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.
Best Answer
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.