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Does the Gaussian have complex poles?

  1. Oct 10, 2012 #1
    I know that

    \int_{- \infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.

    What if I tried evaluating this over the complex plane? If I created a contour that ran along the real axis, then circled around from positive to negative infinity this should not change things as along the curve at infinity my complex function should be 0. So using that contour, I should have

    \oint_C e^{-z^2} dz = 0,

    unless there are some poles in the upper half of the complex plane. But if there aren't, wouldn't this imply [itex]\int_{- \infty}^{\infty} e^{-x^2}dx = 0[/itex], since the integral over the other part of my contour is 0? I'm forced to conclude I am missing something here, but can't quite put my finger on what that might be.
  2. jcsd
  3. Oct 10, 2012 #2


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    So, your closed contour consists of a piece along the real axis, say from -R to R, as R tends to infinity, and then a circular contour of radius R. This circular arc part of the integral is

    $$\int_0^\pi d\theta Ri e^{i\theta} \exp\left(-R^2(\cos(2\theta)+i \sin(2\theta))\right).$$

    This integral is not zero, and your contour calculation shows that it must be ##-\sqrt{\pi}## as ##R \rightarrow \infty##. The reason this not zero in the limit as ##R## goes to infinity is that ##\cos(2\theta)## is not always positive on ##[0,\pi)##, so the ##\exp(-R^2\cos(2\theta))## does not drive the integral to zero.

    Note: it is actually pretty tough to show that ##\int_{-\infty}^\infty dx \exp(-x^2) = \sqrt{\pi}## via contour integral methods. The only example that I've seen which actually accomplishes it had a rather non-trivial integrand.
  4. Oct 11, 2012 #3
    Really I think that would be a sufficient proof that:

    [tex]\lim_{R\to\infty} \int_0^{\pi} e^{-z^2}dz=-\sqrt{\pi},\quad z=Re^{it}[/tex]

    However, it is based on the fact that the function [itex]e^{-z^2}[/itex] is entire and therfore has no poles in the finite complex plane.
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