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I have the following integral:

$$ I = \int_{0}^{t} e^{-\tau}\theta(t-\tau )d\tau $$

where $\theta(t) \left\{\begin{matrix} 0& 0 \leq t < 1\\ 5 & t\geq 1 \end{matrix}\right.$ is the Heaviside step function. $$$$ I need to get the Laplace transform of it. I know there is a rule for it: $ \mathcal{L} \left \{ I \right \} = F(s)G(s), $ but I don't know if its correct the do the transformation as follows:

$$ \mathcal{L} \left \{ e^{-\tau} \right \}= e^{-\tau}\mathcal{L}\left \{ 1\right \}$$
Because the transformation should be in respect of $ t $

Thanks in advance

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    $\begingroup$ When you use the rule $\mathcal{L} \left \{ I \right \} = F(s)G(s),$ you are assuming that $f(t)=e^{-t}$ and $g(t)=\theta(t)$. So you can not pull $e^{-\tau}$ out. Also, your definition of Heavise step function doesn't seem right to me. Shouldn't it be $1$ when $t\ge \tau$? $\endgroup$ – KittyL Dec 18 '16 at 19:34
  • $\begingroup$ You are right, bit it's a particular function in this case. Thanks for the help! Can you put it as an answer please? $\endgroup$ – Joshua Salazar Dec 18 '16 at 19:42
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    $\begingroup$ $\theta$ has nothing to do with $\tau$ in your definition. Are you sure it is exactly like that? Or is it just $\theta(t)=...$? $\endgroup$ – KittyL Dec 18 '16 at 20:01
  • $\begingroup$ I think you're right. So in this case $\mathcal{L} = \mathcal{L} \left \{ \theta(t-\tau) \right \} = \frac{ 5e^{-t} }{s}$, is it right? $\endgroup$ – Joshua Salazar Dec 18 '16 at 20:28
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When you use the rule $\mathcal{L} (I)=F(s)G(s)$, you are assuming that $f(t)=e^{−t}$ and $g(t)=\theta(t)$. So you can not pull $e^{−\tau}$ out.

So in this case $\mathcal{L}(I)=F(s)G(s)$, where $F(s)$ is the Laplace transform of $f(t)=e^{-t}$, and $G(s)$ is the Laplace transform of $\theta(t)$.

Now for $F(s)$, we can use the translation $\mathcal{L}(e^{at}f(t))=F(s-a)$ to obtain $1/(s+1)$. And $G(s)=5/s$. So $\mathcal{L}(I)=5/(s(s+1))$.

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