Expert: Alon Mandes - 4/22/2009

Question
I have a question about a problem that's in the multiple integrals in polar coordinates, but doesn't really seem like it should be here.  The question is: We define the improper integral over the entire R^2 plane as

I = ∫∫ e^-(x^2+y^2) dA over R^2
= lim (a->∞) ∫∫ e^-(x^2+y^2) dA (over Sa), where Sa is the square with vertices (±a,±a).

Use this to show that:

(∫ e^-(x^2) dx)(∫ e^-(y^2) dy) = pi  Both integrals are evaluated from -∞ to ∞.  

This is part b, part a the integral was over Da (a disc centered over the origin with ever growing radius).  Part a was easy, however, I have quite a few problems with part b.  First is that I never seen integrals multiplied, don't know how it's done or reversed.  I also can't integrate e^-(x^2).  I did find that one online, but it involves the error function, which I don't know either.  I think I'm supposed to convert to polar, but don't see how that's possible with what's given.  Can you give me some direction, please?

Answer
∫∫ e^-(x²+y²) dxdy Can also be written as :
∫∫ e^(-x²)*e^(-y²) dxdy . This is called separable function, a
function that we can be written as multiplying of separate x
function & y function. Now, the last integral can be written as :
∫e^(-x²)∫dxe^(-y²)dy=∫e^(-x²)dx*∫dxe^(-y²)dy.
If we define I=∫∫ e^-(x²+y²) dxdy , then I=[ ∫e^(-t²)dt ]² & we
conclude that ∫e^(-t²)dt which represent the Error function is √I.
All we have to do is calculate the double integral :
∫∫ e^-(x²+y²) dxdy . To do so, we need to switch to polar coordinates. Don't forget that the Jacobian is dxdy=pdpdӨ. So,
∫∫ e^-(x²+y²) dxdy = ∫∫ e^-(p²) pdpdӨ = ∫∫ pe^-(p²) dpdӨ =
∫dӨ∫pe^-(p²)dp=∫dӨ (-1/2)e^-(p²) {p : 0 -> ∞}=∫dӨ/2 {Ө:0->2π)
=2π/2=π . Hence,
∫∫ e^-(x²+y²) dxdy = (∫ e^-(x^2) dx)(∫ e^-(y^2) dy) = π .

Q.E.D

Alon.