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Homework Help: Exponential Integral

  1. Nov 20, 2008 #1
    Need help solving this.

    [tex]\int^{\infty}_{-\infty}[/tex] [tex]e^{\frac{-x^{2}}{\sigma^{2}}}[/tex] [tex]e^{-ikx}dx[/tex]

    That's the integral of the product of the exponentials , couldn't get latex to make it look right.

    Supposedly usefull information(I can't see how);
    [tex]\int^{\infty}_{-\infty}[/tex] [tex]e^{\frac{-x^{2}}{\sigma^{2}}}[/tex]dx =[tex]\sigma\sqrt{\pi}[/tex]

    Not looking for answers, just suggestions of methods. I have been trying to expand the complex exponential via the Euler theorem and then use integration by parts to solve, but can't get anywhere.

    If anyone knows of an integration method I can look up to deal with this your help would be appreciated.
  2. jcsd
  3. Nov 20, 2008 #2
    Are you familiar with the method of contour integration to evaluate improper definite integrals?

    If so, consider completing the square in the exponent, so that, if I is your original integral, we have [tex]\Large I = e^{\frac{-\sigma^2k^2}{4}}\lim_{a,b\to \infty} \int^b_{-a} e^{\frac{-1}{\sigma^2}\left(x+\frac{i\sigma^2 k}{2}\right)^2} \, dx[/tex]. Then you can look to close the contour with a rectangle and apply Cauchy's integral theorem.
  4. Nov 20, 2008 #3
    No, I don't know anything about contour integration or Cauchy's integral theorem. I also doubt my professor would expect us to.
    But thanks for the suggestion. I'll look into that at some point, it just won't be useful to me right now.
  5. Nov 20, 2008 #4


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    You only need contour integration to know that the integral of from x=-inf to x=inf of exp(-(x-c)^2), where c is a complex constant, is the same as the integral of exp(-x^2). If you want to know why, it's because exp(-(x-c)^2) has no poles. If you don't want to know why, then just do the completing the square that Unco suggests and do the change of variable u=(x-c) without asking why it works. This might be what your professor expects.
  6. Nov 21, 2008 #5
    Thanks, Dick and Unco. That is all I needed to do. Last night I was too worn out and unnecessarily threw out all of Unco's post at the sight of "contour integration".
    It's always the simple things I miss while trying to make problems much more complicated then they should be.
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