Laplace transform of an integral? 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

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 $\mathcal{L}(I)=5/(s(s+1))$.
