Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Asymptotic Expansion of an Integral

  1. May 20, 2012 #1
    I'm not exactly sure that what I want to do is an asymptotic expansion, but basically I would like to find a power series approximation in s of

    -s [itex]\int[/itex] [itex]\frac{e^{-x^{2}}}{x-s}[/itex] dx

    for large values of s.

    The integral is meant to be from -∞ to +∞.

    I can see that for large s, the denominator becomes approximately -s, and this cancels with the -s out the front of the integral, so that the whole thing is approximately equal to [itex]\sqrt{\pi}[/itex], but this is too much of an approximation. I would like to have some 'error' terms in there as well, after the [itex]\sqrt{\pi}[/itex], except I can't seem to get the right ones using Taylor expansions. Should I be looking to do asymptotic expansions instead? Or is there a way to do it with just power series?

    Thanks a lot!
  2. jcsd
  3. May 20, 2012 #2
    That integral has a non-integrable singularity at x=s. Is it meant to be a Cauchy principal value?
  4. May 21, 2012 #3
    Yes, I think it is. It's expansion is supposed to be

    [itex]\sqrt{\pi}[/itex](1+[itex]\frac{1}{2s^{2}}[/itex]+[itex]\frac{3}{4s^{4}}[/itex]+ higher order terms)

    I'm not really familiar with the Cauchy Principle Value.

  5. May 21, 2012 #4
    To calculate the principal value of an "integral" with singularity at point s, you cut out the interval (s-e,s+e) and integrate over the rest of the line. Then you take the limit as e goes to 0.

    For example, if your function is 1/x and your domain is [-1,1], then you have a non-integrable singularity at 0. So you cut out the interval (-e,e). You integrate over the rest of the domain and get zero in this case 1/x is odd. Then you take the limit as e goes down to zero.

    That aside, you can generate the formula you want by doing a formal calculation that ignores the principal value nature of that integral. You are interested in the asymptotic behavior as s gets large, so substitute t=1/s and now you are interested in the behavior for small t. A power series jumps out naturally, and you can get the coefficients with integration by parts. You can probably combine the formal calculation with the definition of principal value and rigorously justify all the steps.

    Another angle on this problem is that your integral is the Hilbert transform of a function that is its own Fourier transform. That might be exploitable for slick calculations and results. See the wikipedia page on the hilbert transform for more info.
  6. May 22, 2012 #5
    Hi !
    The asymptotic expansion is derived in attachment. The calculus involves the Dawson function.
    This method is equivalent to the method using the Hilbert's transform of exp(-x²), which involves the erfi function. The known relationship between erfi and Dawson functions leads to the same final result.

    Attached Files:

    Last edited: May 22, 2012
  7. May 22, 2012 #6
    Thanks for your help!! I've succeeded in deriving the power series approximation. I don't know about Hilbert Transforms yet, but it looks really nice, like Fourier Transforms or Laplace Transforms. Thanks for introducing it!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook