# Polar Coordinates Improper Integral Proofs

## Homework Statement

(a) we define the improper integral (over the entire plane R2)
$$I=\int\int_{R^2}e^{-(x^2+y^2)}dA=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dy dx=\lim_{a\rightarrow\infty}\int\int_{D_{a}} e^{-(x^2+y^2)} dA$$

where Da is the disk with radius a and center the origin. Show that
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dA=\pi$$

(b)
An equilivent definition of the improper integral in part (a) is
$$\int\int_{R^2}e^{-(x^2+y^2)}dA=\lim_{a\rightarrow\infty}\int\int_{S_{a}} e^{-(x^2+y^2)} dA$$

where Sa is the square with vertices $$(\pm a,\pm a)$$ Use this to show that
$$\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\pi$$

(c) deduce that

$$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$$

(d) By making the change of variable $$t=\sqrt{2}x$$. show that
$$\int_{-\infty}^{\infty}e^{\frac{-x^2}{2}}dx=\sqrt{2\pi}$$

## The Attempt at a Solution

I proved a by doing a change of variables into polar coordinates. However, I'm not quite sure how I would go about proving b.The idea that Sa is a square means I would be using Cartesian coordinates however, those 2 integrals don't have an antiderviative, what should I look at next?

$$\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dy dx$$