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Homework Help: Evaluate the integration

  1. Mar 1, 2013 #1
    1. The problem:
    I came up to this integration and noticed that it contains singularity at the center (x,y)=(0,0).
    [tex] \iint\limits_D \frac{x^2-y^2}{(x^2+y^2)^2} dxdy[/tex]

    Note that the integration domain is a unit circle centered at the origin.

    I was thinking that the maybe the integral could be evaluated through the polar coordinate. So I used the follows:

    2. Relevant equations
    [tex] x^2 +y^2 = r^2 [/tex]
    [tex] x= r \cdot cos(\theta), y= r \cdot sin(\theta) [/tex]

    3. The attempt at a solution
    In that case the integration was transformed to the following integration:
    [tex] \int_0^1 \frac{1}{r}dr \cdot \int_0^{2\pi}(cos^2(\theta)-sin^2(\theta))d\theta[/tex]

    Evaluation of the first term still includes a singular point with r=0, while the second term gives 0. However, looks like there is a finite value for this integration which is not 0.

    Can anybody helps me out on this? Thanks very much.
    Last edited: Mar 1, 2013
  2. jcsd
  3. Mar 1, 2013 #2


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    welcome to pf!

    hi shaleba! welcome to pf! :smile:

    by interchanging x and y (or x and -y) in the original integral, the integral (if it is finite) is obviously 0 for any region symmetric about the lines x = y or x = -y

    in particular, it's 0 for any ring around the origin
    why do you say it looks like there is a finite value?

    since it's the equivalent of 1 - 1 + 1 + 1 + 1 - … , i suppose it depends exactly how you define the integral :confused:
  4. Mar 1, 2013 #3
    Re: welcome to pf!

    Thank you very much for your reply Tim.
    I understand your rationale on the integration. However, I think the key is about the singularity at the center.

    For the rationale you mentioned, the 2d delta function looks like satisfy what you mentioned too, but the integration is not 0.

    The reason I think it's a finite but not 0, is because this integral appears in one paper and looks like not equal to 0. They didn't show details about evaluating it.

  5. Mar 1, 2013 #4


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    what delta function? :confused:
    which book?

    the integral is a two-dimensional limit, and its value depends how you define the approach to that limit

    if you define it the obvious way, the integral is 0

    but you can define it so as to give*the integral any value
  6. Mar 1, 2013 #5

    rude man

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    As I see it:

    Your integration of f(r,θ) wrt theta first gave zero and you're worried because the second integration wrt r blows up at r = 0.

    But look at what happens to your function f(r,θ) = (cos2θ - sin2θ)/r2

    integrated within a small circle about the origin:
    lim r → 0 of f(r,θ)(πr2) = πg(θ) where g(θ) = cos2θ - sin2θ and πg(θ) is always between -π and +π so there is no singularity at r = 0.

    So I would have to second the motion that the integral of f(r,θ) over the specified area is zero.
  7. Mar 2, 2013 #6
    in polar coordinates ##(x^2+y^2)^2=r^4## and the surface element ##dx dy## becomes ##r dr dθ##, so the integrand is ##(cos^2(θ)-sin^2(θ))/r^3=cos(2θ)/r^3##. Hence, the integral is the product of two terms: one depends on ##θ## only, the other depends on ##r## only. Can you find their values? What definition for an improper integral are you using?
  8. Mar 5, 2013 #7
    Thanks for your reply. But looks like from the numerator [tex]x^2 -y^2 [/tex], there arises another [tex]r^2[/tex]
    The [tex]\theta[/tex] term looks like to be 0 while the other term is singular.

  9. Mar 5, 2013 #8
    thanks for reply. the Delta function I mentioned is the Dirac delta function.

  10. Mar 5, 2013 #9


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    no, because the dirac delta function (btw, technically it's a distribution, not a function, but that doesn't matter here) is always positive,

    and you can always change the order of any positive sum (or integral)

    the difficulty with the given function is that it's alternately positive and negative in the four sectors round the origin, so the order in which you take the limits does matter :wink:
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