# Change of limits when integrating with polar coordinates

1. Oct 22, 2009

1. The problem statement, all variables and given/known data

∫ e^($$\pi$$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^$$\pi$$(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^($$\pi$$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 $$\pi$$/2 and -$$\pi$$/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. Oct 22, 2009

### foxjwill

3. Oct 23, 2009

### HallsofIvy

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

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