$\mathcal L \left( t^2 e^t \operatorname{u}(t-6)\right)$

Question

Find: $$\mathcal L \left( t^2 e^t \operatorname{u}(t-6)\right)$$

My attempt

$$\begin{array}{rcl} \mathcal L \left( F(t-a) \operatorname{u}(t-a) \right) &=& e^{-as} F(s) \\ \mathcal L \left( t f(t) \right) &=& \dfrac{-\mathrm d}{\mathrm ds} f(s) \\ \mathcal L \left( t^2 f(t) \right) &=& \dfrac{-\mathrm d^2}{\mathrm ds^2} f(s) \\ \end{array}$$

Then:

$$\begin{array}{cl} &\mathcal L \left( t^2 e^t \operatorname{u}(t-6)\right) \\ &\dfrac{-\mathrm d^2}{\mathrm ds^2} \left(e^t \operatorname{u}(t-6)\right) \\ \implies&e^t \operatorname{u}(t-6) \\ \end{array}$$

$$\mathcal L\left(e^t \operatorname{u}(t-6)\right) = e^{-6s} \frac 1 {s-1}$$

is my work correct ??

Answer 1

Well, we have:

$$\mathscr{L}_t\left[t^2\cdot e^t\cdot\text{u}\left(t-6\right)\right]_{\left(\text{s}\right)}=\left(-1\right)^2\cdot\frac{\text{d}^2}{\text{d}\text{s}^2}\left(\mathscr{L}_t\left[e^t\cdot\text{u}\left(t-6\right)\right]_{\left(\text{s}\right)}\right)=$$ $$\left(-1\right)^2\cdot\lim_{\text{p}\to\text{s}-1}\frac{\text{d}^2}{\text{d}\text{p}^2}\left(\mathscr{L}_t\left[\text{u}\left(t-6\right)\right]_{\left(\text{p}\right)}\right)=\left(-1\right)^2\cdot\lim_{\text{p}\to\text{s}-1}\frac{\text{d}^2}{\text{d}\text{p}^2}\left(\frac{e^{-6\text{p}}}{\text{p}}\right)\tag1$$