(c) Utilize o teorema da integral no domínio do tempo: \frac{F(s)}{s}=\mathcal{L}\left\{\int_{0}^{t} f(\tau) d \tau\right\} para obter as transformadas inversas de:
Substituindo e avaliando de 0 a t (após alguma álgebra direta) obtém-se
\int_0^t v(\tau)\,d\tau =\frac{e^{\alpha t}}{D^2}\big((\alpha^{2}-\beta^{2})\sin\beta t -2\alpha\beta\cos\beta t\big)+\frac{2\alpha\beta}{D^2}.
Então
J_2=t\frac{e^{\alpha t}}{D}(\alpha\sin\beta t-\beta\cos\beta t) -\left[\frac{e^{\alpha t}}{D^2}\big((\alpha^{2}-\beta^{2})\sin\beta t -2\alpha\beta\cos\beta t\big)+\frac{2\alpha\beta}{D^2}\right].
Agora volte a I:=\int_{0}^{t}(t-\tau)e^{\alpha \tau}\sin(\beta \tau)\,d\tau = tJ_1 - J_2. Ao montar tJ_1-J_2 observam-se cancelamentos no termo com t e^{\alpha t}(\alpha\sin-\beta\cos), e sobra
I=\frac{t\beta}{D}+\frac{e^{\alpha t}}{D^2}\big((\alpha^{2}-\beta^{2})\sin\beta t -2\alpha\beta\cos\beta t\big)+\frac{2\alpha\beta}{D^2}.
Finalmente, lembrando do fator \dfrac{1}{\beta} que multiplica I para obter \mathcal{L}^{-1}\{H(s)\}, temos
\mathcal{L}^{-1}\{H(s)\}(t) =\frac{1}{\beta}I =\frac{t}{D}+\frac{2\alpha}{D^{2}}+\frac{e^{\alpha t}}{\beta D^{2}}\big((\alpha^{2}-\beta^{2})\sin\beta t -2\alpha\beta\cos\beta t\big),
onde novamente D=\alpha^{2}+\beta^{2}. Verificação rápida: para t=0 a expressão dá zero, que é o valor correto de \mathcal{L}^{-1}\{H(s)\}(0).
Resposta
\mathcal{L}^{-1}\{G(s)\}(t) =\frac{1}{\alpha^{2}+\beta^{2}}\left(\frac{\alpha}{\beta}e^{\alpha t}\sin\beta t - e^{\alpha t}\cos\beta t + 1\right)
\mathcal{L}^{-1}\{H(s)\}(t) =\frac{t}{\alpha^{2}+\beta^{2}}+\frac{2\alpha}{(\alpha^{2}+\beta^{2})^{2}}+\frac{e^{\alpha t}}{\beta(\alpha^{2}+\beta^{2})^{2}}\big((\alpha^{2}-\beta^{2})\sin\beta t -2\alpha\beta\cos\beta t\big)