 # Find the Laplace transforms of the functions given in problem f(t)=sin pi t text{ if } 2leq tleq3 , f(t)=0 text{ if } t<2 text{ or if } t>3 lwfrgin 2020-11-23 Answered
Find the Laplace transforms of the functions given in problem

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Step 1
Given function is
$$\begin{cases}\sin \pi t & 2\leq t\leq3\0 & 23\end{cases}$$
The Laplace transform of function f(t) is defined by
$L\left[f\left(t\right)\right]={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$={\int }_{0}^{2}{e}^{-st}f\left(t\right)dt+{\int }_{2}^{3}{e}^{-st}f\left(t\right)dt+{\int }_{3}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$

$={\int }_{2}^{3}{e}^{-st}\mathrm{sin}\left(\pi t\right)dt$
Step 2
Now, we firstly evaluate the indefinite integral by parts.
$I=\int {e}^{-st}\mathrm{sin}\left(\pi t\right)dt=\mathrm{sin}\left(\pi t\right)\int {e}^{-st}dt-\int \left(\left(\frac{d}{dt}\right)\mathrm{sin}\left(\pi t\right)\int {e}^{-st}dt\right)dt$
$I=-\frac{{e}^{-st}}{s}\mathrm{sin}\left(\pi t\right)-\int \pi \mathrm{cos}\left(\pi t\right)\left(-\frac{{e}^{-st}}{s}\right)dt$
$I=-\frac{{e}^{-st}}{s}\mathrm{sin}\left(\pi t\right)+\frac{\pi }{s}\int {e}^{-st}\mathrm{cos}\left(\pi t\right)dt$
Again , we integrate by parts.
$I=-\frac{{e}^{-st}}{s}\mathrm{sin}\left(\pi t\right)+\frac{\pi }{s}\left[\mathrm{cos}\left(\pi t\right)\int {e}^{-st}dt-\int \left(\frac{d}{dt}\mathrm{cos}\pi t\int {e}^{-st}dt\right)dt\right]$
$I=-\frac{{e}^{-st}}{s}\mathrm{sin}\left(\pi t\right)+\frac{\pi }{s}\left[-\frac{{e}^{-st}}{s}\mathrm{cos}\left(\pi t\right)-\int \left(-\pi \mathrm{sin}\left(\pi t\right)\right)\left(-\frac{{e}^{-st}}{s}\right)dt\right]$
$I=-\frac{{e}^{-st}}{s}\mathrm{sin}\left(\pi t\right)+\frac{\pi }{s}\left[-\frac{{e}^{-st}}{s}\mathrm{cos}\left(\pi t\right)-\frac{\pi }{s}\int {e}^{-st}\mathrm{sin}\left(\pi t\right)dt\right]$
$I=-\frac{{e}^{-st}}{s}\mathrm{sin}\left(\pi t\right)-\frac{\pi {e}^{-st}}{{s}^{2}}\mathrm{cos}\left(\pi t\right)-\frac{{\pi }^{2}}{{s}^{2}}I$
$I+\frac{{\pi }^{2}}{{s}^{2}}I=-\frac{{e}^{-st}}{s}\mathrm{sin}\left(\pi t\right)-\frac{\pi {e}^{-st}}{{s}^{2}}\mathrm{cos}\left(\pi t\right)$
$I=-\frac{{e}^{-st}}{{\pi }^{2}+{s}^{2}}\left[s\left(\mathrm{sin}\left(\pi t\right)\right)+\pi \mathrm{cos}\left(\pi t\right)\right]$
Step 3
Now, we evaluate the Laplace transform of given function.
$L\left[f\left(t\right)\right]={\left[-\frac{{e}^{-st}}{{\pi }^{2}+{s}^{2}}\left[s\left(\mathrm{sin}\left(\pi t\right)\right)+\pi \mathrm{cos}\left(\pi t\right)\right]\right]}_{2}^{3}$