I've run into the following integral, and I'm not sure how to evaluate it.
$$F(k)=\int d^{2}\mathbf{x}\left(e^{-\frac{\left(x-2a\right)^{2}}{4w^{2}}}+e^{-\frac{\left(x+2a\right)^{2}}{4w^{2}}}+2e^{-\frac{x^{2}}{4w^{2}}}\right)\left(e^{-\frac{\left(y-2b\right)^{2}}{4w^{2}}}+e^{-\frac{\left(y+2b\right)^{2}}{4w^{2}}}+2e^{-\frac{y^{2}}{4w^{2}}}\right)J_{0}(k\mid\mathbf{x}\mid)$$
Where I've written x to have components $x$ and $y$, and $a$, $b$ and $w$ are all positive constants.(Edit: the integral is over all space)
The reason I believe this can be solved is because I have the following expression which should hopefully be the result of evaluating this integral.
$$F(k)=16\pi e^{-w^{2}k^{2}}(1+J_{0}(2bk)+J_{0}(2ak)+J_{0}(2k\sqrt{a^{2}+b^{2}})) $$
I'm trying the verify the topmost method as a way to obtain this result, which is why I already have a tentative answer.
I can see, for example, how to get the $1$ term in the tentative solution, so I have some hope this can be done.
If I were to evaluate it in polar coordinates, a useful integral would be:
$$\intop_{0}^{2\pi}d\theta e^{\alpha\cos(\theta)+\beta\sin(\theta)}$$
I know $\intop_{0}^{2\pi}d\theta e^{\alpha\cos(\theta)}=2\pi I_0(\alpha)$, which is both reassuring and worrying, as it makes me wonder if the top integral is missing $i$.
Does anyone have any idea of how to go about evaluating either of these integrals?
To compute the second integral, let $r=\sqrt{\alpha^2+\beta^2}$ and observe that we may write $(\alpha,\beta)=(r \cos\phi,r\sin\phi)$ for some particular $\phi$. The sum-to-products formula then gives $$\alpha \cos\theta+\beta\sin\theta = r \cos\phi\cos\theta+r\sin \phi \sin\theta = r\cos(\theta-\phi).$$ So the integral becomes $$\int_0^{2\pi}\,d\theta e^{\alpha \cos\theta+\beta\sin\theta} = \int_0^{2\pi}\,d\theta e^{r \cos(\theta-\phi)}.$$ Since the integrand is periodic, we can take $\theta\to\theta+\phi$ to eliminate $\phi$ without changing the limits of integration and the answer is simply $2\pi I_0(r) = 2\pi I_0(\sqrt{a^2+b^2}).$