Laplace Transform Example: $\mathcal{L}({e}^{it})={\left[\frac{{e}^{(i-s)t}}{i-s}\right]}_{0}^{\mathrm{\infty}}=\frac{1}{s-i}$?

cimithe4c
2022-10-15
Answered

Laplace Transform Example: $\mathcal{L}({e}^{it})={\left[\frac{{e}^{(i-s)t}}{i-s}\right]}_{0}^{\mathrm{\infty}}=\frac{1}{s-i}$?

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RamPatWeese2w

Answered 2022-10-16
Author has **15** answers

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.

$$\begin{array}{rl}\mathcal{L}\mathcal{(}{\mathcal{e}}^{\mathcal{i}\mathcal{t}}\mathcal{)}& ={\int}_{0}^{\mathrm{\infty}}{e}^{(i-s)t}\mathrm{d}t\\ & ={\left[{\displaystyle \frac{{e}^{(i-s)t}}{i-s}}\right]}_{0}^{\mathrm{\infty}}\\ & =\underset{\to 0}{\underset{\u23df}{\underset{t\to \mathrm{\infty}}{lim}{\displaystyle \frac{{e}^{it}}{{e}^{st}(i-s)}}}}-{\displaystyle \frac{1}{i-s}}\\ & \to {\displaystyle \frac{1}{s-i}}\end{array}$$

$$\begin{array}{rl}\mathcal{L}\mathcal{(}{\mathcal{e}}^{\mathcal{i}\mathcal{t}}\mathcal{)}& ={\int}_{0}^{\mathrm{\infty}}{e}^{(i-s)t}\mathrm{d}t\\ & ={\left[{\displaystyle \frac{{e}^{(i-s)t}}{i-s}}\right]}_{0}^{\mathrm{\infty}}\\ & =\underset{\to 0}{\underset{\u23df}{\underset{t\to \mathrm{\infty}}{lim}{\displaystyle \frac{{e}^{it}}{{e}^{st}(i-s)}}}}-{\displaystyle \frac{1}{i-s}}\\ & \to {\displaystyle \frac{1}{s-i}}\end{array}$$

duandaTed05

Answered 2022-10-17
Author has **6** answers

Note that for any real t and complex s, we have

$$\left|{e}^{(i-s)t}\right|={e}^{\mathrm{Re}((i-s)t)}={e}^{-at},$$

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

$$\underset{t\to \mathrm{\infty}}{lim}\frac{{e}^{(i-s)t}}{i-s}=0.$$

$$\left|{e}^{(i-s)t}\right|={e}^{\mathrm{Re}((i-s)t)}={e}^{-at},$$

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

$$\underset{t\to \mathrm{\infty}}{lim}\frac{{e}^{(i-s)t}}{i-s}=0.$$

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