skystasvs
2022-09-13
Answered

What is Laplace transform of $\frac{{e}^{-t}}{t}\mathrm{sin}3t\mathrm{sin}2t$

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nirosoh9

Answered 2022-09-14
Author has **16** answers

$$\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t)=\frac{{e}^{-t}}{2t}\mathrm{cos}(t)-\frac{{e}^{-t}}{2t}\mathrm{cos}(5t)$$

Let $f(t)={e}^{-t}\mathrm{cos}(t)$ and $g(t)={e}^{-t}\mathrm{cos}(5t)$, we get

$$\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t)=\frac{1}{2}(\frac{f(t)}{t}-\frac{g(t)}{t})$$

Using the property that

$$L(\frac{f(t)}{t})=\underset{s}{\overset{\mathrm{\infty}}{\int}}F(u)\text{}du$$

we get

$$L(\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t))=\frac{1}{2}[\underset{s}{\overset{\mathrm{\infty}}{\int}}F(u)\text{}du-\underset{s}{\overset{\mathrm{\infty}}{\int}}G(u)\text{}du]$$

where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$. Using $L({e}^{-at}\mathrm{cos}(wt))=\frac{s+a}{(s+a{)}^{2}+{w}^{2}}$, we can say

$$F(u)=\frac{s+1}{(s+1{)}^{2}+1}$$

$$G(u)=\frac{s+1}{(s+1{)}^{2}+25}$$

Computing the integrals:

$$\underset{s}{\overset{\mathrm{\infty}}{\int}}F(u)\text{}du=\underset{s}{\overset{\mathrm{\infty}}{\int}}\frac{u+1}{(u+1{)}^{2}+1}\text{}du=\frac{1}{2}[\underset{u\to \mathrm{\infty}}{lim}\mathrm{ln}((u+1{)}^{2}+1)-\mathrm{ln}((s+1{)}^{2}+1)]$$

$$\underset{s}{\overset{\mathrm{\infty}}{\int}}G(u)\text{}du=\underset{s}{\overset{\mathrm{\infty}}{\int}}\frac{u+1}{(u+1{)}^{2}+25}\text{}du=\frac{1}{2}[\underset{u\to \mathrm{\infty}}{lim}\mathrm{ln}((u+1{)}^{2}+25)-\mathrm{ln}((s+1{)}^{2}+25)]$$

Denote the limits as A,B, then:

$$L(\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t))=\frac{1}{4}[\mathrm{ln}((s+1{)}^{2}+25)-\mathrm{ln}((s+1{)}^{2}+1)+A-B]$$

which is

$$L(\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t))=\frac{1}{4}[\mathrm{ln}\frac{(s+1{)}^{2}+25}{(s+1{)}^{2}+1}+\underset{u\to \mathrm{\infty}}{lim}\mathrm{ln}\frac{(u+1{)}^{2}+25}{(u+1{)}^{2}+1}]=\frac{1}{4}\mathrm{ln}\frac{(s+1{)}^{2}+25}{(s+1{)}^{2}+1}$$

Let $f(t)={e}^{-t}\mathrm{cos}(t)$ and $g(t)={e}^{-t}\mathrm{cos}(5t)$, we get

$$\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t)=\frac{1}{2}(\frac{f(t)}{t}-\frac{g(t)}{t})$$

Using the property that

$$L(\frac{f(t)}{t})=\underset{s}{\overset{\mathrm{\infty}}{\int}}F(u)\text{}du$$

we get

$$L(\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t))=\frac{1}{2}[\underset{s}{\overset{\mathrm{\infty}}{\int}}F(u)\text{}du-\underset{s}{\overset{\mathrm{\infty}}{\int}}G(u)\text{}du]$$

where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$. Using $L({e}^{-at}\mathrm{cos}(wt))=\frac{s+a}{(s+a{)}^{2}+{w}^{2}}$, we can say

$$F(u)=\frac{s+1}{(s+1{)}^{2}+1}$$

$$G(u)=\frac{s+1}{(s+1{)}^{2}+25}$$

Computing the integrals:

$$\underset{s}{\overset{\mathrm{\infty}}{\int}}F(u)\text{}du=\underset{s}{\overset{\mathrm{\infty}}{\int}}\frac{u+1}{(u+1{)}^{2}+1}\text{}du=\frac{1}{2}[\underset{u\to \mathrm{\infty}}{lim}\mathrm{ln}((u+1{)}^{2}+1)-\mathrm{ln}((s+1{)}^{2}+1)]$$

$$\underset{s}{\overset{\mathrm{\infty}}{\int}}G(u)\text{}du=\underset{s}{\overset{\mathrm{\infty}}{\int}}\frac{u+1}{(u+1{)}^{2}+25}\text{}du=\frac{1}{2}[\underset{u\to \mathrm{\infty}}{lim}\mathrm{ln}((u+1{)}^{2}+25)-\mathrm{ln}((s+1{)}^{2}+25)]$$

