When taking Laplace transforms such as
$$\int_{t_0}^\infty e^{-st},$$
we subsequently get
$$-\dfrac{1}{s} \left[ e^{-st} \right]^\infty_{t_0}.$$
Now, my textbook author will just claim that $e^{-s \infty} = 0$, leaving us with the solution $\dfrac{1}{s} e^{-s t_0}$. However, since $s$ is a complex number, this seems exceedingly handy-wavy to me. I find this unacceptable, and I want to understand the mathematics of what's going on here.
In my research, I have encountered the concept of radius of convergence, and I suspect that this has something to do with what's going on here. However, doing a search for keywords, my textbook does not address the radius of convergence in any direct context of the Laplace transform, but rather later in the context of the Taylor/Laurent series.
I would greatly appreciate it if people could please take the time to explain what's going on here. Please note that I have not studied complex analysis – just pieces of it – so please provide careful explanations.
Best Answer
As you said
$$\int_{t_0}^\infty e^{-st}dt = \lim_{r \to +\infty}-\dfrac{1}{s} \left[ e^{-st} \right]^r_{t_0}$$
Now, the RHS is convergent if and only if $\mbox{Re}(s) >0$.
Added details
$$\lim_{r \to +\infty}-\dfrac{1}{s} \left[ e^{-st} \right]^r_{t_0}= \dfrac{1}{s} e^{-s t_0} - \lim_{r \to +\infty}\dfrac{1}{s} e^{-sr} \\ =\dfrac{1}{s} e^{-s t_0}- \lim_{r \to +\infty}\dfrac{1}{s} e^{-\mbox{Re}{(s)}r} \left(\cos( -\mbox{Im}(s)r)+i \sin(-\mbox{Im}(s)r) \right) $$
Now, if
In this case $$\lim_{r \to +\infty}-\dfrac{1}{s} \left[ e^{-st} \right]^r_{t_0}=\dfrac{1}{s} e^{-s t_0}$$
This yields the formula $$\int_{t_0}^\infty e^{-st}dt =\dfrac{1}{s} e^{-s t_0} \qquad \mbox{ for } \mbox{Re}(s) >0 \,.$$
Remember that, whenever we speak about the Laplace trasnform, the formula we get holds over the region of convergence. It is always important to emphasize/state what this region is, but this is often lost in a side note.