How do the Laplace Transforms vary between the two following functions? What I am really asking is if I calculated the following Laplace Transform correctly...
$$\mathcal{L}\{(t-1)u(t-1)\}=\frac{e^{-s}}{s^2}$$
...what should the Laplace Transform be for this function:
$$\mathcal{L}\{(t)u(t-1)\}=?$$
The one sided Laplace transform is defined as $$ \mathcal{L}\{f(t)\} = \int_0^{\infty}f(t)e^{-st}dt $$ and the Unit step function is defined as $$ \mathcal{U}(t)= \begin{cases} 1, & t>0\\ 0, & t<0 \end{cases} $$ so the $$ \mathcal{L}\{t\mathcal{U}(t-1)\} = \int_0^{\infty}t\mathcal{U}(t-1)e^{-st}dt = \int_1^{\infty}te^{-st}dt = \mbox{???} $$