I am having trouble showing the following, which shows up from coherence theory:

$\frac{\pi b^2}{\alpha^2}(1-J_0^2(\alpha b)-J_1^2(\alpha b))=\int_0^{2\pi}\int_0^b\int_0^b r_1r_2\frac{J_1\left (\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}\right )}{\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}} dr_1dr_2d\theta$

Where $J_n$ is the nth order Bessel function of the first kind. The result is so nice, but I can't find a way to show it. Can anyone provide some help in showing this equality?


1 Answer 1


In answer to my own question, the equality can be shown as follows. First, we realize that $\int_0^{2\pi}\int_0^b\int_0^b r_1r_2\frac{J_1\left (\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}\right )}{\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta)}} dr_1dr_2d\theta\\=\frac{1}{2\pi}\int_0^{2\pi}\int_0^{2\pi}\int_0^b\int_0^b r_1r_2\frac{J_1\left (\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_1-\theta_2)}\right )}{\alpha\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_1-\theta_2)}} dr_1dr_2d\theta_1d\theta_2$

From here we notice that $\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_1-\theta_2)}$ is the distance between two particular points in the circle of radius $b$. Let that distance be denoted as $L$.

Then we have $\frac{1}{2\pi}\iint_{A_1}\iint_{A_2}\frac{J_1(\alpha L)}{\alpha L} dA_1 dA_2$.

We can write this instead as $\frac{(\pi b^2)^2}{2\pi}\int_0^{2b}\frac{J_1(\alpha L)}{\alpha L}p(L) dL$, where $p(L)$ is the probability density of choosing two points of distance $L$ within a circle of radius $b$. This probability density is well known. See for instance equation (5) of Ricardo García-Pelayo 2005 J. Phys. A: Math. Gen. 38 3475 doi:10.1088/0305-4470/38/16/001 and references therein.

We have $p(L)=\frac{L}{\pi b^2}\left ( 4\cos^{-1}[d/(2b)]-L\sqrt{4b^2-L^2}/b^2 \right ), 0\leq L \leq 2b$

The integral can now be easily evaluated and gives the required equality.

This technique is usable any time the integrand is a function of only one parmater, like L in this case.


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