My question concerns the interval of convergence for the Laplace transform of the Bessel function of the first kind or order zero, namely $J_0(t)$, when found from its well-known Maclaurin series expansion of: $$J_0 (t) = \sum_{n = 0}^\infty \frac{(-1)^n t^{2n}}{2^{2n}(n!)^2},$$ which is valid for all $t$.
Under certain conditions on the coefficient $a_n$ found in a Maclaurin series expansion, finding its corresponding Laplace transform term-by-term is permissible (for a statement of those conditions, see, for example, P1.3.1 on page 21 of Laplace Transforms and Application by E. J. Watson (1981)). The Maclaurin series expansion for $J_0(t)$ satisfied these conditions. So in finding $\mathcal{L}\{J_0(t)\}(s)$ we have: \begin{align*} \mathcal{L}\{J_0(t)\}(s) &= \mathcal{L} \left \{\sum_{n = 0}^\infty \frac{(-1)^n t^{2n}}{2^{2n} (n!)^2} \right \}(s)\\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{2^{2n}(n!)^2} \mathcal{L}\{t^{2n}\}(s)\\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{2^{2n}(n!)^2} \cdot \frac{(2n)!}{s^{2n + 1}}\\ &= \frac{1}{s} \sum_{n = 0}^\infty \frac{(-1)^n}{2^{2n}} \binom{2n}{n} \frac{1}{s^{2n}}\\ &= \frac{1}{s} \frac{1}{\sqrt{1 + 1/s^2}} = \frac{1}{\sqrt{s^2 + 1}}, \end{align*} and is valid for $s > 1$. Note here the following Maclaurin series expansion has been used: $$\frac{1}{\sqrt{1 + x^2}} = \sum_{n = 0}^\infty \binom{2n}{n} \frac{(-1)^n}{2^{2n}} x^{2n}, \quad |x| < 1,$$ and we have set $x = \frac{1}{s}$.
It is however known that the Laplace transform for $J_0(t)$ is valid for all $s > 0$. My question then is this:
How can the interval of convergence for $\mathcal{L}\{J_0(t)\}(s)$ as found above be extended to $0 < s \leqslant 1$?