So I know that $$L(t^2) = \frac{2}{s^3}$$ but why is is that, if we use the convolution theorem for $1 \ast t^2$, we get $$L(1*t^2) = \frac{1}{s}*\frac{2}{s^3}$$
Isn't $1 \ast t^2$ equal to $t^2$? Or am I missing something in the definition? Sorry if it's a dumb question but I just really can't seem to understand.
(edit: corrected the fraction)
(edit: sorry about that, I just realized I definitely didn't have a good grasp on the definition!)
$$\mathcal {L}(1*t^2) = \mathcal {L} (1) \mathcal {L}(t^2)=\frac{1}{s}\times \dfrac {2!}{s^3}=\dfrac 2 {s^4}$$ Note that the convolution theorem gives us : $$f(t)=1 *t^2 =\int^t_0 1 \times \tau ^2 d\tau$$ $$f(t)=\dfrac {\tau ^3}3 \bigg |_0^t=\dfrac {t^3}3$$ $$\mathcal {L}(f(t))=\dfrac 13 \dfrac {3!}{s^4}=\dfrac 2 {s^4}$$