I was watching the video: https://www.youtube.com/watch?v=IYOzTt-gB8A.
I thought we must have some condition for the $f$. When we say that the evaluation at $\infty$ of $\int_0^t F(u)du\dfrac{e^{-st}}{-s}$ is $0$, what we are supposing?
For existence of transform, we suppose $|F(t)|<Ke^{st}$ for bigg $t$. And if we have $F(t)=e^{st}$, for instance, is it OK? If yes, we have
$$\int_0^t F(u)du\dfrac{e^{-st}}{-s}=\int_0^t -\dfrac{1}{s}dt=\infty$$...?
What are the conditions to the existence?
Many thanks in advance.
$$\mathcal{L} \left\{\int_0^t F(\tau)\, d\tau \right\} \textrm{ exists when } |F(t)| \le Ke^{at}.$$
For example, if $F(t)=e^{at}$, then
$$\mathcal{L} \left\{\int_0^t F(\tau)\, d\tau \right\}=\int_0^\infty e^{-st} \left\{\int_0^t e^{a\tau}\, d\tau \right\} dt=\int_0^\infty e^{-st}\left[ \frac{e^{at}-1}{a} \right]\, dt$$ which converges for $\operatorname{Re}s>a.$