Intuitively how can the Laplace transform be injective? You are taking an integral with limits $0$ and $\infty$. So you don't care about the function before $0$. Define $g(x)=x^2$ for $x>0$ and $g(x)=0$ for $x \leq 0$ Define $f(x)=x^2$ for $x>0$ and $f(x)=-x$ for $x \leq 0$
$f$ and $g$ must have the same Laplace transform by definition right? Yet they are not equal. Can someone please explain?