$\mathcal{L}\{f(t)\} = \displaystyle\int_0^{+\infty} f(t)e^{-st}\ dt = \displaystyle\int_0^{+\infty} te^{-st}\ dt = \left.-\dfrac{te^{-st}}{s}\right|_0^{+\infty} + \dfrac{1}{s}\displaystyle\int_0^{+\infty} e^{-st}\ dt =$
$= \left.-\dfrac{te^{-st}}{s}\right|_0^{+\infty} - \left.\dfrac{e^{-st}}{s^2}\right|_0^{+\infty}$, que converge para $s > 0$.
Logo $\fbox{$\mathcal{L}\{t\} = \dfrac{1}{s^2},\ s > 0$}$.
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