I was going through an Electrical Engineering textbook and came across the following proof for one of the properties of the Unilateral Laplace transform.
Integration property of the unilateral Laplace transform: Link for the proof from the book.
In the proof, it is stated that:
$$ e^{-st} \rightarrow 0 \mbox{ as } t \rightarrow \infty \ \ \ (i)$$
and therefore the term:
$$ -\frac{e^{-st}}{s}\int_{0-}^{t}{f(\tau) \ d\tau} = 0 \mbox{, as } t \rightarrow \infty \ \ \ (ii)$$
But my doubt is: Isn't there a case where the real part of $s$, gets cancelled by the function obtained after the integration of $f(\tau)$?
If that happens then we can't guarantee that (ii) would hold true right - i.e. the LHS in (ii) would not be zero right?
So, isn't the proof that has been provided wrong?