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Convergence Problem

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider the integrals [tex]\int_1^\infty \frac{k}{x^2+k^p\cos^2x}dm(x)[/tex], where m is the Lebesgue measure. For what p do the integrands have an integrable majorant? For what p do the integrals tend to 0?


    2. Relevant equations



    3. The attempt at a solution

    Pick some large constant C. For [tex]x> C k^{p/2}[/tex], the denominator is approximately x2, so the integral is at least as big as

    [tex]k\int_{Ck^{p/2}}^\infty \frac{dx}{x^2} = \frac{1}{C k^{p/2-1}}[/tex].
    So, when p/2<1, (so p<2) the integral diverges.

    When p≥2, that is not a problem, so we need to look at
    [tex]\int_1^{Ck^{p/2}} \frac{k}{x^2+k^p \cos^2 x} dx[/tex].

    Now, substitute x=[tex]k^{p/2}u[/tex]. The integral becomes

    [tex]k^{1+p/2} \int_{k^{-p/2}}^1 \frac{du}{k^p u^2 + k^p \cos^2 k^{p/2} u} = k^{1-p/2}
    \int_{k^{-p/2}}^1 \frac{du}{u^2 + \cos^2 k^{p/2} u}.[/tex]
    Now, the integral is has no singularity at 0, but I'm not sure where to go from here.
     
    Last edited: Dec 11, 2013
  2. jcsd
  3. Dec 11, 2013 #2

    jbunniii

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    There is no problem with the integral from ##1## to ##C k^{p/2}##:
    $$\int_{1}^{C k^{p/2}} \frac{k}{x^2 + k^p \cos^2 x} dx \leq \int_{1}^{C k^{p/2}} \frac{k}{x^2} dx$$
    So you just need a bound for ##1/x^2## on the interval ##[1, C k^{p/2}]##.
     
  4. Dec 11, 2013 #3
    Ok, I think I can manage that. Thanks!
     
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