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Measurability/Lebesgue Integration

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    2n0plj9.jpg

    2. Relevant equations

    For part (a):
    28a6mbp.jpg

    3. The attempt at a solution

    I think I've done all of this question except for the very last part of (c).

    For (a) I've found the integral [tex]\int \frac{x^2-y^2}{(x^2+y^2)^2}\;dx[/tex] using Wolfram Alpha but how would I integrate it bare handedly?

    Since the double integral wrt x then y does not equal the double integral wrt y then x, f is not Lebesgue integrable over [0,1]x[0,1]. However if 0<a<b then f is Lebesgue integrable over [a,b]x[a,b] and the integral is 0 by Fubini.

    The last part of (c) is what I'm not sure about: deducing that [tex]\int f = |S^+| - |S^-|.[/tex] Am I going about it the right way? We can write [itex]f=f^+ - f^-[/itex] where [itex]f^+ = \max (f,0)[/itex] and [itex]f^- = -\min (f,0)[/itex] (i.e. the +ve and -ve parts of f respectively).

    Then we know if [itex]f\in L^1 (\mathbb{R})[/itex] then [itex]f^+ , f^- \in L^1 (\mathbb{R})[/itex] and [itex]f^+ , f^- \geq 0[/itex].

    Now is the following correct? [tex]\int f = \int (f^+ - f^-) = \int f^+ - \int f^- = |S^+| - |S^-|[/tex]
     
  2. jcsd
  3. Mar 28, 2012 #2

    morphism

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    Homework Helper

    If guess you could notice that it might be d/dx of something over (x^2+y^2) (because of the quotient rule), and then guess what the numerator should be.

    Or, you could proceed as follows: $$\begin{align*} \int \frac{x^2-y^2}{(x^2+y^2)^2}\;dx &= \int \frac{x^2}{(x^2+y^2)^2}\;dx - y^2 \int \frac{1}{(x^2+y^2)^2}\;dx \\
    &= \int \frac{1}{x^2+y^2} - \frac{y^2}{(x^2+y^2)^2}\;dx - y^2 \int \frac{1}{(x^2+y^2)^2}\;dx \\
    &= \int \frac{dx}{x^2+y^2} - 2y^2 \int \frac{1}{(x^2+y^2)^2}\;dx. \end{align*}$$

    To deal with the second integral on the RHS, all you need to know is how to integrate
    $$ \int \frac{dv}{(v^2+1)^2}. $$ But this is easy (though slightly tedious). E.g. start off with the sub v=tan(t).

    Yes, of course. This follows from the previous part applied to ##f^+## and ##f^-##.
     
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