Laplace transform of a simple equation (Simple question)

[LagrangeEuler]

Lets consider very simple equation $x''(t)=0$ for $x(0)=0$, $x'(0)=0$. By employing Laplace transform I will get
$$s^2X(s)=0$$ where $X(s)$ is Laplace transform of $x(t)$. Why then this is equivalent to
$$X(s)=0$$
why we do not consider $s=0$?

 

[anuttarasammyak]

The solution is x(t)=0. Laplace transform of 0 is also 0.

 

[LagrangeEuler]

I know that. I am asking you why we do not consider that for $s=0$, $X(s)$ can be different from zero?

 

[anuttarasammyak]

I do not think our problem is to get value of s to satisfy
$$s^2 X(s) = 0$$
because s is parameter transformed which varies from 0 to infinity in usual.
We should have interest on Laplace transform X(s) and observe that it returns zero for any s.

 

[pasmith]

I am asking you why we do not consider that for $s = 0$, $X(s)$ can be different from zero?

The definition $$X(s) \equiv \int_0^\infty x(t)e^{-st}\,dt$$ is not valid for every $s \in \mathbb{C}$ (in general it's only valid for $s$ with sufficiently large and positive real part). Where that definition is not valid, we must define $X(s)$ by analytic continuation from the domain where it is valid. In this case, the equation $s^2X(s) = 0$ obtained from integrating $x''(t)e^{-st}$ is only valid for $\operatorname{Re}(s) > 0$. The unique analytic extension of $X$ to $\operatorname{Re}(s) \leq 0$ then gives $X(s) \equiv 0$ everywhere.