From mathematica, I get $\mathscr{L^{-1}}(\frac{K_0(as)}{s}) = Log[\frac{t}{a} + \sqrt{-1+\frac{t^2}{a^2}}]$ when $a \leq t$.
Then we know $ \mathscr{L^{-1}} (s F(s)) = \frac{d}{dt} f(t) + \mathscr{L^{-1}} (f(0))$.
So I get $\mathscr{L^{-1}}(K_0(as)) = \frac{d}{dt} Log[\frac{t}{a} + \sqrt{-1+\frac{t^2}{a^2}}] + \mathscr{L^{-1}} ( Log[i]) = \frac{1}{\sqrt{t^2-a^2}} + \frac{i \pi}{2} \delta(t)$ . But we know $\mathscr{L^{-1}}(K_0(as)) =\frac{1}{\sqrt{t^2-a^2}} $ for $a \leq t$. I do not understand why the $\mathscr{L^{-1}} (f(0)) = \frac{i \pi}{2} \delta(t)$ part just disappeared.
And I am thinking about this question because I want to get $\mathscr{L^{-1}}(sK_0(as))$ using the same method. However, I noticed there is no $\mathscr{L^{-1}}(sK_0(as))$ in the laplace table or using mathematica. So my second question is that why can't I use the same method to calculate $\mathscr{L^{-1}}(sK_0(as))$ . (Because if I can, then I believe mathematica should be able to get it). Thank you for any help.