Given a discontinuous linear graph with a range of (0,1), how would go about finding the Laplace transform?
Okay well Here is an example I guess
$\displaystyle f(t)=\begin{cases} e^t, 0 \le t\le 1 \\ e, t > 1\end{cases}$
The Lapalce transform is
$\displaystyle \mathcal{L}(f)=\int_{0}^{\infty}e^{-st}dt=\int_{0}^{1}e^{-st}f(t)dt+\int_{1}^{\infty}e^{-st}f(t)dt$
Now using the definition of f gives
$\displaystyle \mathcal{L}(f)=\int_{0}^{1}e^{-st}e^{t}dt+\int_{1}^{\infty}e^{-st}\cdot 1dt$
Now just evaluate each of these integrals to get
$\displaystyle \mathcal{L}(f)=\int_{0}^{1}e^{-(s-1)t}dt+\int_{1}^{\infty}e^{-st}dt=-\frac{1}{s-1}e^{-(s-1)t}\bigg|_{0}^{1}-\frac{1}{s}e^{-st}\bigg|_{1}^{\infty}=$
$\displaystyle -\frac{e^{-(s-1)}}{s-1}+\frac{1}{s-1}+\frac{1}{s}e^{-s}$