Let $A = \{ \alpha \in \mathbb{R}^2 : \frac{1}{2}|k| \leqslant |\alpha| \leqslant 2|k| \}$, where $k \in \mathbb{R}^2$ is non-zero, and $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^2$. I am trying to either compute or bound the follow integral in terms of $|k|$:

$$\displaystyle \int_{A} |\alpha|^{-3/2}|k-\alpha|^{-3/2} \ \mathrm{d}\alpha$$

What I would like to obtain is something that is $O(|k|^{-1 - \epsilon})$, for some $\epsilon > 0$. However, I can't manage to obtain a bound like this. The best I am able to manage is $O(|k|^{-1})$. If we notice that on $A$, the norm $|k - \alpha|$ varies from $0$ to $3|k|$, and using $|\alpha|^{-3/2} \leqslant C|k|^{-3/2}$, the integral becomes

$$\displaystyle C|k|^{-3/2} \int_{0}^{3|k|} \int_{|k - \alpha| = r}|k - \alpha|^{-3/2} \ \mathrm{d}S \ \mathrm{d}r = C|k|^{-3/2}\int_{0}^{3|k|}r\cdot r^{-3/2} \ \mathrm{d}r = O(|k|^{-1}),$$

and this estimate is not what I want. Can anyone manage to get a better estimate of the form $O(|k|^{-1 - \epsilon})$?

  • 2
    $\begingroup$ No, the integral is precisely $C\cdot \lvert k\rvert^{-1}$. $\endgroup$ – Daniel Fischer Feb 4 '17 at 15:07

Obviously we may assume that $k=(r,0)$ with $r>0$, then the given integral takes the following form:

$$ \int_{0}^{2\pi}\int_{r/2}^{2r}\frac{1}{\sqrt{\rho}\left(\rho^2-2r\rho\cos\theta+r^2\right)^{3/4}}\,d\rho\, d\theta\\ = \frac{1}{r}\,\underbrace{\int_{0}^{2\pi}\int_{1/2}^{2}\frac{1}{\sqrt{u}\left(u^2-2u \cos\theta+1\right)^{3/4}}\,du\, d\theta}_{\text{bounded}}$$ hence the integral behaves exactly like $\frac{1}{|k|}$.

  • $\begingroup$ Thanks, this is very helpful. How do you know that the double integral is bounded? $\endgroup$ – user363087 Feb 4 '17 at 16:15
  • 1
    $\begingroup$ @user363087: because $u^2-2u\cos\theta+1$ is bounded between $(u-1)^2$ and $(u+1)^2$. If you perform such replacements you get a bounded double integral. $\endgroup$ – Jack D'Aurizio Feb 4 '17 at 16:19

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