Integrals related to the modified Bessel functions of the second kind 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.
 A: 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 $F4$.
