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Polar Coordinates Improper Integral Proofs

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    (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

    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

    (c) deduce that


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

    3. 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?
  2. jcsd
  3. Nov 3, 2011 #2
    update, I proved everything, however I'm not sure If my proof for b is what they're asking for. I said that

    [tex]\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[/tex]

    and since I had proved that the rhs of the equation equals pi it must be true that the left hand side of the equation equals pi as well. Dosen't a square of infinite dimensions have the same area as a polar rectangle where 0≤θ≤2∏ and 0≤r≤infinity. Does this satisfy the question being asked?
  4. Nov 7, 2011 #3
    That's exactly what I did for part b. I have this assigned so I need some help on this problem as well.

    I don't know how to do part c yet. I took the equation from b, and divided by the e^(-(y^2)) integral and all I would have to do is prove that that integral equals √∏. I tried to say that -y^2=x^2-r^2 so if you replace the e^(-(y^2)) integral with e^(x^2)/e^(r^2) (which I'm not sure if you can do), then idk I'm kinda stuck.
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