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Asymptotic form of Fourier type integral

  1. Jun 18, 2011 #1
    I have an integral like

    [tex]F(\lambda)=\int_{-\infty}^\infty e^{i\lambda x} f(x) dx,[/tex]
    where [itex]\lambda[/itex] is a real parameter and [itex]f(x)[/itex] is an integrable function of x. I am looking for a method to calculate an approximate form of [itex]F(\lambda)[/itex] for very small [itex]|\lambda|[/itex]. Methods like stationary phases or steepest descent can sometimes be used to calculate similar asymptotic expressions for large values of the parameter, but I am not sure how to proceed in case [itex]\lambda[/itex] is small.

    Thanks.
     
  2. jcsd
  3. Jun 18, 2011 #2

    Mute

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    A couple of ideas:

    You could try changing variables. For example, letting [itex]y = \lambda x[/itex] gives

    [tex]F(\lambda) = \frac{1}{\lambda}\int_{-\infty}^{\infty} dy e^{iy} f\left(\frac{y}{\lambda}\right).[/tex]

    If you know the large argument asymptotics of f(x), you could then expand f(y/lambda) in an asymptotic series and hopefully the resulting integrals for that are simpler. There is an issue with the y = 0 contribution to the integral - however, since it's just a point I don't think it contributes.

    The other option is to expand

    [tex]F(\lambda) \approx F(0) + F'(0)\lambda[/tex]

    where

    [tex]F(0) = \int_{-\infty}^\infty dx~f(x)[/tex]
    [tex]F'(0) = i\int_{-\infty}^{\infty} dx~x f(x)[/tex]

    This assumes of course that you can do these integrals and that they exist.
     
  4. Jun 19, 2011 #3
    Thanks, but the problem with the first suggestion is that f(x) only admits a Taylor expansion for large values of its argument, so the expansion of [itex]f(y/\lambda)[/itex] is not integrable from [itex]-\infty[/itex] to [itex]+\infty[/itex]. To give its exact form, f(x) is given by

    [tex]f(x)=\frac{(a^2+x^2)^{-b}}{x+ic},[/tex]
    where [itex]a,b,c>0[/itex] and [itex]b<<1[/itex].

    Unfortunately the second suggestion doesn't work either, since [itex]x f(x)[/itex] is not integrable.
     
  5. Jun 19, 2011 #4

    Mute

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    The expansion you would need to do is a large-argument expansion, because y/lambda would be large. It looks like the leading contribution from a large y/lambda expansion would be [itex](\lambda/y)^{1-2b}[/itex], although wolframalpha claims the Fourier transform of that is zero for 0 < Re(b) < 1/2. Hopefully the leading correction to the asymptotic series for [itex]f(y/\lambda)[/itex] is non-zero and integrable.

    Yes, if [itex]F(\lambda)[/itex] is not an analytic function of lambda, this won't work.
     
  6. Jun 19, 2011 #5
    I am probably missing something but, [itex]y/\lambda[/itex] is not large on the whole interval of integration: no matter how small [itex]\lambda[/itex] is, there will always be an interval around [itex]y=0[/itex] over which [itex]y/\lambda[/itex] is small. In this particular case, the Taylor expansion of the integrand is only valid for large enough [itex]y/\lambda[/itex]; to be exact, for [itex]y/\lambda>\max(a,c)[/itex]. Even if we wanted to integrate this expansion, the leading term, which I think is actually [itex](y/\lambda)^{-1-2b}[/itex], is not integrable over [itex]\mathbb{R}[/itex], due to its behavior near the origin.
     
  7. Jun 19, 2011 #6
    Whether you can use stationary phase/steepest descent depends on the derivatives of f(x) in relation to lambda. You might still be able to use it.

    What's your f?
     
  8. Jun 19, 2011 #7
    Hi, my f is
    [tex]f(x)=\frac{(a^2+x^2)^{-b}}{x+ic}.[/tex]
     
  9. Jun 19, 2011 #8
    For some rational values of b you might find a closed form solution via complex contour integration.

    My gut tells me you are going to see the gamma function and simple confluent hypergeometric functions but only for some b.

    Have you got access to a copy of Gradshteyn & Ryzhik? Or Abramowitz & Stegun? Your library should have these, and frankly so should you.
     
  10. Jun 19, 2011 #9
    Also, you could do a Taylor series using the first two or three terms of the exponential, leave f(x) as it is. You will probably find closed form integrals for those, especially in the two references I gave.
     
  11. Jun 19, 2011 #10
    b is usually not rational, and I don't think it's possible to solve this integral by contour integration for irrational b. Expanding the exponential in a Taylor series doesn't work, unfortunately (see posts 3-4).
     
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