- #1
themadhatter1
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Homework Statement
(a) we define the improper integral (over the entire plane R2)
[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 Da 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 Sa 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 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?