Answer to Question #127131 in Differential Equations for Samhan

Question

Find the Laplace transform of the piecewise continuous function
f(x) = {1, 0 ≤ t < 1}
{-3e^-t, t ≥ 1}

Answer 1

$F(s)=\int\limits_0^\infty e^{-t\cdot s}\cdot f(t)\,dt=\\ =\int\limits_0^1 e^{-t\cdot s}\cdot f(t)\,dt +\int\limits_1^\infty e^{-t\cdot s}\cdot f(t)\,dt=\\ =\int\limits_0^1 e^{-t\cdot s}\cdot 1\cdot\,dt +\int\limits_1^\infty e^{-t\cdot s}\cdot (-3\cdot e^{-t})\,dt=\\ =\frac{1-e^{-s}}{s}-\frac{3\cdot e^{-s-1}}{s+1} ;$