I am considering the following integral:

$$\displaystyle \int \limits_{|k| \geqslant 1} \left( \int \limits_{|\alpha| \geqslant 1} \frac{J_{1}^{2}(\rho |\alpha|)}{|\alpha|^2} \ \mathrm{d}\alpha \right) \left( \int \limits_{|\alpha| \geqslant 1} \frac{J_{1}^{2}(\rho |k - \alpha|)}{|k - \alpha|^2} \ \mathrm{d}\alpha \right) \ \mathrm{d}k,$$

where $k \in \mathbb{R}^2$ is non-zero, $\alpha$ varies over $\mathbb{R}^2$, $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^2$, $\rho$ is treated as a large constant, and $J_{\nu}$ denotes the Bessel function of the first kind. My question concerns the first of the two inner integrals. I would like the first one to evaluate to something which looks like $O(|k|^{-1})$. Therefore, I have said the following: for any $k$, choose $\epsilon > 0$ such that $0 < \epsilon |k|^2 \leqslant 1$. Then we may estimate the inner integral as:

$$\displaystyle \int \limits_{|\alpha| \geqslant 1} \frac{J_{1}^{2}(\rho |\alpha|)}{|\alpha|^2} \ \mathrm{d}\alpha \leqslant \int \limits_{|\alpha| \geqslant \epsilon |k|^2} \frac{J_{1}^{2}(\rho |\alpha|)}{|\alpha|^2} \ \mathrm{d}\alpha.$$

The integral on the right then evaluates to $O(|k|^{-2})$, which means that the whole integral converges (the second inner integral evaluates to give a sum of squares of Bessel functions, which, when multiplied by $O(|k|^{-2})$, converges). However, this seems somewhat cheap: could we not then simply take $\epsilon > 0$ such that $0 < \epsilon|k|^t \leqslant 1$ for any $t$? Certainly for any $k$ such an $\epsilon$ exists, but I'm concerned that I'm breaking some obvious rule by estimating the whole integral in this way. Does anyone see a problem with this?

  • $\begingroup$ what you are saying (in the plain words of a poor physicist) that $\int_b^{\infty}dxf^2(x)\geq\int_a^{\infty}dxf^2(x)$ for any $a>b>0$. This seems to be a valid statement since you are adding only positve quantities ..right? $\endgroup$ – tired Feb 10 '17 at 12:15
  • $\begingroup$ the only thing i see here is that $k$ in the outer integral becomes arbitrary large so you have the constraint $\epsilon\sim o(k^{-2})$. Does this cange anything? $\endgroup$ – tired Feb 10 '17 at 12:21
  • 1
    $\begingroup$ to ensure that $\epsilon k^2<1$ for any $k>1$ we can choose $\epsilon=1/k^{2+\delta}$ for $\delta>0$. performing your standard estimates i think the integral should be $\sim 1/(\epsilon k^2)=1/k^{-\delta}$ which obviously can't be turned in something good enough so that the outer integral converges... $\endgroup$ – tired Feb 10 '17 at 12:49

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