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. to find ℒ{f(t)}. (Write your answer as a function of s.) f(t) = cos(t), 0 ≤ t < π 0, t ≥ π ℒ{f(t)} = (s > 0)

Answer 1

`f(t)=\begin{cases}\cos t&\text{for }0\le t<\pi\\0&\text{for }t\ge\pi\end{cases}`

Write
`f(t)` in terms of the step function
`u(t)`:

`f(t)=(u(t)-u(t-\pi))\cos t`

where the step function is defined by

`u(t)=\begin{cases}1&\text{for }t\ge0\\0&\text{for }t<0\end{cases}`

The Laplace transform is then

`\mathcal L_s\{f(t)\}=F(s)=\displaystyle\int_0^\infty f(t)e^(-st)\,\mathrm dt=\int_0^\pi\cos t\,e^(-st)\,\mathrm dt`

`\implies F(s)=((e^(-\pi s)+1)s)/(s^2+1)`

(if you're stuck on the integral, try integrating by parts)