Find the Laplace of the given function using the definition
$f(t)=\left\{\begin{array}{rll} -1, & \text { se } 0 \leq t<1 \\ 1, & \text { se } t \geq 1 \end{array}\right.$
I know what the answer is according to a sheet that I have of common transforms but I am not 100% on how to get there using the definition
I know that if the question were to be:
f(t) = -1[H(t-0) - H(t-1)] + 1H(t-1)
then it's a simple
$\int_0^{\infty} e^{-s t} * (-1[H(t-0) - H(t-1)] + 1H(t-1))$
but, how to solve this question?
what I've been trying:
$\begin{aligned} & \text { I) } \int_0^1 e^{-s t} u(t) d t=-\int_0^{-1} e^{-s t} d t=\left.\frac{1}{s} e^{-s t}\right|_0 ^1=\frac{1}{s} e^{-s*1}-\frac{1 e^{0*t}}{S}=\frac{e^{-s}-1}{S} \\ & \text { II) } \int_1^{\infty} e^{-s t} u(t) d t=\lim _{A \rightarrow \infty}\left(-\frac{e^{-S A}}{S}+\frac{e^{-S \cdot 0}}{S}\right)=\frac{1}{S} \\ & \rho\{f(t)\}=\frac{e^{-s}}{s} \\ & \end{aligned}$