- #1

themadhatter1

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## Homework Statement

(a) we define the improper integral (over the entire plane R

^{2})

[tex]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[/tex]

where D

_{a}is the disk with radius a and center the origin. Show that

[tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dA=\pi[/tex]

(b)

An equilivent definition of the improper integral in part (a) is

[tex]\int\int_{R^2}e^{-(x^2+y^2)}dA=\lim_{a\rightarrow\infty}\int\int_{S_{a}} e^{-(x^2+y^2)} dA [/tex]

where S

_{a}is the square with vertices [tex](\pm a,\pm a)[/tex] Use this to show that

[tex]\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy=\pi[/tex]

(c) deduce that

[tex]\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}[/tex]

(d) By making the change of variable [tex]t=\sqrt{2}x[/tex]. show that

[tex]\int_{-\infty}^{\infty}e^{\frac{-x^2}{2}}dx=\sqrt{2\pi}[/tex]

## Homework Equations

## 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 S

_{a}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?