Inverse Laplace transform of function with square roots.

Question

Let $F\colon \mathbb{C}\to \mathbb{C}$ be defined as $$F(s) = \frac{1}{2+s+\sqrt{4s+s^2}}.$$ My question is, what is the inverse Laplace transform of $F$? From solving similar problems I have a feeling that Bessel functions are involved. My first guess would be $f(t) = \exp(-2t)J_0(2t)$, but that is incorrect. Any ideas?

Answer 1

Hint: the inverse Laplace transform of: $$ f(s) = \frac{1}{s+\sqrt{s^2-4}}=\frac{s-\sqrt{s^2-4}}{4} $$ is given by:

$$ (\mathcal{L}^{-1} f)(x) = \color{red}{\frac{I_1(2x)}{2x}}=\frac{1}{2}\sum_{n\geq 0}\frac{x^{2n}}{n!(n+1)!}\tag{1} $$

where $I_1$ is a modified Bessel function of the first kind, and it is probably easier to show that the Laplace transform of the RHS of $(1)$ is $f(s)$ by exploiting the Taylor series of $I_1$ and the generalized binomial theorem / the generating function for Catalan numbers. From $(1)$ it follows that:

$$ (\mathcal{L}^{-1} F)(t) = \color{red}{\frac{I_1(2t)}{2t}e^{-2t}}\tag{2} $$