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

$f(t)=\begin{cases}\cos t&\text{for }0\le t$

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

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

The Laplace transform is then

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

$\implies F(s)=\dfrac{(e^{-\pi s}+1)s}{s^2+1}$

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