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

Answer:

Laplace transform of f(t)=t cost is, $\frac{s^2-1}{(s+1)^2}$.

Step-by-step explanation:

Given,

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

To find by Laplace integral method,

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

$=\int_{0}^{\infty}te^{-st}\times\frac{e^{it}+e^(-it}}{2}dt$

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

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

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

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

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

$=\frac{s^2-1}{(s+1)^2}$