Did I laplace tranform this unit step function correctly?

Question

$$L(t^3U(t-2))$$ $$L((t-2)^3U(t-2))$$ $$e^{-2s}L((t-2)^3)$$ $$e^{-2s}L(t^3-6t^2+12t-8)$$ $$e^{-2s}(\frac{3}{s^4}-\frac{12}{s^3}+\frac{12}{s^2}-\frac{8}{s})$$

There's my procedure above, is this correct? if not..where did I go wrong?

Answer 1

The analysis in the OP is flawed. Note that we can write

$$\begin{align} \mathscr{L}(t^3u(t-2))&=\int_2^\infty t^3 e^{-st}\,dt\\\\ &=\int_0^\infty (t+2)^3e^{-s(t+2)}\,dt\\\\ &=e^{-2s}\int_0^\infty (t+2)^3 e^{-st}\,dt\\\\ &=e^{-2s} \mathscr{L}((t+2)^3) \end{align}$$

Now, simply use $(t+2)^3=t^3+6t^2+12t+8$ along with $\mathscr{L}(t^n)=\frac{n!}{s^{n+1}}$.