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When is an integral infinite help

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data
    The problem is this:

    Let f : R^2 -> R be

    [tex] f(x,y) = \frac{xy}{(x^2+y^2)^2}\qquad (x,y)\neq(0,0) [/tex]

    With [itex] f(0,0) = 0 [/itex].

    Question:
    Is [itex] f [/itex] integrable with respect to the Lebesguemeasure [itex] m_2 [/itex] om the set [itex] [-1,1]\times[-1,1] [/itex]?

    3. The attempt at a solution

    Well [itex]f[/itex] is integrable if and only is

    [tex] \int_{[-1,1]^2} \vert f\vert\,dm_2(x,y) < \infty[/tex]

    The integral is infinite, and so f isn't integrable.
    The thing that makes it confusing, is that in the next problem it says,
    Show that both the double integrals exist

    [tex] \int_{[-1,1]}\left(\int_{[-1,1]} f(x,y)\,dm(x)\right)dm(y)[/tex]
    [tex] \int_{[-1,1]}\left(\int_{[-1,1]} f(x,y)\,dm(y)\right)dm(x)[/tex]

    But how can they exist if f isn't integrable?


    EDIT:
    Hmm I had made an error. It did seem kinda wired that the integral of a positive would give me something like [itex] \infty-\infty[/itex]. The integral is infinite though. The expression I had found would only hold for y > 0.
     
    Last edited: Oct 6, 2007
  2. jcsd
  3. Oct 6, 2007 #2
    First question : Dont you think that the integral should have two boundaries? its from [-1,1] to what? please try to be more specific so we can help
     
  4. Oct 6, 2007 #3
    Actually [-1,1] is not a point it's the interval from -1 to 1. That's just another way of writting the integral of the function f(x,y) from -1 to 1 with respect to x (hence I have writtin m(x), the Lebesgue measure with respect x).
     
  5. Oct 6, 2007 #4

    HallsofIvy

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    I assume that the notation "[-1,1]2" means [itex]-1\le x\le 1[/itex] and [itex]-1\le y\le 1[/itex]. A function is as long as the set of points on which it is NOT continuous is a measurable set. For what points is this function not continuous?
     
  6. Oct 6, 2007 #5
    Yep, that's the square I'm talking about.

    What is the function if it's not defined on a measurable set? Integrable or measurable?
    The function is measurable though, since as f is only discontinuous at the point (0,0), and we just define f(0,0)=0, and make f defined on the whole square.
     
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