I'm currently trying to estimate this integral:

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

where $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^d$, $\rho \gg 1$ is independent of $k \in \mathbb{R}^2 \backslash \{ 0 \}$ and $\alpha$ (and so can be treated as a constant), and $J_{\nu}$ denotes the Bessel function of the first kind. I want to make some kind of coordinate change that writes $|k-\alpha|$ in a nicer way. For instance, suppose I let $\alpha_1 = \tau \cos \theta$, $\alpha_2 = \tau \sin \theta$, so that $|\alpha| = \tau$. I can also assume without loss of generality that $(k_1, k_2) = (r, 0)$, for $r > 0$, so that $|k| = r$. Under this substitution, the integral becomes

$$\displaystyle \int_{0}^{2\pi} \int_{0}^{\infty}\frac{J_{1}(\rho\tau)J_{1}\left(\rho\sqrt{(r - \tau \cos \theta)^2 + \tau^2 \sin^{2}\theta)}\right)}{\sqrt{(r - \tau \cos \theta)^2 + \tau^2 \sin^{2}\theta)}} \ \mathrm{d}\tau \ \mathrm{d}\theta$$

$$\displaystyle \int_{0}^{2\pi} \int_{0}^{\infty}\frac{J_{1}(\rho\tau)J_{1}\left(\rho\sqrt{r^2 - 2r\tau\cos\theta + \tau^2}\right)}{\sqrt{r^2 - 2r\tau\cos\theta + \tau^2}} \ \mathrm{d}\tau \ \mathrm{d}\theta,$$

and then letting $\tau = ur$ gives

$$\displaystyle \int_{0}^{2\pi} \int_{0}^{\infty}\frac{J_{1}(\rho u r)J_{1}\left(\rho r\sqrt{u^2 - 2u\cos\theta + 1}\right)}{\sqrt{u^2 - 2u\cos\theta + 1}} \ \mathrm{d}u \ \mathrm{d}\theta.$$

This is quite difficult to deal with. Is there a better substitution I can make that can let me write both $|k - \alpha|$ and $|\alpha|$ in a nicer way? Is it possible to write $|k-\alpha|$ in the form $r|\alpha|$, say? Or is that nonsensical?

  • 1
    $\begingroup$ Are we free to assume $\rho$ is a positive real number? $\endgroup$ – David H Feb 8 '17 at 18:14
  • $\begingroup$ Yes, it can be treated as a large, positive real number. $\endgroup$ – user363087 Feb 8 '17 at 18:15

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.