# 2D orthogonality of Bessel functions

In 1D, the orthogonality relation for zeroth-order Bessel functions of the first kind is

$$\int_0^1xJ_0(xu_{0n})J_0(xu_{0m})dx=0.5J_1^2(u_{0m})\delta_{mn}$$

where $$u_{0n}$$ is the $$n$$-th root of $$J_0(x)$$.

Consider now an azimuthally symmetric Bessel function, $$J_0(\rho u_{0n})$$, where $$\rho=\sqrt{x^2+y^2}$$. What is the equivalent orthogonality relation? I tried something like

$$\int_0^1\int_0^1\sqrt{x^2+y^2}J_0\left(\sqrt{x^2+y^2}u_{0n}\right)J_0\left(\sqrt{x^2+y^2}u_{0m}\right)dxdy=0.5J_1^2(u_{0m})\delta_{mn}$$

but this doesn't work. P.s. for various reasons, I do not wish to replace $$\rho=\sqrt{x^2+y^2}$$ in the above integral (later I need to replace one of the two Bessel functions with an arbitrary function to compute an overlap integral, but this is beyond the scope of this question).

## 1 Answer

When changing from polar to Cartesian coordinates, the integral changes like $$\int_0^{2\pi}d\theta\int_0^1\rho J_0(\rho u_{0n})J_0(\rho u_{0n})d\rho=\int_0^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}J_0(\sqrt{x^2+y^2} u_{0n})J_0(\sqrt{x^2+y^2} u_{0n})dydx.$$ The key here is that you have to make sure you are integrating over the same region in both coordinate systems.

We then have $$\int_0^1\rho J_0(\rho u_{0n})J_0(\rho u_{0n})d\rho=\frac{1}{2\pi}\int_0^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}J_0(\sqrt{x^2+y^2} u_{0n})J_0(\sqrt{x^2+y^2} u_{0n})dydx.$$