# Find the laplace transform of the following function f(t)=tu_2(t)

Find the laplace transform of the following function
$f\left(t\right)=t{u}_{2}\left(t\right)$
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$\text{Step 1}$

$\text{Step 2}$

$L\left(f\left(t\right)\right)={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$={\int }_{0}^{2}f\left(t\right){e}^{-st}dt+{\int }_{2}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$=0+{\int }_{2}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$={\int }_{2}^{\mathrm{\infty }}t{e}^{-st}dt$
$={\left[t\int {e}^{-st}dt\right]}_{2}^{\mathrm{\infty }}-{\int }_{2}^{\mathrm{\infty }}\left[\frac{d}{dt}\cdot t\int {e}^{-st}dt\right]dt$
$={\left[-\frac{t}{s}{e}^{-st}\right]}_{2}^{\mathrm{\infty }}+\frac{1}{s}{\int }_{2}^{\mathrm{\infty }}{e}^{-st}dt$
$=\frac{2}{s}{e}^{-2s}-\frac{1}{{s}^{2}}{\left[{e}^{-st}\right]}_{2}^{\mathrm{\infty }}$
$={e}^{-2s}\left(\frac{2}{s}+\frac{1}{{s}^{2}}\right)=\frac{{e}^{-2s}}{{s}^{2}}\left(2s+1\right)$