I am trying to analytically evaluate
$$f(a,b)=\int_0^{2\pi} K_0(\sqrt{a^2+b^2-2ab\cos(\theta)}) d \theta$$
where $K_0$ is a modified Bessel function of the second kind and $a,b>0$. I happen to know the solution by a different method, it is
$$f(a,b)=2\pi I_0(\min \{ a,b \}) K_0(\max \{ a,b \})$$
where $I_0$ is a modified Bessel function of the first kind. But I don't know how to evaluate the integral directly, and CAS's seem to choke on it (though they can check the result above numerically just fine). It seems like it is probably useful to use the identity
$$K_0(x)=\int_0^\infty \frac{\cos(x t)}{\sqrt{t^2+1}} dt$$
and then interchange the order of integration, but then I still need to be able to compute
$$\int_0^{2\pi} \cos(\sqrt{a^2+b^2-2ab\cos(\theta)}) d \theta$$
for arbitrary $a,b>0$, before proceeding to the $dt$ integral.
Note that the problem is significantly simpler when $a=b$ because in this case $\sqrt{a^2+b^2-2ab\cos(\theta)}=2a\sin(\theta/2)$; in this case the cosine integral is $2\pi J_0(2at)$, which results in the easier problem
$$2\pi \int_0^\infty \frac{J_0(2at)}{\sqrt{t^2+1}}.$$
But in the case $a \neq b$ I don't even see how to begin.
