1
$\begingroup$

This was asked in MSE, here, but the answer was not satisfactory.

I want to compute the asymptotic behavior of the integral $$ f(K,a)=\int_0^1 (1-x)^Ke^{iKa\frac{x}{1-x}}x^2dx$$ when $K$ is large and $0<a<1$. I tried two different approaches.

1) My first idea was that the exponential, a fast-oscillating function around $x=1$, is killed by the $(1-x)^K$, and the integral should be dominated by the vicinity of $x=0$. Therefore, I put $x=y/K$ and approximate $(1-y/K)^K\approx e^{-y}$ and $\frac{x}{1-x}\approx \frac{y}{K}$ to get

$$f(K,a)\approx \frac{1}{K^3}\int_0^\infty e^{-y+iay}y^2dy=\frac{2}{K^3(1-ia)^3}.$$

2) On the other hand, the stationary phase approximation should be valid. If I write $$f(K,a)=\int_0^1 e^{KS(x)}x^2dx,$$ with $S(x)=\log(1-x)+iax/(1-x)$, the equation $S'(x_0)=0$ gives $x_0=1-ia$. Second derivative is $S''(x_0)=-1/a^2$. Hence, this idea leads to $$f(K,a)\approx e^{KS(x_0)}x_0^2\sqrt{\frac{\pi a^2}{K}}=(ia)^Ke^{K(1-ia)}(1-ia)^2a\sqrt{\frac{\pi}{K}}.$$

These two results are completely different! I need help understanding this.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ Your method 1) can be made precise pretty easily, by bounding the part of the integral away from 0, and then applying Taylor expansions. On the other hand, your method 2) is very hand-wavy. There is no reason to think the main contribution should come from a stationary point with Re(x) = 1, since the integrand is very small except in a tiny neighborhood of x=0. $\endgroup$
    – Matt Young
    Apr 6, 2018 at 16:15
  • $\begingroup$ @MattYoung I favor method 1) as well, but I would like to know what is precisely the obstacle to applying a stationary phase approximation $\endgroup$
    – thedude
    Apr 6, 2018 at 16:24
  • $\begingroup$ The obstacle is that the stationary point is not close to $x=0$ (where the mass of the integrand is concentrated). You can think of stationary phase as applying a quadratic approximation to the phase centered at the stationary point. $\endgroup$
    – Matt Young
    Apr 6, 2018 at 16:46
  • 1
    $\begingroup$ Another obstacle is that there is no reason to climb to the stationary point when going from $0$ to $1$: you are in a deep valley all the way, so the "mountain path" approach makes no sense. You could easily detect it yourself by noticing that the answer is predicted to be of order $e^K$, which is much larger than all the values of the integrand. $\endgroup$
    – fedja
    Apr 6, 2018 at 17:19
  • $\begingroup$ @fedja "makes no sense" is a bit strong, no? I can see that using the stationary point might not be the best approach, but at least it should give the right answer, no? Why does it break down? Am I not allowed to deform the contour due to some singularity? $\endgroup$
    – thedude
    Apr 6, 2018 at 17:26

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.