Question

DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform of f, provided that the integral converges. Find ℒ{f(t)}. (Write your answer as a function of s.) f(t) = t cos t ℒ{f(t)} = (s > 0)

Answer 1

Laplace transform of f(t)=t cost is,
`(s^2-1)/((s+1)^2)`.

Explanation:

Given,

`f(t)=t\cos t, t\geq 0`

To find by Laplace integral method,

`L\{f(t)\}=\int_(0)^(\infty)e^(-st)t\cos tdt`

`=\int_(0)^(\infty)te^(-st)*\frac{e^(it)+e^(-it}}{2}dt`

`=\textit{Real part of} \int_(0)^(\infty)te^(-s+i)tdt`

`=\textit{Real part of} \{\Big[t(e^((-s+i)t))/(-s+i)\Big]_(0)^(\infty)-\int_(0)^(\infty)(e^((-s+i)t))/(-s+i)dt\}`

`=\textit{Real part of} \{-(1)/(-s+i)\Big[(e^((-s+i)t))/(-s+i)\Big]_(0)^(\infty)\}`

`=\textit{Real part of}\Big[(1)/((-s+i)^2)\Big]`

`=\textit{Real part of}\Big[(s^2+2i+1)/((s^2+1)^2)\Big]`

`=(s^2-1)/((s+1)^2)`