# What is Laplace transform of (e^(-t)/(t) sin 3t sin 2t

What is Laplace transform of $\frac{{e}^{-t}}{t}\mathrm{sin}3t\mathrm{sin}2t$
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nirosoh9
$\frac{{e}^{-t}}{t}\mathrm{sin}\left(3t\right)\mathrm{sin}\left(2t\right)=\frac{{e}^{-t}}{2t}\mathrm{cos}\left(t\right)-\frac{{e}^{-t}}{2t}\mathrm{cos}\left(5t\right)$
Let $f\left(t\right)={e}^{-t}\mathrm{cos}\left(t\right)$ and $g\left(t\right)={e}^{-t}\mathrm{cos}\left(5t\right)$, we get
$\frac{{e}^{-t}}{t}\mathrm{sin}\left(3t\right)\mathrm{sin}\left(2t\right)=\frac{1}{2}\left(\frac{f\left(t\right)}{t}-\frac{g\left(t\right)}{t}\right)$
Using the property that

we get

where $F\left(s\right),G\left(s\right)$ are Laplace Transforms of $f\left(t\right),g\left(t\right)$. Using $L\left({e}^{-at}\mathrm{cos}\left(wt\right)\right)=\frac{s+a}{\left(s+a{\right)}^{2}+{w}^{2}}$, we can say
$F\left(u\right)=\frac{s+1}{\left(s+1{\right)}^{2}+1}$
$G\left(u\right)=\frac{s+1}{\left(s+1{\right)}^{2}+25}$
Computing the integrals:

Denote the limits as A,B, then:
$L\left(\frac{{e}^{-t}}{t}\mathrm{sin}\left(3t\right)\mathrm{sin}\left(2t\right)\right)=\frac{1}{4}\left[\mathrm{ln}\left(\left(s+1{\right)}^{2}+25\right)-\mathrm{ln}\left(\left(s+1{\right)}^{2}+1\right)+A-B\right]$
which is
$L\left(\frac{{e}^{-t}}{t}\mathrm{sin}\left(3t\right)\mathrm{sin}\left(2t\right)\right)=\frac{1}{4}\left[\mathrm{ln}\frac{\left(s+1{\right)}^{2}+25}{\left(s+1{\right)}^{2}+1}+\underset{u\to \mathrm{\infty }}{lim}\mathrm{ln}\frac{\left(u+1{\right)}^{2}+25}{\left(u+1{\right)}^{2}+1}\right]=\frac{1}{4}\mathrm{ln}\frac{\left(s+1{\right)}^{2}+25}{\left(s+1{\right)}^{2}+1}$
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Makayla Reilly
With integration of the formula $\mathcal{L}\left({e}^{ct}\right)=\frac{1}{s-c}$ respect to $c$ we find
$\mathcal{L}\left(\frac{{e}^{ct}}{t}\right)=\mathrm{ln}\frac{1}{s-c}$
then by $\mathrm{sin}\alpha =\frac{{e}^{i\alpha }-{e}^{-i\alpha }}{2i}$ we write
$\begin{array}{rl}\mathcal{L}\left(\frac{{e}^{-t}\mathrm{sin}3t\mathrm{sin}2t}{t}\right)& =-\frac{1}{4}\mathcal{L}\left(\frac{{e}^{\left(5i-1\right)t}-{e}^{\left(-i-1\right)t}-{e}^{\left(i-1\right)t}+{e}^{\left(-5i-1\right)t}}{t}\right)\\ & =-\frac{1}{4}\mathcal{L}\left(\mathrm{ln}\frac{1}{s-\left(5i-1\right)}-\mathrm{ln}\frac{1}{s-\left(-i-1\right)}-\mathrm{ln}\frac{1}{s-\left(i-1\right)}+\mathrm{ln}\frac{1}{s-\left(-5i-1\right)}\right)\\ & =-\frac{1}{4}\mathrm{ln}\frac{\left(s+1{\right)}^{2}+1}{\left(s+1{\right)}^{2}+25}\end{array}$