Laplace Transform Example: $\mathcal{L}(e^{it}) = \left[ \frac{e^{(i - s)t}}{i - s} \right]_0^\infty = \frac{1}{s - i}$?

Question

My textbook on Laplace transforms gives the following example:

$$\begin{align} \mathcal{L}(e^{it}) &= \int_0^\infty e^{-st} e^{it} \ dt \\ &= \int_0^\infty e^{t(i - s)} \ dt \\ &= \left[ \dfrac{e^{(i - s)t}}{i - s} \right]_0^\infty \\ &= \dfrac{1}{s - i} \\ &= \dfrac{s}{s^2 + 1} + i\dfrac{1}{s^2 + 1} \end{align}$$

How did the author get from $\left[ \dfrac{e^{(i - s)t}}{i - s} \right]_0^\infty$ to $\dfrac{1}{s - i}$? We actually get $\left[ \dfrac{e^{(i - s)t}}{i - s} \right]_0^\infty = \dfrac{e^{(i - s)\infty}}{i - s} - \dfrac{1}{s - i}$, and I don't see how it's valid to just discard $\dfrac{e^{(i - s)\infty}}{i - s}$.

I would greatly appreciate it if people could please take the time to explain what's going on here.

Answer 1

Note that De Moivre's formula tells you that $e^{it}$ for any $t\in \mathbb{R}$ is bounded because $\sin t$ and $\cos t$ are bounded.

$$\begin{aligned}\mathcal{L(e^{it})}&=\int_{0}^{\infty}e^{(i-s)t}\mathrm dt\\&=\left[\dfrac{e^{(i-s)t}}{i-s}\right]_{0}^{\infty}\\&=\underbrace{\lim_{t\to \infty}\dfrac{e^{it}}{e^{st}(i-s)}}_{\to 0}-\dfrac{1}{i-s}\\&\to \dfrac{1}{s-i}\end{aligned}$$

Answer 2

$\newcommand{\Re}{\operatorname{Re}}$Note that for any real $t$ and complex $s$, we have $$\left|e^{(i-s)t}\right| = e^{\Re((i-s)t)} = e^{-at},$$ where $a=\Re(s)$. (Recall in general that $\left|e^z\right| = e^{\Re(z)}$ for any complex $z$.) So you can see that if $a=\Re(s) > 0$, we have $\lim\limits_{t\to\infty}\left|e^{(i-s)t}\right| = 0$, and so $$\lim\limits_{t\to\infty}\frac{e^{(i-s)t}}{i-s}=0.$$