This question is very much related to this (one).
Let $F\colon \mathbb{C}\to \mathbb{C}$ be defined as $$F(s) = \frac{1}{4+3s+\sqrt{s(4+s)}}.$$ My question is what is the inverse Laplace transform of $F$? Like the previous link, I suspect that it has something to to with Bessel functions. I have used Mathematica in a trial and error process and haven't made any progress.
I appreciate any help.
We have:
$$G(s)=F(s-2) = \frac{1}{3s-2+\sqrt{s^2-4}} $$ so, in order to tackle this problem like the previous one, we should find the coefficients of the Taylor series at $x=0$ of: $$ H(x) = G\left(\frac{1}{x}\right) = \frac{x}{3-2x+\sqrt{1-4x^2}}=\sum_{n\geq 1}g_n x^n\tag{1} $$ to deduce (have also a look at Ramanujan's master theorem): $$ (\mathcal{L}^{-1}G)(t) = \sum_{n\geq 1}\frac{g_n t^{n-1}}{(n-1)!},\qquad (\mathcal{L}^{-1} F)(t)=e^{-2t}\sum_{n\geq 0}\frac{g_{n+1} t^n}{n!}\tag{2}$$ The main issue here is that while the Taylor coefficients of $\frac{x}{1-\sqrt{1-4x^2}}$ are pretty easy to find in terms of Catalan numbers, the coefficients $g_n$ appearing in $(1)$ and $(2)$ are a bit more involved.
Moreover, $H(x)$ has two singularities located at $\frac{3\pm i\sqrt{7}}{4}$, hence the RHS of $(2)$ is expected to converge (much) slower than the Taylor series of a Bessel function.