I have only had a one hour lecture on these, so it is hard to think about these.
Show that $\displaystyle \cal{L} $ $\displaystyle [f'(t)] = sF(s) - f(0)$
Integrating by parts we obtain...
$\displaystyle \mathcal {L} \{f^{'}(t)\}= \int_{0}^{\infty} e^{-st}\cdot f^{'}(t)\cdot dt =$
$\displaystyle = |e^{-st}\cdot f(t)|_{0}^{\infty} + s\cdot \int_{0}^{\infty} e^{-st}\cdot f(t)\cdot dt =$
$\displaystyle s\cdot \int_{0}^{\infty} e^{-st}\cdot f(t)\cdot dt - f(0)= s\cdot F(s) - f(0)$ (1)
... where we use the fact that if $\displaystyle \mathcal{L} \{f(t)\}$ exists, then is...
$\displaystyle \lim_{p \rightarrow \infty} e^{-sp} f(p) =0$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Applying the residue theorem we find...
$\displaystyle F(s)=\frac{s-1}{(s-3)\cdot (s+1)} = \frac{r(3)}{s-3} + \frac{r(-1)}{s+1}$ (1)
... where...
$\displaystyle r(3)= \lim_{s \rightarrow 3} F(s)\cdot (s-3)= \frac{1}{2}$
$\displaystyle r(-1)= \lim_{s \rightarrow -1} F(s)\cdot (s+1)= \frac{1}{2}$ (2)
... so that is...
$\displaystyle \mathcal {L}^{-1} \{\frac{s-1}{(s-3)\cdot (s+1)} \}= \frac{1}{2}\cdot e^{-t} + \frac{1}{2}\cdot e^{3t}$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
for the inverse transform use partial fractions to split the expression up. you can check if i did it correctly but i got (1/2)/(s-3) + (1/2)/(s+1). so you should be able to recognize these as laplace transforms of the exponential function. so you should get (1/2)e^(3t) + (1/2)e^(-t).