How can we prove the correctness of the integration property of the Laplace transform?

Question

I was going through an Electrical Engineering textbook for understanding the Laplace transform and came across the following proof for one of the properties of the Unilateral Laplace transform.

Integration property of the unilateral Laplace transform:

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?

Answer 1

You are right that the argument in the proof is not correct, or at least misleading. The fact that $e^{-st}$ becomes zero as $t\to\infty$ is true for any $s$ with $\textrm{Re}\{s\}>0$. However, the integral

$$\int_0^{\infty}f(\tau)d\tau$$

might not exist, so we can't just compute the limit by claiming that the first term becomes zero.

We have to compute

$$\lim_{t\to\infty}-\frac{e^{-st}}{s}\int_0^{t}f(\tau)d\tau\tag{1}$$

The limit $(1)$ only equals zero for $s$ inside the region of convergence (ROC) of $f(t)$, i.e., we don't just require that $e^{-st}\to 0$ for $t\to\infty$, but we require it to decay sufficiently fast to compensate for the growth of the integral.