# Laplace transform involving step function $$\displaystyle{f{{\left({t}\right)}}}={\frac{{{\sin{{\left({2}{t}\right)}}}}}{{{e}^{{{2}{t}}}}}}+{t}{\left\lbrace\cdot\right\rbrace}{u}{\left({t}-{4}\right)}$$

Laplace transform involving step function
$f\left(t\right)=\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t·u\left(t-4\right)$

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srasloavfv
$\text{F}\left(\text{s}\right)={L}_{t}{\left[\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t\theta \left(t-4\right)\right]}_{\left(\text{s}\right)}\phantom{\rule{0.222em}{0ex}}={\int }_{0}^{\mathrm{\infty }}{e}^{-\text{s}t}\left(\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t\theta \left(t-4\right)\right)dt$
Where $\theta \left(x\right)$ is the Heaviside step function.
Now:
$\text{F}\left(\text{s}\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-\text{s}t}\cdot \frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}dt+{\int }_{0}^{\mathrm{\infty }}t{e}^{-\text{s}t}\theta \left(t-4\right)dt=$
${\int }_{0}^{\mathrm{\infty }}{e}^{-t\left(2+\text{s}\right)}\mathrm{sin}\left(2t\right)dt+{\int }_{4}^{\mathrm{\infty }}t{e}^{-st}dt$
So, we get:
1. When $R\left[\text{s}\right]>-2$
${\int }_{0}^{\mathrm{\infty }}{e}^{-t\left(2+s\right)}\mathrm{sin}\left(2t\right)dt=\frac{2}{{s}^{2}+4\left(2+s\right)}$
2. When $R\left[\text{s}\right]>0$
${\int }_{4}^{\mathrm{\infty }}t{e}^{-st}dt=\frac{{e}^{-4s}\left(1+4s\right)}{{s}^{2}}$
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lildeutsch11xq2j

You can use the formula for finding the laplace transform of secondary part-
$L\left\{f\left(t\right)u\left(t-a\right)\right\}$$=L\frac{\left\{f\left(t+a\right)\right\}×{e}^{-as}}{s}$

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