I'm investigating the following double integral:

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

where $\rho \gg 1$ is some (very) large constant, $k \neq 0$ is a vector in $\mathbb{R}^2$, $\alpha \in \mathbb{R}^2$, $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^2$, and $J_{\nu}$ denotes the Bessel function of the first kind.

Note that, for sufficiently large, positive $z$, we have $|J_{\nu}(z)| \leqslant C_{\nu}|z|^{-1/2}$. This lets us bound the integral by

$$\displaystyle \int_{|k| \geqslant \rho} \int_{|\alpha| \geqslant \rho} \frac{1}{|k-\alpha|^3} \ \mathrm{d}\alpha \ \mathrm{d}k.$$

I would like to know whether or not this integral converges for any $\rho$ sufficiently large. I suppose this could be done by making the substitution $\tau = |k - \alpha|$, but I'm not sure how the limits of the integrals (particularly the lower limits) behave under this change of coordinates. Can anyone help?

  • 1
    $\begingroup$ An observation: the integral does not converge when $\rho = 0$. This can be seen by making the substitution $\vec{k} = \vec{q} + \frac{1}{2} \vec{r}$ and $\vec{\alpha} = \vec{q} - \frac{1}{2} \vec{r}$. In terms of these variables, the integrand only depends on $\vec{r}$, and so the integral over the $\vec{q}$-plane will diverge. (The Jacobian for this transformation, BTW, is exactly 1.) I suspect that since this divergence occurs due to the infinite volume of integration over $\vec{q}$, it will persist when we take the $\rho \neq 0$ case: ... $\endgroup$ – Michael Seifert Feb 17 '17 at 1:49
  • 1
    $\begingroup$ ... for any fixed value of $\vec{r}$, the allowed volume of $\vec{q}$-space will still be infinite. $\endgroup$ – Michael Seifert Feb 17 '17 at 1:53
  • $\begingroup$ I see, thanks. I suspected that we would indeed have divergence. Your coordinate substitution was very helpful: if I am not mistaken, you have proved the convergence of my integral here, taking $\alpha = s$, $k = t + s$. Though I might be wrong. $\endgroup$ – user363087 Feb 17 '17 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.