# This laplace Transform are true L(e^(it))=[(e^((i−s)t))/(i−s)]_0^(oo)=(1)/(s−i)?

Laplace Transform Example: $\mathcal{L}\left({e}^{it}\right)={\left[\frac{{e}^{\left(i-s\right)t}}{i-s}\right]}_{0}^{\mathrm{\infty }}=\frac{1}{s-i}$?
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RamPatWeese2w
Note that De Moivre's formula tells you that ${e}^{it}$ for any $t\in \mathbb{R}$ is bounded because $\mathrm{sin}t$ and $\mathrm{cos}t$ are bounded.
$\begin{array}{rl}\mathcal{L}\mathcal{\left(}{\mathcal{e}}^{\mathcal{i}\mathcal{t}}\mathcal{\right)}& ={\int }_{0}^{\mathrm{\infty }}{e}^{\left(i-s\right)t}\mathrm{d}t\\ & ={\left[\frac{{e}^{\left(i-s\right)t}}{i-s}\right]}_{0}^{\mathrm{\infty }}\\ & =\underset{\to 0}{\underset{⏟}{\underset{t\to \mathrm{\infty }}{lim}\frac{{e}^{it}}{{e}^{st}\left(i-s\right)}}}-\frac{1}{i-s}\\ & \to \frac{1}{s-i}\end{array}$
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duandaTed05
Note that for any real t and complex s, we have
$|{e}^{\left(i-s\right)t}|={e}^{\mathrm{Re}\left(\left(i-s\right)t\right)}={e}^{-at},$
where $a=\mathrm{Re}\left(s\right)$. (Recall in general that $|{e}^{z}|={e}^{\mathrm{Re}\left(z\right)}$ for any complex z.) So you can see that if $a=\mathrm{Re}\left(s\right)>0$, we have $\underset{t\to \mathrm{\infty }}{lim}|{e}^{\left(i-s\right)t}|=0$, and so
$\underset{t\to \mathrm{\infty }}{lim}\frac{{e}^{\left(i-s\right)t}}{i-s}=0.$