Laplace Transform (and Radius of Convergence?): $e^{-s \infty} = 0$?

Question

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.

Answer 1

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 $\mbox{Re}(s) >0$.

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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

• $\mbox{Re}{(s)} >0$ then $\lim_{r \to \infty} e^{-\mbox{Re}{(s)}r} =0$ and by the squeeze theorem $$ \lim_{r \to +\infty}-\dfrac{1}{s} \left[ e^{-sr} \right]^r_{t_0} = \dfrac{1}{s} e^{-s t_0}$$
• $\mbox{Re}{(s)} =0$ then $$-\dfrac{1}{s} \left[ e^{-st} \right]^r_{t_0}= \left(\cos( -\mbox{Im}(s)r)+i \sin(-\mbox{Im}(s)r) \right)$$ does not have a limit at $r =\infty$.
• $\mbox{Re}{(s)} < 0$ then $\lim_{r \to \infty} e^{-\mbox{Re}{(s)}r} =\infty$ and the limit $$ \lim_{r \to +\infty} - \dfrac{1}{s} \left[ e^{-st} \right]^r_{t_0}$$ cannot be finite since $$\lim_{r \to +\infty}\left| -\dfrac{1}{s} \left[ e^{-st} \right]^r_{t_0} \right| = \infty$$

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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.