# Logarithmic singularities are locally square integrable

1. Nov 30, 2013

### lmedin02

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

Last edited: Nov 30, 2013