Erdelyi et.al "Table of integral transforms, vol. I" on p. 50 cites the following integral $$ \int_0^\infty\frac{K_0(x)K_0(\alpha x)}{K_0(\beta x)}\cos xy\phantom{.}dx, $$ but instead of printing the result, they just gave a reference to the following article http://link.springer.com/article/10.1007/BF01655994
Can anybody who has an access to this journal look it up and post the answer here (not the whole derivation, but just the answer, so knowledge of German is not needed. if there are Hankel functions $H_0^{(1)}$ instead of $K_0$ it doesn't matterr)? Thanks.
I don't read German too well, but I think that the reason the formula is not given in Erdelyi et.al. is simply that it is not a simple closed formula. In the paper cited, I only see (after 5 minutes of browsing) the integral expressed as a series expansion (it is the cosine that is expanded), and it is not exactly the integral you look for (see below). Also, it is Hankel functions and not modified Bessel functions that are discussed.
Let me give some short details on the integral that vaguely looks like the one you ask for.
Formula (19) reads. $$ \varphi_2=-\frac{q}{4\pi\epsilon_0}\frac{1}{h}i\int_0^{+\infty} \frac{H_0^{(1)}(i\alpha)H_0^{(1)}(i\alpha\rho/h)}{H_0^{(1)}(i\alpha\rho_m/h)}\cos(\alpha x/h)\,d\alpha $$ Then, quickly, they put $\rho=h$ and as far as I understand they continue with this throughout. This means, I'm afraid, that the article does not really include the type of integral that you look for. Let me anyway write some of the stuff in that article.
In formula (21), $\Theta_0$ is introduced as $$ \Theta_0(\rho_m/h)=i\int_0^{+\infty}\frac{[H_0^{(1)}(i\alpha)]^2}{H_0^{(1)}(i\alpha\rho_m/h)}\,d\alpha, $$ and on the bottom of page 82 it is said
Then it is calculated that $\Theta_0(1)=1$ (formula (24)) and $\Theta_0(0)=0$ (formula (26)).
After that, a series representation of $\varphi_2$ is obtained via a Maclaurin expansion of the cosine. The result (formula (34)) is $$ \varphi_2=-\frac{q}{4\pi\epsilon_0}\frac{1}{h}\biggl[\Theta_0-\sqrt{\frac{\rho_m/h}{1-\rho_m/(2h)}}\biggl\{\frac{\frac{3}{4}\frac{1}{2}\Bigl(\frac{x}{h}\Bigr)^2}{\Bigl(2-\frac{\rho_m}{h}\Bigr)^2}- \frac{\frac{7\cdot 5\cdot 3\cdot 1}{8\cdot 6\cdot 4\cdot 2}\Bigl(\frac{x}{h}\Bigr)^4}{\Bigl(2-\frac{\rho_m}{h}\Bigr)^4} +-\cdots\biggr\}\biggr] $$
Note that there is no $\rho$ here ($\rho_m$ is something different)...
I don't have the motivation to dig deeper into this at the moment. I hope you got the information you looked for. Otherwise, please comment and I'll see if I can put some more light on the issue...