Given that $$L\left[ {f\left( t \right)} \right]\, = \,$$ $${{s + 2} \over {{s^2} + 1}},$$ $$$L\left[ {g\left( t \right)} \right] = {{{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}},$$$ $$$h\left( t \right) = \int\limits_0^t {f\left( \tau \right)\,g\left( {t - \tau } \right)\,d\tau ,} $$$ $$L\left[ {h\left( t \right)} \right]$$ is

$$If\,\,L\left[ {f\left( t \right)} \right]\, = \,F\left( s \right),$$ then $$L\left[ {f\left( {t - T} \right)} \right]$$ is equal to

If L$$\left[ {f\left( t \right)} \right]$$ = $$\omega /\left( {{s^2} + {\omega ^2}} \right),$$ then the value of $$\matrix{ {Lim\,f\,\left( t \right)} \cr {t \to \infty } \cr } $$

The Laplace Transform of e_at .cos$$\left( {\alpha t} \right).u\left( t \right)$$ is equal to