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Gaussian integral to polar coordinates - limit help?

  1. May 29, 2012 #1
    I'm trying my very best to understand it, but really, I just couldn't get it. I read four books now, and some 6 pdf files and they don't give me a clear cut answer :(

    Alright, so this integral;
    ∫e-x2dx from -∞ to ∞, when converted to polar integral, limits become from 0 to 2∏ for the outer integral, then 0 to ∞ for the inner integral.

    What I don't understand is, why is it if the original integral is ∫e-x2dx from 0 to ∞, the outer integral's limits become ∏/2? Why is it not ∏?

    Further, what if I need to integrate the same function, say from 0 to 1? Will polar integration help me? If it does, what will happen?

    Thanks and more power guys.
     
  2. jcsd
  3. May 29, 2012 #2
    Oh, to add: I know the "apparent magic" that is supposed to happen here (The double integral thing by assigning the entire integral to a since variable, say I then squaring it, then the dummy variable thing, then convert dxdy to rdrd(theta)), but what I really don't get is the polar conversion of the limit. Thanks. :D
     
  4. May 29, 2012 #3
    So the region is all of the plane, so you started with the rectangular limits, x and y each go from negative to positive infinity.

    In polar coordinates, that (entire plane) corresponds to r going from 0 to infinity, while theta goes from 0 to 2pi.
     
  5. May 29, 2012 #4
    So if your original integral goes from 0 to infty, then you square it and get an integral over the part of the plane where both x and y go from 0 to infty. In other words, it is just over the first quadrant. And your calculated integral is 1/4 what you would have computed over the whole plane. So when you take the square root, you get 1/2 of the original integral that went from -infty to infty as expected.
     
  6. May 31, 2012 #5
    Thanks, I got it now why was it ∏/2.

    I couldn't get the explanation why did it became 2∏ when integrating from -∞ to ∞, though.

    Is it because, if my original integral goes from -∞ to ∞, then I square it, it would generate an area from -y to y axis, and -x to x axis? And since that is basically the entire Cartesian plane, I have to use 2∏ which is also the entire Polar plane? I don't know, I'm just building it from the ∏/2 example.

    A little follow up too, will this double integration w/ polar coordinates can solve integrals like, say, ∫e-x2dx from -6 to 3? I wanted accurate answers, but if it's too much hassle I guess I'm better off with the trapezoidal technique with very small stepsizes. Thanks guys.
     
  7. May 31, 2012 #6
    Hi !
    (details of integrations in attachment)
     

    Attached Files:

  8. May 31, 2012 #7
    Of course, in case of a finite domain of integration, if it is not circular, but rectangular, the polar coordinates are of no use : Integration would be much too complicated.
    Even with cartesian coordinates, the result cannot be expressed in terms of a finite number of elementary functions. It requiers infinite series or, on closed form, a special function namely the erf function (see attachment)
    Numerical values for the erf function are provided by numerical calculus softwares.
    Efficient algorithms can be found on the web. For example, an algorithm for erf computation is shown on page 14 of the paper "Régressions et équations intégrales" :
    http://www.scribd.com/JJacquelin/documents
     

    Attached Files:

    • Erf.JPG
      Erf.JPG
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    Last edited: May 31, 2012
  9. May 31, 2012 #8
    Oh! Thanks very much for the help. :D

    And yeah, just to confirm, is my thinking "correct"?

    Is it because, if my original integral goes from -∞ to ∞, then I square it, it would generate an area from -y to y axis, and -x to x axis? And since that is basically the entire Cartesian plane, I have to use 2∏ which is also the entire Polar plane? I don't know, I'm just building it from the ∏/2 example.

    I just wanted to be sure. :D
     
  10. May 31, 2012 #9
    Yes, it is correct.
     
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