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 <br /> 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&#039;&#039;(t)e^{-st} is only valid for \operatorname{Re}(s) &gt; 0. The unique analytic extension of X to \operatorname{Re}(s) \leq 0 then gives X(s) \equiv 0 everywhere.