Proof with Double Integral with two variables

Question

I need to prove that the following statement is true:

$$\iint_D e^{-x^2-y^2}dxdy=4\left(\int_0^Re^{-t^2}dt\right)^2$$ where $ D = \{|x| \leq R, |y| \leq R\} $ and $ 0 \leq R.$

I've tried exchanging the variables to polar coordinates:

$$x = t\cos\theta, \quad y = t\sin\theta$$ so $$t^2 = x^2 +y^2.$$

but I was unable to find the new integration limits for $t$ and $\theta$.

Any help with advancing the proof would be greatly appreciated!

Answer 1

The region is a square. Converting to polar on this problem won't help.

$\int_{-R}^R \int_{-R}^{R} e^{-x^2-y^2} \ dx\ dy$

$\int_{-R}^R \int_{-R}^{R} e^{-x^2}e^{-y^2} \ dx\ dy$
We can pull out the $e^{-y^2}$ factor.
$\int_{-R}^R e^{-y^2}\int_{-R}^{R} e^{-x^2} \ dx\ dy$
Since there is no dependece on $y$ in the inside integral we can do this.
$\int_{-R}^R e^{-y^2}\ dy\int_{-R}^{R} e^{-x^2} \ dx$

Variable names are arbitrary. The integrals are the same. $(\int_{-R}^{R} e^{-t^2} \ dt)^2$ And since the integrand is symmetric... $(2\int_{0}^{R} e^{-t^2} \ dt)^2$

While this is all we have been asked to show, we are ready to talk about polar coversion.

Let $R$ go to infinity. Whether we want to model the entire xy plane an infinite square, or a disk of infinite radius, it won't matter.

$\lim_\limits{R\to\infty}\int_{-R}^R \int_{-R}^{R} e^{-x^2-y^2} \ dx\ dy = \lim_\limits{R\to\infty}\int_{0}^{2\pi} \int_{0}^{R} re^{-r^2} \ dr\ d\theta$

$\lim_\limits{R\to\infty}4(\int_{0}^R \int_{0}^{R} e^{-t^2} \ dt)^2 = \lim_\limits{R\to\infty}\pi (e^0 - e^{-R^2}) = \pi\\ \int_{0}^\infty \int_{0}^{R} e^{-t^2} \ dt = \frac {\sqrt\pi}{2}$

Answer 2

You won't get a nice answer, regardless. But instead of integrating over the square (and multiplying by $4$), integrate over half that square, the triangle $\{(x,y): 0\le y\le x\le R\}$, and, using symmetry, multiply the answer by $8$. In polar coordinates, this region is given by $$0\le\theta\le\pi/4, \quad 0\le r\le R\sec\theta.$$ Thus, your integral becomes $$\int_0^{\pi/4}\int_0^{R\sec\theta} e^{-r^2}\,r\,dr\,d\theta = \frac12\int_0^{\pi/4} \left(1-e^{-R^2\sec^2\theta}\right)d\theta,$$ and now we are stuck unless we take the limit as $R\to\infty$.