By one-to-one I mean, can two unequal functions have the same Laplace transform?
To demonstrate where this is coming from, assume that that we have a function : $F(s) = e^{-(s+7)}$
If we want to get the inverse laplace transform we can either apply the first shifting property of Laplace transforms as follow : $$f(t) = e^{-7t}\delta(t-1)$$
Or apply the linearity property directly as follows : $$\because F(s) = e^{-7}e^{-s} \therefore f(t) = e^{-7}\delta(t-1) $$ Where $\delta(t)$ is the Dirac Delta Distribution.
Notice that we get two different functions with each approach, which makes an assumption that the inverse Laplace transform is not one-to-one.
Similarly, we can see that a function has two different Laplace transforms.
Second : It can be proved that the Laplace transform of the function $f(t) = \sin t/t$ is $F(s) = \pi/2 - \arctan s$
Assuming that $G(s) = -\arctan s$
By differentiating both sides : $$G'(s) = {-1\over{s² + 1}}$$
By taking the inverse laplace transform to both sides:
$$\therefore -tg(t) = -\sin t \therefore g(t) = {\sin t \over t}$$
So we can see that F and G are two different functions but they have different inverse Laplace transforms, which adds to the original assumption that Laplace transform is not one-to-one.
Can someone explain, is there anything missing or wrong? Or Laplace transform is not one-to-one actually?