# Given L(t^2)=(2)/(s^3) but why is is that, if we use the convolution theorem for 1 ** t^2, we get L(1 ** t^2)=1/s ** (2)/(s^3)

I know that
$L\left({t}^{2}\right)=\frac{2}{{s}^{3}}$
but why is is that, if we use the convolution theorem for $1\ast {t}^{2}$, we get
$L\left(1\ast {t}^{2}\right)=\frac{1}{s}\ast \frac{2}{{s}^{3}}$
Isn't $1\ast {t}^{2}$ equal to ${t}^{2}$?
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$\mathcal{L}\left(1\ast {t}^{2}\right)=\mathcal{L}\left(1\right)\mathcal{L}\left({t}^{2}\right)=\frac{1}{s}×\frac{2!}{{s}^{3}}=\frac{2}{{s}^{4}}$
Note that the convolution theorem gives us :
$f\left(t\right)=1\ast {t}^{2}={\int }_{0}^{t}1×{\tau }^{2}d\tau$
$f\left(t\right)=\frac{{\tau }^{3}}{3}{|}_{0}^{t}=\frac{{t}^{3}}{3}$
$\mathcal{L}\left(f\left(t\right)\right)=\frac{1}{3}\frac{3!}{{s}^{4}}=\frac{2}{{s}^{4}}$