I am given with the expression $$F(s)= K_{v}(\sqrt{as})I_{v}(\sqrt{bs})$$
where
$I_{v}(\sqrt{as})$ is modified Bessel function of the first kind of order $v$.
$K_{v}(\sqrt{as})$ is modified Bessel function of the second kind of order $v$.
I want to find out the inverse laplace transform of above expression.
Given the following function in Laplace domain. $$F(s) = K_v(\sqrt{as}+\sqrt{bs})I_v(\sqrt{as}-\sqrt{bs})$$ The corresponding inverse transformation is as follows. $$f(t) = \mathscr{L}^{-1}[{F(s)}]= \frac{1}{2t}\exp\left(-\,{a + b \over 2t}\right){\rm I}_{v}\left(\,{a - b \over 2t}\right)$$
This form of solution is given in Bateman, Harry (1954) Tables of Integral Transforms [Volumes I ] - Modified Bessel Functions of Other Arguments (56).
I need to know how to prove this transformation.
Only comment.
With CAS (Maple) I have:
$$\mathcal{L}_s^{-1}\left[K_v\left(\sqrt{a s}\right) I_v\left(\sqrt{b s}\right)\right](t)=\frac{\exp \left(-\frac{a+b}{4 t}\right) I_v\left(\frac{\sqrt{a b}}{2 t}\right)}{2 t}$$ for: $a>0,b>0,t>0$
You can find this transform in book on page: 247
