Double integral, positive and negative infinity as bounds

Question

I'm not really sure how to approach this problem, at first I tried substituting a for positive infinity and b for negative infinity but got stuck when I tried to substitute these in after integrating for x.

Answer 1

This is quite a well-known integral - as it stands now, there is not much hope to integrate in Euclidean coordinates, since $\int e^{-x^2}d x$ has no elementary antiderivative.

But think about possibly switching to a different coordinate system. Is there one which simplifies things?

Answer 2

Based on your comment I'm assuming you've changed the integral to:

$$\int_0^{2\pi}\int_{0}^{\infty}r\ e^{-r^2}\ dr\ d\theta=2\pi\int_{0}^{\infty}r\ e^{-r^2}\ dr$$

To find this integral you can consider the limit of:

$$\lim_{n\to\infty}2\pi\int_0^{n}r\ e^{-r^2}\ dr$$

let $u=r^2$ so $du=2r\ dr$

$$=2\pi\lim_{n\to\infty}\int_0^{n}-\frac{1}{2}\ e^{-u}\ du$$

$$=2\pi\lim_{n\to\infty}\left(-\frac{1}{2}\ e^{-u}\right)_0^{n}$$

$$=2\pi\lim_{n\to\infty}\left(-\frac{1}{2}e^{-n}+\frac{1}{2}\right)$$

$$=2\pi\times\frac{1}{2}$$

$$=\pi$$

Answer 3

Note that $$\int_{-\infty}^{\infty}e^{-x^{2}}dx=2\int_{0}^{\infty}e^{-x^{2}}dx\overset{x^{2}=u}{=}\int_{0}^{\infty}u^{-1/2}e^{-u}du=\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} $$ so $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}-y^{2}}dxdy=\int_{-\infty}^{\infty}e^{-y^{2}}\int_{-\infty}^{\infty}e^{-x^{2}}dxdy=\pi.$$