Why do Bessel functions of the first kind come up in the following 2-dimensional Fourier transform?

Question

The following equation is given for a function $\gamma$:

$\gamma = \pm \delta \left[\int \frac{d^2 p}{(2\pi)^2}qp(1-\hat{q}\cdot \hat{p})^2 \pi a^2 e^{-|q-p|^2 a^2/4}\right]^{1/2}$

where q and p are two dimensional vectors (hats signifying normalization), and everything else is a constant. In a paper I'm reading the resulting expression is equal to:

$\gamma = \pm (\delta/a)\sqrt{2}\frac{1}{\zeta}\left[e^{-\zeta^2/4} \int_0^\infty du u^2 e^{-u^2/\zeta^2}[3I_0 (u) - 2I_1 (u) + I_2 (u)]\right]^{1/2}$

with $I_n$ a modified Bessel function of the first kind, $\zeta = qa$, and $u$ is a variable. I am unsure where to begin in showing this is true. I have a feeling using the expression of $I_n$ in terms of Chebyshev polynomials is a hint for obtaining $[3I_0 (u) - 2I_1 (u) + I_2 (u)]$, but I don't see how to get rid of the dot product in the initial expression above. Help is much appreciated!

Answer 1

Write both $\mathbf q$ and $\bf p$ in polar coordinates. You then have $$\gamma = \pm \delta \left[\int \frac{d p d\phi_p}{(2\pi)^2}qp^2(1-\cos(\phi_q -\phi_p))^2 \pi a^2 e^{- (q^2+ p^2 -2 qp \cos(\phi_q -\phi_q)) a^2/4}\right]^{1/2}.$$ As the integrand only depends on $\phi_q - \phi_p$, we can instead of integrating over $\phi_q$ as well integrate over $\phi_q - \phi_p$. The Bessel function enter via $$\int_{0}^{2\pi} \frac{dx}{2\pi} e^{\alpha \cos x} = I_0(\alpha)$$ $$\int_{0}^{2\pi} \frac{dx}{2\pi} \cos x\, e^{\alpha \cos x} = I_1(\alpha)$$ $$\int_{0}^{2\pi} \frac{dx}{2\pi} \cos^2 x\, e^{\alpha \cos x} = I_2(\alpha) + \frac{I_1(\alpha)}{\alpha}.$$