Checking region of validity for a standard Laplace transform In these notes, http://www.math.psu.edu/papikian/Kreh.pdf, Theorem 2.14 it states that
$$\mathcal{L}[J_0](s)=\frac{1}{\sqrt{1+s^2}} $$
which I suppose is equivalent to
$$ \int_0^{\infty}J_0(x)e^{-sx}dx=\frac{1}{\sqrt{1+s^2}}$$
but it makes not reference to what values of $s$ and $x$ this is valid for. I think $x \in \mathbb{R}$, $x \geq 0$ but can $s$ belong to all of $\mathbb{C}$?
I know that $J_0$ is an entire function if this helps.
 A: $J_0(x)$ is bounded on $\mathbb R$, 
so the integral defining the Laplace transform will converge absolutely for $\text{Re}(s) > 0$.  
From the asymptotics $J_0(x) = \sqrt{2/(\pi x)} \sin(x + \pi/4) + O(x^{-3/2})$ as $x \to \infty$, the integral diverges for $\text{Re}(s) < 0$.   For $\text{Re}(s) = 0$, say $s = i\omega$, it converges (conditionally) if 
$\int_1^\infty x^{-1/2} e^{i (\omega \pm 1) x}\; dx$ both do.
Now integration by parts gives 
$$ \int_1^R x^{-1/2} e^{i t x}\; dx =  \dfrac{ie^{it}}{t} - \dfrac{i e^{itR}}{t\sqrt{R}} - \dfrac{i}{2t} \int_1^R x^{-3/2} e^{itx}\; dx $$
and since $x^{-3/2} e^{itx}$ is integrable on $[1,\infty)$ this has a limit as $R \to \infty$ for real $t \ne 0$.  Thus for $\text{Re}(s) = 0$, the integral defining the Laplace transform converges except at $s = \pm i$.
For $s = i$, we can explicitly calculate
$$ \int_0^R J_0(x) e^{-ix}\; dx = R\; (J_0(R) + i J_1(R)) e^{-iR}$$
and the asymptotics of $J_0(R)$ and $J_1(R)$ shows that this diverges as $R \to \infty$.  Similarly for $s=-i$.
