Respuesta :
Answer:
Laplace transform of f(t)=t cost is, [tex]\frac{s^2-1}{(s+1)^2}[/tex].
Step-by-step explanation:
Given,
[tex]f(t)=t\cos t, t\geq 0[/tex]
To find by Laplace integral method,
[tex]L\{f(t)\}=\int_{0}^{\infty}e^{-st}t\cos tdt[/tex]
[tex]=\int_{0}^{\infty}te^{-st}\times\frac{e^{it}+e^(-it}}{2}dt[/tex]
[tex]=\textit{Real part of} \int_{0}^{\infty}te^{-s+i}tdt[/tex]
[tex]=\textit{Real part of} \{\Big[t\frac{e^{(-s+i)t}}{-s+i}\Big]_{0}^{\infty}-\int_{0}^{\infty}\frac{e^{(-s+i)t}}{-s+i}dt\}[/tex]
[tex]=\textit{Real part of} \{-\frac{1}{-s+i}\Big[\frac{e^{(-s+i)t}}{-s+i}\Big]_{0}^{\infty}\}[/tex]
[tex]=\textit{Real part of}\Big[\frac{1}{(-s+i)^2}\Big][/tex]
[tex]=\textit{Real part of}\Big[\frac{s^2+2i+1}{(s^2+1)^2}\Big][/tex]
[tex]=\frac{s^2-1}{(s+1)^2}[/tex]
