I have seen a proof of $\mathcal{L}^{-1}(0) =0$ here: Inverse Laplace Transform of zero, but I would like to prove it by definition.
We have $\int_0^\infty e^{-st}f(t)dt=0$. Here, I would like to say that this implies that $e^{-st}f(t)=0$, and since $e^{-st}\neq0$, we have that $f(t)=0$.
However, I'm struggling with justifying that $\int_0^\infty f(x)dx=0$ implies $f(x)=0$.
I have tried to think of functions that have the same "positive" and "negative" area under the graph for $x \in [0, \infty)$, and I have only been able to think of $sin(x)$ and $cos(x)$, for which the integral $\int_0^\infty f(x)dx$ is not defined, so my intuition tells me that $f(x)$ should indeed be indentically $0$.
Is my implication correct? Why/ why not?