Given the integral :
$\int_{0}^{\infty}\frac{xK_{\mu}(x(a+b))}{K_{\mu}(xa)}J_0(cx)dx$
where $K_{\mu}(.)$ is the modified Bessel function of the second kind of order $\mu$ and $J_0(.)$ is the Bessel function of order $0$. $a,b,c$ and $\mu$ are positive real constants.
Is it possible to explicitly calculate the value of this integral? I guess the first step to this would be to proving it's convergence and then moving forward. But I am not sure where should I start with this. Any sort of help with this is much appreciated.
For convergence ... it seems $$ \lim_{x \to 0^+}\frac{xK_{\mu}(x(a+b))}{K_{\mu}(xa)}J_0(cx) = \left(\frac{a}{a+b}\right)^\mu $$ exists. And as $x \to +\infty$, $$ \frac{xK_{\mu}(x(a+b))}{K_{\mu}(xa)}J_0(cx)\sim \left( {\frac {\sqrt {a}\sin \left( cx \right) }{\sqrt {a+b}\sqrt { \pi}\sqrt {c}}}+{\frac {\sqrt {a}\cos \left( cx \right) }{\sqrt {a+b} \sqrt {\pi}\sqrt {c}}} \right) \frac{1}{\sqrt{x}} $$ so I guess that it converges (but only conditionally) on that end.