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Gaussian Integral Using Residues

  1. Aug 18, 2005 #1
    I am trying to to the Gaussian integral using contour integration.

    What terrible mistake have I made.

    I = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x

    I consider the following integral:

    \int_C \mathrm{e}^{-z^2} \mathrm{d}z

    where [itex]C[/itex] is the half-circle (of infinite radius) in the upper-half plane.
    There are no singularities in the upper-half plane. So,

    \int_C \mathrm{e}^{-z^2} \mathrm{d}z = 0

    Now, I can break this integral up into two parts.

    0 = \int_C \mathrm{e}^{-z^2} \mathrm{d}z = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x + \int_{\theta=0}^{\theta=\pi} \mathrm{e}^{-z^2} \mathrm{d}z


    \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{d}x &= - \int_{\theta=0}^{\theta=\pi} \mathrm{e}^{-z^2} \mathrm{d}z =
    -\lim_{R\rightarrow \infty} \int_{0}^{\pi} \mathrm{e}^{-R^2 \exp(2\mathrm{i}\theta)}\cdot \mathrm{i} R \mathrm{e}^{\mathrm{i}\theta} \mathrm{d}\theta \stackrel{?}{=} \sqrt{\pi}

    I know the answer should be [itex]\sqrt{\pi}[/itex]...so the RHS must give this result! But it does not...at least, I have convinced myself that the RHS actually goes to zero. The exponential dominates the linear term, so in the limit as R goes to infinity, the RHS equals zero.

    How can this be...it seems I have shown that the gaussian integral is zero.

    I should cry.
  2. jcsd
  3. Aug 18, 2005 #2


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    The terrible mistake have you made is the Gaussian diverges to infinity on the arc. You could show that all the integrals of the Gaussian from 0 to points r*exp(i theta) are equal when sin(theta)<=cos(theta). This won't answer any thing though as you would still need to calculate one such integral. I feel that this integral is not best done by contour integration (though it is possible and I have seen it done). There are other easier ways.
  4. Aug 18, 2005 #3
    ...which becomes painfully obvious when I include the effect of the oscillations. It is true that R->infinity makes exp(-R^2 exp(2*I*theta)) very large...but the oscillation causes it to go back and forth between +infinity and -infinity.

    It seems that no matter which contour I pick (as far as infinite semi-circles go), I will have this issue. At iR and -iR, e^(-z^2) will have trouble. Thus, it seems that my contour CANNOT be anything that goes to infinity in the imaginary part.

    I know that the Guassian integral can be done easily in polar coordinates...that doesn't concern me. I _want_ to do it when complex integration. So how do you do it?
  5. Aug 19, 2005 #4


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    This page about statistics
    has an article called Information on the History of the Normal Law
    in which the desired integral is found 7 ways. #7 uses contour integration
    In specific the function f(z)=exp(pi i z^2)/sin(pi z) is integrated on the parallelogram with vertices +-1/2+-Rexp(pi i/4) where R goes to infinity.
    Also the oct 98' issue of the American Mathematical Monthly has an article on this integral (actually an integral that this one is a special case of).
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