Denote the limits as A,B, then:

$$L(\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t))=\frac{1}{4}[\mathrm{ln}((s+1{)}^{2}+25)-\mathrm{ln}((s+1{)}^{2}+1)+A-B]$$

which is

$$L(\frac{{e}^{-t}}{t}\mathrm{sin}(3t)\mathrm{sin}(2t))=\frac{1}{4}[\mathrm{ln}\frac{(s+1{)}^{2}+25}{(s+1{)}^{2}+1}+\underset{u\to \mathrm{\infty}}{lim}\mathrm{ln}\frac{(u+1{)}^{2}+25}{(u+1{)}^{2}+1}]=\frac{1}{4}\mathrm{ln}\frac{(s+1{)}^{2}+25}{(s+1{)}^{2}+1}$$

Makayla Reilly

Answered 2022-09-15
Author has **3** answers

With integration of the formula $\mathcal{L}\left({e}^{ct}\right)={\displaystyle \frac{1}{s-c}}$ respect to $c$ we find

$$\mathcal{L}\left({\displaystyle \frac{{e}^{ct}}{t}}\right)=\mathrm{ln}{\displaystyle \frac{1}{s-c}}$$

then by $\mathrm{sin}\alpha ={\displaystyle \frac{{e}^{i\alpha}-{e}^{-i\alpha}}{2i}}$ we write

$$\begin{array}{rl}\mathcal{L}\left({\displaystyle \frac{{e}^{-t}\mathrm{sin}3t\mathrm{sin}2t}{t}}\right)& =-{\displaystyle \frac{1}{4}}\mathcal{L}\left({\displaystyle \frac{{e}^{(5i-1)t}-{e}^{(-i-1)t}-{e}^{(i-1)t}+{e}^{(-5i-1)t}}{t}}\right)\\ & =-{\displaystyle \frac{1}{4}}\mathcal{L}(\mathrm{ln}{\displaystyle \frac{1}{s-(5i-1)}}-\mathrm{ln}{\displaystyle \frac{1}{s-(-i-1)}}-\mathrm{ln}{\displaystyle \frac{1}{s-(i-1)}}+\mathrm{ln}{\displaystyle \frac{1}{s-(-5i-1)}})\\ & ={-{\displaystyle \frac{1}{4}}\mathrm{ln}{\displaystyle \frac{(s+1{)}^{2}+1}{(s+1{)}^{2}+25}}}\end{array}$$

$$\mathcal{L}\left({\displaystyle \frac{{e}^{ct}}{t}}\right)=\mathrm{ln}{\displaystyle \frac{1}{s-c}}$$

then by $\mathrm{sin}\alpha ={\displaystyle \frac{{e}^{i\alpha}-{e}^{-i\alpha}}{2i}}$ we write

$$\begin{array}{rl}\mathcal{L}\left({\displaystyle \frac{{e}^{-t}\mathrm{sin}3t\mathrm{sin}2t}{t}}\right)& =-{\displaystyle \frac{1}{4}}\mathcal{L}\left({\displaystyle \frac{{e}^{(5i-1)t}-{e}^{(-i-1)t}-{e}^{(i-1)t}+{e}^{(-5i-1)t}}{t}}\right)\\ & =-{\displaystyle \frac{1}{4}}\mathcal{L}(\mathrm{ln}{\displaystyle \frac{1}{s-(5i-1)}}-\mathrm{ln}{\displaystyle \frac{1}{s-(-i-1)}}-\mathrm{ln}{\displaystyle \frac{1}{s-(i-1)}}+\mathrm{ln}{\displaystyle \frac{1}{s-(-5i-1)}})\\ & ={-{\displaystyle \frac{1}{4}}\mathrm{ln}{\displaystyle \frac{(s+1{)}^{2}+1}{(s+1{)}^{2}+25}}}\end{array}$$

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