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Non-trivial Gaussian integral

  1. Aug 6, 2015 #1
    Hi everyone,

    in the course of trying to solve a rather complicated statistics problem, I stumbled upon a few difficult integrals. The most difficult looks like:

    [tex] I(k,a,b,c) = \int_{-\infty}^{\infty} dx\, \frac{e^{i k x} e^{-\frac{x^2}{2}} x}{(a + 2 i x)(b+2 i x)(c+2 i x)} [/tex]

    where [tex] a,b,c [/tex] are real positive numbers and [tex] k [/tex] is a real number. This integral cannot be done by simple contour integration because of the Gaussian factor. From the context, I expect error functions to appear in the solution.

    Any clever ideas?

  2. jcsd
  3. Aug 6, 2015 #2


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    I think you can do this with a few standard tricks. First, assume that we can do partial fraction decomposition, so that we can consider three integrals of the form
    $$I_{k,a} = \int_{-\infty}^{\infty} dx\, \frac{e^{i k x} e^{-\frac{x^2}{2}} x}{ x - ia/2 }.$$
    Next, use the Schwinger parametrization
    $$ \frac{1}{x - ia/2 } = -i \int_0^\infty du\, e^{i u(x - i a/2)},$$
    $$I_{k,a} = -i e^{a/2} \int_0^\infty du\, \int_{-\infty}^{\infty} dx\, e^{i (u+k) x} e^{-\frac{x^2}{2}} x.$$
    Next, the factor of ##x## can be handled by replacing it with a derivative ##-i d/dk##, so
    $$I_{k,a} = -e^{a/2} \frac{d}{dk} \int_0^\infty du\, \int_{-\infty}^{\infty} dx\, e^{i (u+k) x} e^{-\frac{x^2}{2}} .$$
    The Gaussian integral is done by completing the square:
    $$I_{k,a} = - e^{a/2} \frac{d}{dk} \int_0^\infty du\, \sqrt{2\pi} e^{-(u+k)^2/2 } .$$
    Finally, we redefine variable to ##u'=u+k## and use the fundamental theorem of calculus to find
    $$I_{k,a} = \sqrt{2\pi} e^{(-k^2 + a)/2} .$$
    This has the right behavior in the limit as ##k,a\rightarrow 0##.

    Note that you can skip the partial fractions and just do a more complicated Schwinger parameterization in your original integral, but I'm not sure that would save too much effort.
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