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Concerning the Gaussion integral in polar coordinates

  1. Mar 9, 2008 #1
    I'm looking at the proof for the Gaussion integral in polar coordinates and I don´t understand why theta reaches from 0 to 2pi in the integral since you can´t get a negative value out of an exponential function (and therefor the exponential function is never in the 3rd and fourth quadrant, which spans pi to 2pi). Here's the proof

    http://upload.wikimedia.org/math/8/5/b/85bb26bab98e69735c439dcfee9807d6.png part 1

    http://upload.wikimedia.org/math/f/1/0/f10de92bc974482b9f714bdaa7fda10d.png part 2
  2. jcsd
  3. Mar 9, 2008 #2


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    Hi naggy! :smile:

    (btw, if you type alt-p, it prints π)

    It has nothing to do with the integrand (in this case, an exponential).

    It's only conerned with the limits of integration, in changing from ∫∫dxdy to ∫∫drdtheta.

    ∫∫dxdy was over the whole plane.

    So ∫∫drdtheta must be also. That means over all r and all theta.

    r can't be negative, so theta has to go all the way round, from 0 to 2π. :smile:
  4. Mar 9, 2008 #3
    When you change it to a double integral, you go from [tex]R^2[/tex] to [tex]R^3[/tex], and [tex]e^{-x^2}[/tex] and [tex]e^{-y^2}[/tex] are defined for all x and all y, so in effect both exponential functions are in all two dimensional quadrants.
  5. Mar 9, 2008 #4


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    It is the value of the exponential function that is never never negative, not the variables.
  6. Mar 9, 2008 #5


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    … integrating over neverland …

    naggy, he means [tex]R^2[/tex] (… you knew that, didn't you? … :smile:)
    Do you mean never never never never negative, or never never never never never negative? :confused:
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