Respuesta :
[tex]f(t)=\begin{cases}\cos t&\text{for }0\le t<\pi\\0&\text{otherwise}\end{cases}[/tex]
The Laplace transform is then
[tex]\mathcal L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_0^\pi e^{-st}\cos t\,\mathrm dt[/tex]
Let [tex]I[/tex] denote the integral we want to compute. Integrating by parts, setting
[tex]u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt[/tex]
[tex]\mathrm dv=\cos t\,\mathrm dt\implies v=\sin t[/tex]
gives
[tex]\displaystyle I=e^{-st}\sin t\bigg|_{t=0}^{t=\pi}+s\int_0^\pi e^{-st}\sin t\,\mathrm dt[/tex]
Integrate by parts again, setting
[tex]u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt[/tex]
[tex]\mathrm dv=\sin t\,\mathrm dt\implies v=-\cos t[/tex]
Then
[tex]\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)[/tex]
[tex]I=e^{-st}(\sin t-s\cos t)\bigg|_{t=0}^{t=\pi}-s^2I[/tex]
[tex](s^2+1)I=s(e^{-\pi s}+1)[/tex]
[tex]I=\dfrac s{s^2+1}(e^{-\pi s}+1)[/tex]
The Laplace Transform of the piecewise Function is [tex]\mathcal {L}\{f(t)\} = \frac{s\cdot (e^{-s\cdot \pi + 1})}{s^{2}+1}[/tex].
Given that Function is continuous and integrable. The Laplace Transform is defined by the following expression:
[tex]\mathcal {L} \{f(t)\} = \int\limits^{+\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt[/tex], [tex]t \ge 0[/tex] (1)
We have the following piecewise Function:
[tex]f(t) = \left \{ {{\cos t,\,0 \le t \le \pi} \atop {0,\,t\ge \pi}} \right.[/tex] (2)
By (1) and using Integral Properties, the Laplace Transform of the piecewise Function is:
[tex]\mathcal{L}\{f(t)\} = \int\limits^{\pi}_{0} {e^{-s\cdot t}\cdot \cos t} \, dt + \int\limits^{+\infty}_{\pi} {e^{-s\cdot t}\cdot 0} \, dt[/tex]
[tex]\mathcal{L}\{f(t)\} = \int\limits^{\pi}_{0} {e^{-s\cdot t}\cdot \cos t} \, dt[/tex] (3)
By Part Integration, we have the following outcome:
[tex]u = e^{-s\cdot t}[/tex], [tex]dv = \cos t\,dt[/tex], [tex]du = -s\cdot e^{-s\cdot t}[/tex], [tex]v = \sin t[/tex]
[tex]\mathcal {L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |\limits ^{+\infty}_{0} + s\int\limits^\pi_0 {e^{-s\cdot t}\cdot \sin t} \, dt[/tex] (4)
[tex]m = e^{-s\cdot t},\, dn = \sin t\,dt,\, dm = -s\cdot e^{-s\cdot t},\,n = -\cos t[/tex]
[tex]\mathcal{L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |_{0}^{\pi}-s\cdot e^{-s\cdot t}\cdot \cos t|_{0}^{\pi}-s^{2}\int\limits^\pi_0 {e^{-s\cdot t}\cdot \cos t} \, dt[/tex]
[tex]\mathcal{L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |_{0}^{\pi}-s\cdot e^{-s\cdot t}\cdot \cos t|_{0}^{\pi}-s^{2}\cdot \mathcal{L}\{f(t)\}[/tex]
[tex](s^{2}+1)\cdot \mathcal{L}\{f(t)\} = e^{-s\cdot t}\cdot \sin t |_{0}^{\pi}-s\cdot e^{-s\cdot t}\cdot \cos t|_{0}^{\pi}[/tex]
[tex]\mathcal {L}\{f(t)\} = \frac{s\cdot (e^{-s\cdot \pi + 1})}{s^{2}+1}[/tex]
The Laplace Transform of the piecewise Function is [tex]\mathcal {L}\{f(t)\} = \frac{s\cdot (e^{-s\cdot \pi + 1})}{s^{2}+1}[/tex].
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