Change of limits when integrating with polar coordinates

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SUMMARY

The discussion centers on the integration of the function ∫ e^{-\pi x^2} dx over the limits -∞ to ∞, transitioning to polar coordinates for evaluation. The correct transformation involves changing the limits for r from 0 to ∞ and θ from 0 to 2π. The integral does not converge if e^{\pi x^2} is used; instead, e^{-\pi x^2} must be applied. The double integral in polar coordinates is simplified by restricting it to the first quadrant, where r remains from 0 to ∞ and θ is adjusted to 0 to π/2.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with polar coordinates and their application
  • Knowledge of Gaussian integrals and convergence criteria
  • Experience with integration techniques in multivariable calculus
NEXT STEPS
  • Study the properties of Gaussian integrals, specifically ∫ e^{-\pi x^2} dx
  • Learn about the transformation of coordinates in multiple integrals
  • Explore the derivation and application of polar coordinates in integration
  • Investigate convergence criteria for improper integrals
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Students and educators in calculus, mathematicians working with multivariable integrals, and anyone interested in the application of polar coordinates in integration problems.

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Homework Statement




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


Homework Equations



∫∫ dxdy = ∫∫ rdrdθ



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!
 
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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].<br /> <br /> 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].[/itex]
 

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