Question

Use Definition 7.1.1, DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = ∞ 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

Answer 1

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

The Laplace transform is then

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

Let $I$ denote the integral we want to compute. Integrating by parts, setting

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=\cos t\,\mathrm dt\implies v=\sin t$

gives

$\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\int_0^\pi e^{-st}\sin t\,\mathrm dt$

Integrate by parts again, setting

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$

$\mathrm dv=\sin t\,\mathrm dt\implies v=-\cos t$

Then

$\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\left(-e^{-st}\cos t\bigg|_{t=0}^{t=\pi}-s\int_0^\pi e^{-st}\cos t\,\mathrm dt\right)$

$I=e^{-st}(\sin t-s\cos t)\bigg|_{t=0}^{t=\pi}-s^2I$

$(s^2+1)I=s(e^{-\pi s}+1)$

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