The problem arises from calculating the spatial dependence of the electric fields of a disk charge (charge uniformly distributed on a disk in the x-y plane). I can find the field in the x-y plane is scaled by: $$ F_1(b,r)=\frac{1}{2\pi} \int_0^b dx \int_0^{2\pi} d\theta x K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right) $$ with $b$ the radius of the disk and $r$ the distance from the center of the disk.The modifield Bessel function $K_0(x)$ comes from the Fourier transform of the point-charge's field in the z-direction. The non-trivial solusion of $F_1$ is: $$ F_2(b,r)=1-b K_1(b) I_0(r) $$ for $r \leq b$ and $$ F_2(b,r)=b K_0(r) I_1(b) $$ for $ r > b$. But I cannot figure out how to achieve $F_2$ from $F_1$.
Here $K_0(x)$, $K_1(x)$, $I_0(x)$, and $I_1(x)$ are the modified Bessel function of the second and first kinds, respectively.
Later I found that the problem can be reduced to finding the solution of $$ F_3(b,r)=\int_0^{2\pi} d\theta K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right) $$ The non-trivial solution of $F_3$ is $$ F_4(b,r) = 2\pi \left[ \theta(r-b) K_0(r) I_0(b) + \theta(b-r) K_0(b) I_0(r) \right] $$ with $ \theta(x)$ the Heaviside step function.
One can numerically test the identity of $F_1$ and $F_2$, and also of $F_3$ and $F_4$. I tried to find the answer from handbooks of integrals and special functions but was not successful.
Clearly, $F_1$ and $F_3$ are well-defined problems of mathematics, independent of the physical problems they are extracted from. So I raise the question here and hope someone could help me find the techniques necessary to do the integrals.
This result can be derived from the Graff addition theorem for the modified Bessel function (eq. (8) in Watson, XI, 11.3) which reads, for $K_0$ \begin{equation} K_0\left(\sqrt{ Z^2+z^2-2Zz\cos\phi} \right)=\sum_{m=-\infty}^\infty K_m(Z)I_m(z)\cos m\phi \end{equation} which is valid for $\left|z\right|<\left|Z\right|$.
For $0\le r\le b$, we first decompose the integral $F_1(b,r)$ into \begin{align} F_1(b,r)&=\frac{1}{2\pi} \int_0^b \int_0^{2\pi} K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right)\,xdxd\theta \\ &=\frac{1}{2\pi} \int_0^{2\pi} d\theta\left[\int_0^r+\int_r^b\right]K_0\left(\sqrt{r^2+x^2-2rx\cos{\theta}}\right)\,xdx \end{align} For the first $x-$integral, we choose $z=x,Z=r$ in the Graff representation and $z=r,Z=x$ for the second. Only the terms with $m=0$ survive after the angular integration, then \begin{equation} F_1(b,r)=K_0(r)\int_0^rI_0(x)\,xdx+I_0(r)\int_r^bK_0(x)\,xdx \end{equation} These integrals are readily done (DLMF): \begin{equation} F_1(b,r)=rK_0(r)I_1(r)-bI_0(r)K_1(b)+rI_0(r)K_1(r) \end{equation} From the Wronskian of the modified Bessel functions, \begin{equation} I_{\nu}\left(z\right)K_{\nu+1}\left(z\right)+I_{\nu+1}\left(z\right)K_{\nu}\left(z\right)=1/z \end{equation} We deduce \begin{equation} F_1(b,r)=1-bI_0(r)K_1(b)=F_2(b,r) \end{equation} The same decomposition gives also the identity between $F_3$ and $F4$.