Math Help - Double integral proof.

1. Double integral proof.

Prove that $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2} dx \ dy=\lim_{b \rightarrow \infty} \int_{-b}^b \int_{-b}^b e^{-x^2-y^2}dx \ dy=4 \left( \int_0^{\infty} e^{-x^2}dx \right)^2$.
I'm not sure how to prove the first equality. It's intuitively obvious, but I don't know how to go about proving it

I've assumed the first equality to be true. This suggests that the region of integration is a square, so the integrals can be switched around pretty easily:

$\lim_{b \rightarrow \infty}\int_{-b}^b \int_{-b}^b e^{-x^2-y^2} dy \ dx$

$\lim_{b \rightarrow \infty} \int_{-b}^b \left[ -\frac{1}{2y}e^{-x^2-y^2} \right]_{-b}^b \ dx$

$\lim_{b \rightarrow \infty} \int_{-b}^b -\frac{1}{2b}e^{-x^2-b^2}-\frac{1}{2b}e^{-x^2-b^2} \ dx$

The problem with the step above is that it becomes: $\int_{-\infty}^{\infty} 0 \ dx$ which looks hopelessly wrong.

Can anyone point me in the right direction?

2. Originally Posted by Showcase_22
I'm not sure how to prove the first equality. It's intuitively obvious, but I don't know how to go about proving it

I've assumed the first equality to be true. This suggests that the region of integration is a square, so the integrals can be switched around pretty easily:

$\lim_{b \rightarrow \infty}\int_{-b}^b \int_{-b}^b e^{-x^2-y^2} dy \ dx$

$\lim_{b \rightarrow \infty} \int_{-b}^b \left[ -\frac{1}{2y}e^{-x^2-y^2} \right]_{-b}^b \ dx$

$\lim_{b \rightarrow \infty} \int_{-b}^b -\frac{1}{2b}e^{-x^2-b^2}-\frac{1}{2b}e^{-x^2-b^2} \ dx$

The problem with the step above is that it becomes: $\int_{-\infty}^{\infty} 0 \ dx$ which looks hopelessly wrong.

Can anyone point me in the right direction?
yes, the first equality is pretty much by definition. i'd just assume it is true. after that, your biggest mistake is thinking that you can integrate like that. your integral is wrong (it would probably do you good to look up how to (and when you can) integrate exponential functions. to integrate this, switch to polar coordinates