I would like to evaluate analytically the improper integral $$ f(z,r,\rho,\nu) = \int_0^\infty \sin(k z)\, K_{i\nu}(k \rho)\, K_{i\nu}(k r)\, dk \, , $$ where $K_{i\nu}$ is the modified Bessel function of the second kind with imaginary order. Here, $\nu \ge 0$. This integral arises in solving a boundary value problem using the Kontorovich–Lebedev transform.
When cosine replaces sine, the integral is known and can be found in the classical reference by Gradshteyn and Ryzhik, Eq. ET I 50(51) (6.672 page 719). It is given by $$ \int_0^\infty \cos(k z)\, K_{i\nu}(k \rho)\, K_{i\nu}(k r)\, dk = \frac{\pi^2}{4} \frac{\operatorname{sech}(\pi\nu)}{\sqrt{\rho r}} \, P_{i\nu-\frac{1}{2}} (\xi) \, , \tag{A} $$ with $$ \xi = \frac{\rho^2+r^2+z^2}{2\rho r} \, . $$
Here are the steps I followed. I used the integral representation of the modified Bessel function for real argument $x>0$, namely $$ K_{i\nu}(x) = \int_0^\infty e^{-x\cosh u}\cos(\nu u)\,du \, , $$ and substituted this representation for both Bessel functions into $I(z,r,\rho,\nu)$. By noting that $$ \int_0^\infty \sin(kz)\, e^{-a k}\, dk = \frac{z}{a^2+z^2}, \qquad a>0 \, , $$ the resulting double integral reads $$ f(z,r,\rho,\nu) = z \int_0^\infty \int_0^\infty \frac{\cos(\nu u)\cos(\nu t)}{(\rho\cosh u + r\cosh t)^2 + z^2}\,du\,dt. $$
Any suggestions on how to proceed further? I’d appreciate your help, thanks.