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Let $A$ = {$\alpha \in \mathbb{R}^2 : |\alpha| \leqslant 3|k|\},$ where $k$ is some non-zero vector in $\mathbb{R}^2$ and $|\alpha|$ denotes the Euclidean norm, i.e. $|\alpha| = \sqrt{\alpha_1^2 + \alpha_2^2}$. I want to compute this integral:

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

where $\rho$ is a constant not depending on $k$ nor $\alpha$, and $J_{\nu}$ is the Bessel function of the first kind.

I'd like to know if the following manipulation is valid, or if I'm missing any scalar factors (or any other details). Since $A$ is a closed disk, we can integrate over the radius and over the circumference, like so:

$$\displaystyle \int_{0}^{3|k|}\int_{|k - \alpha| = r} \frac{J_{1}(\rho |k - \alpha|)}{|k-\alpha|} \ \mathrm{d}S \ \mathrm{d}r = \int_{0}^{3|k|} \frac{2\pi r J_{1}(\rho r)}{r} \ \mathrm{d}r$$

$$\displaystyle = 2\pi \int_{0}^{3|k|}J_{1}(\rho r) \ \mathrm{d}r = \frac{2\pi}{\rho}(1 - J_{0}(3\rho |k|)),$$

where we used the result $\int_{0}^{a}J_{1}(x) \ \mathrm{d}x = 1 - J_{0}(a)$. Is everything correct, or have I missed a factor of $\rho$ somewhere?

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