Find Laplace Transform using unit step function and t-shifting. (5.3-38)
Sophia Terry 
Please check my work. Did I calculate the following Laplace Transform correctly?
$$\mathcal{L}\{t^2u(t-1)\}=\mathcal{L}\{(t-1)^2u(t-1)\}+\mathcal{L}\{u(t-1)\}$$
$$=\frac{2e^{-s}}{s^3}+\frac{e^{-s}}{s}$$
$\endgroup$2 Answers
$\begingroup$Your answer is incorrect. It is easier to resort to definition of the Laplace transform. $$\int_0^{\infty} t^2\mathcal{U}(t-1)e^{-st}dt = \int_1^{\infty} t^2e^{-st}dt$$ Note that $\frac{\partial^2}{\partial s^2}e^{-st} = t^2e^{-st}$. $$ \frac{\partial^2}{\partial s^2}\int_1^{\infty} e^{-st}dt = \frac{d^2}{ds^2}\frac{e^{-s}}{s} = \frac{2 e^{-s}}{s^3}+\frac{2 e^{-s}}{s^2}+\frac{e^{-s}}{s} $$
$\endgroup$ 3 $\begingroup$Replace $t$ with $t + 1$ and you can see the answer is no:
$$\mathcal{L}\{(t+1)^2u(t)\} = \mathcal{L}\{t^2 u(t)\} + 2t\mathcal{L}\{u(t)\} + \mathcal{L}\{u(t)\} \neq \mathcal{L}\{t^2u(t)\}+\mathcal{L}\{u(t)\}$$
$\endgroup$ 2
