Exercício 6.44. Se f : [0;1)! R, a transformada de Laplace de f(x) é a função L(s) de�nida pela integral imprópria a L(s):= Z 1 0 e�sxf(x) dx ; s �...

Vamos calcular as transformadas de Laplace das funções dadas: 1. k (constante) L(s) = ∫[0,1) e^(-sx) k dx L(s) = k ∫[0,1) e^(-sx) dx L(s) = k [e^(-sx)/(-s)]_0^1 L(s) = k [(e^(-s) - 1)/(-s)] 2. x L(s) = ∫[0,1) e^(-sx) x dx L(s) = [-xe^(-sx)/s]_0^1 + (1/s) ∫[0,1) e^(-sx) dx L(s) = [-xe^(-s)/s + 1/s (-e^(-sx)/s)]_0^1 L(s) = [-e^(-s)/s + (e^(-s) - 1)/s^2] 3. sen(x) L(s) = ∫[0,1) e^(-sx) sen(x) dx L(s) = (1/2i) ∫[0,1) e^(-sx) (e^(ix) - e^(-ix)) dx L(s) = (1/2i) [∫[0,1) e^(-(s-i)x) dx - ∫[0,1) e^(-(s+i)x) dx] L(s) = (1/2i) [(1/(s-i)) - (1/(s+i))] [e^(-(s-i)x) - e^(-(s+i)x))]_0^1 L(s) = (1/2i) [(1/(s-i)) - (1/(s+i))] [e^(-s) (e^(ix) - e^(-ix)) - (e^(-ix) - e^(ix)))] L(s) = (1/2i) [(1/(s-i)) - (1/(s+i))] [e^(-s) (2i sen(x) - 2 cos(x))] L(s) = (sen(x)/s^2 - x cos(x)/s) 4. e^(-x) L(s) = ∫[0,1) e^(-sx) e^(-x) dx L(s) = ∫[0,1) e^(-(s+1)x) dx L(s) = [e^(-(s+1)x)/(-(s+1))]_0^1 L(s) = [(e^(-(s+1)) - 1)/(-(s+1))]