From a sum of cosines to an integral of Bessel function

Question

In a physics paper (pubs.acs.org/doi/10.1021/j100210a011), I see the following transformation:

$$\sum_q \frac{2[1-\cos(\textbf{q} \cdot \textbf{r})]}{q^2} =\frac{1}{\pi} \int_0^{+\infty}[1-J_0(qr)]\frac{dq}{q}$$

in which $\textbf{q}$ is a wave vector (spatial frequencies in 2D), $\textbf{r}$ is a 2D position vector on an undulating surface, and $J_0$ is the Bessel function of the first kind. $q$ is the magnitude of the wave vector and $r$ is the magnitude of the position vector.

I do not understand how it is possible to derive such an equation. Does someone have a clue? How does this Bessel function appear? Why is there a $\pi$ on the right hand side? Also there seems to be an inconsistency in dimension as the left hand side has the dimension of squared distance whereas the right hand side has no dimension.

Answer 1

So this is a bit of physics notation. The sum over wave vectors is short hand for an integral over $n$-dimensional reciprocal space, $$\sum_{\mathbf{q}}\mapsto \int\frac{d^n \mathbf{q}}{(2\pi)^n}.$$ Then the integral follows for $n=2$, in polar coordinates, $$(2\pi)^{-2}\int_0^\infty qdq\int_0^{2\pi}d\phi \, \frac{2[1-\cos(qr\cos\phi)]}{q^2} =\int_{0}^{\infty}[1-J_0(qr)]\frac{dq}{\pi q}.$$

Now the OP refers to a 3D integral, rather than a 2D integral, but that cannot be correct, for $n=3$ the answer would be $$(2\pi)^{-3}\int_0^\infty q^2dq\int_0^{2\pi}d\phi\int_0^\pi\sin\theta d\theta \, \frac{2[1-\cos(qr\cos\theta)]}{q^2}=\int_0^\infty \left(1-\frac{\sin q r}{q r}\right)\,\frac{dq}{\pi^2}.$$