Answer
$$F\left( s \right) = \frac{1}{{{s^2}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( t \right) = t \cr
& {\text{Find the Laplace Transform of the function }} \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}f\left( t \right)} dt, \cr
& {\text{Substitute }}f\left( t \right) = t \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}\left( t \right)} dt \cr
& F\left( s \right) = \int_0^\infty {t{e^{ - st}}} dt \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \int_0^b {t{e^{ - st}}} dt \cr
& {\text{Integrating by parts we obtain }} \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \left[ { - \frac{t}{s}{e^{ - st}} - \frac{1}{{{s^2}}}{e^{ - st}}} \right]_0^b \cr
& F\left( s \right) = - \frac{1}{{{s^2}}}\mathop {\lim }\limits_{b \to \infty } \left[ {ts{e^{ - st}} + {e^{ - st}}} \right]_0^b \cr
& F\left( s \right) = - \frac{1}{{{s^2}}}\mathop {\lim }\limits_{b \to \infty } \left[ {\left( {bs{e^{ - sb}} + {e^{ - sb}}} \right) - \left( {0{e^0} + {e^0}} \right)} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty ,{\text{ }}s > 0 \cr
& F\left( s \right) = - \frac{1}{{{s^2}}}\left[ {0 - 1} \right] \cr
& F\left( s \right) = - \frac{1}{{{s^2}}}\left[ {0 - 1} \right] \cr
& F\left( s \right) = \frac{1}{{{s^2}}} \cr} $$
