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