end{align}$

Stationary phase approximation

The method used below is described in chapter 8 of "Asymptotic Approximations of Integrals", by R. Wong. This will be a heuristic version to leading order, adapted to this example, of what he does rigorously to all orders in a more general setting.

Setting $h(t) = \int_0^1 dx \int_0^x dy\ \delta(x^2-y^2 -t)$, we have:

$\begin{align} I(\lambda) = \int_0^1 e^{i \lambda t} h(t) \end{align}$

One finds: $\begin{align} h(t) = -\frac{1}{4} \ln t + (\text{analytic in } t) \end{align}$

Then:

$\begin{align} I(\lambda) \sim \int_0^\infty dt\ e^{i \lambda t} \left( - \frac{1}{4} \ln t \right) = \frac{i}{4} \frac{\ln \lambda}{\lambda} + \frac{1}{4}\left( \frac{\pi}{2} + i \gamma \right)\frac{1}{\lambda} \end{align}$

We can get the error estimate by noting that $\int_1^\infty dt\ e^{i \lambda t} \ln t$ and $\int_0^1 dt\ e^{i \lambda t}\left( \text{analytic in } t \right)$ are both $O\left(\frac{1}{\lambda}\right)$ as $\lambda\to\infty$.

marlow
  • 793