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Logarithmic singularities are locally square integrable

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    I will like to show that the function [tex]f:\mathbb{R}^2\rightarrow \mathbb{R}[/tex] defined by
    [tex]f(x)=\ln\bigg(1+\dfrac{\mu}{|x-x_0|^2}\bigg),\quad\mu>0[/tex] is in [tex]L^2(\mathbb{R}^2).[/tex]
    2. Relevant equations
    A function is in [tex]L^2(\mathbb{R}^2)[/tex] if its norm its finite, i.e.,
    [tex]||f||_{L^2(\mathbb{R}^2)}^2=\int_{\mathbb{R}^2}{|f|^2 }<\infty[/tex]
    3. The attempt at a solution I attempted to calculate the integral directly using polar coordinates. Consider the following attempt,
    [tex]\int_{\mathbb{R}^2}{\bigg(\ln\bigg(1+\dfrac{\mu}{|x-x_0|^2}\bigg)\bigg)^2}dx=2\pi\int_0^{\infty}{\bigg(\ln\bigg(1+\dfrac{\mu}{r^2}\bigg)\bigg)^2}rdr=2\pi\int_{r\geq R}{\bigg(\ln\bigg(1+\dfrac{\mu}{r^2}\bigg)\bigg)^2}rdr+2\pi\int_{r\leq R}{\bigg(\ln\bigg(1+\dfrac{\mu}{r^2}\bigg)\bigg)^2}rdr[/tex]
    Where above I can take [tex]R>\sqrt{\mu}[/tex] to ensure convergence of the power series of [tex]\ln\bigg(1+\dfrac{\mu}{r^2}\bigg)[/tex] and hence the convergence of the integral over the interval [tex]r\geq R [/tex] since [tex]\ln\bigg(1+\dfrac{\mu}{r^2}\bigg)\sim O\bigg(\dfrac{1}{r^2}\bigg)\in L^2(R,\infty)[/tex] The same argument does not apply to the integral over [tex]r\leq R .[/tex] Any suggestions on how to attack the second integral or the problem. Thanks.
     
    Last edited: Nov 30, 2013
  2. jcsd
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