Inverse Laplace transform of $K_0 \left(r \sqrt{s^2-1}\right)$ This question is about inverse Laplace transform $\mathscr{L}^{-1}:s\rightarrow t$.  Although I was not able to find appropriate contour to invert $K_0 \left(r s\right)$, I somehow know that
$$\mathscr{L}^{-1}\{K_0 \left(r s\right)\}=\frac{\theta(t-r)}{\sqrt{t^2-r^2}}.$$
How do one show that statement rigorously? The main part of my question is: Is it possible, similarly, to express a similar inverse fourier transform
$$\mathscr{L}^{-1}\left\{K_0 \left(r \sqrt{s^2-1}\right)\right\}$$
in terms of elementary functions? Thank you for suggestions. ($K_0$ is the modified Bessel function of second kind, $\theta$ is just Heaviside theta, $r>0$ is a positive real parameter).

Important note: The function $K_0$ in the second laplace transform is ill-defined for $s\in (0,1)$ and is ment to represent only its real part, equivalently, using common identities for Bessel functions,
$$K_0 \left(r \sqrt{s^2-1}\right) = -\frac{\pi}{2}Y_0\left(r\sqrt{1 - s^2}\right)$$
which extends the domain of the original function to $s\in (0,1)$.

With CAS help:
$$\mathcal{L}_s^{-1}\left[K_0\left(r \sqrt{s^2-1}\right)\right](t)=\frac{\theta (t-r) \cosh \left(\sqrt{t^2-r^2}\right)}{\sqrt{t^2-r^2}}$$
for $r>0$
Maple code:
`assuming`([inttrans:-invlaplace(BesselK(0, r*sqrt(s^2-1)), s, t)], [r > 0])

#Heaviside(t-r)*cosh(sqrt(-r^2+t^2))/sqrt(-r^2+t^2)

