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Homework Help: Change of limits when integrating with polar coordinates

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    ∫ e^([tex]\pi[/tex]x^2) dx, with limits -∞ to ∞

    2. Relevant equations

    ∫∫ dxdy = ∫∫ rdrdθ

    3. The attempt at a solution

    Hi, here's what I've done so far:

    Introduce a dummy variable y to get

    ∫∫ e^[tex]\pi[/tex](x^2 + y^2) dxdy, with limits -∞ to ∞ for both dx and dy

    Introduce polar coordinates: x^2 + y^2 = r^2

    The equation becomes:

    ∫∫ e^([tex]\pi[/tex]r^2) rdrdθ

    But I don't know how to change the limits. Am I right in that the r limits stay the same and the θ limits change to [tex]\pi[/tex]/2 and -[tex]\pi[/tex]/2?

    If this is right, when I integrate the first part, I end up with 0. Is this correct?
    Thanks for any help.

    P.S. I don't know why the pi is higher than the other figures, but it's meant to be at the same level!
  2. jcsd
  3. Oct 22, 2009 #2
  4. Oct 23, 2009 #3


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    Science Advisor

    Well, first of all, the integral you give will not converge. [itex]e^{\pi x^2}[/itex] go to infinity to fast at each end. I am going to assume you meant [itex]e^{-\pi x^2}[/itex]
    Since your integration includes the entire plane, you need for r to go from 0 to [itex]\infty[/itex] and [itex]\theta[/b] to go from 0 to [itex]2\pi[/itex].

    But you might find it easier to use the fact that [itex]\int_{-\infty}^\infty e^{-\pi x^2}dx= 2\int_0^\infty e^{-\pi x^2}dx[/itex] so that the double integral is restricted to the first quadrant. r still goes from 0 to [itex]\infty[/itex] but [itex]\theta[/itex] goes from 0 to [itex]\pi/2[/itex].
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