# Find the Laplace transform of f(t)=t int_0^t tau e^(-tau) L(f)(s)= ??

Find the Laplace transform of
$f\left(t\right)=t{\int }_{0}^{t}\tau {e}^{-\tau }$
$L\left(f\right)\left(s\right)$=??
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Peutiedw
First write the function as
$f\left(t\right)=t{\int }_{0}^{t}\tau {e}^{-\tau }=tg\left(t\right),$
then you can use the following properties of the Laplace transform :
1) $\mathcal{L}\left({\int }_{0}^{t}h\left(\tau \right)d\tau \right)\left(s\right)=\frac{H\left(s\right)}{s},$
where H(s) is the Laplace transform of h.
2) $\mathcal{L}\left({t}^{n}g\left(t\right)\right)=\left(-1{\right)}^{n}{G}^{\left(n\right)}\left(s\right),$
where G(s) is the Laplace transform of g(t).
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Dana Russo
${\int }_{0}^{t}d\tau \phantom{\rule{thinmathspace}{0ex}}\tau \phantom{\rule{thinmathspace}{0ex}}{e}^{-\tau }={\left[-\tau \phantom{\rule{thinmathspace}{0ex}}{e}^{-\tau }\right]}_{0}^{t}+{\int }_{0}^{t}d\tau \phantom{\rule{thinmathspace}{0ex}}{e}^{-\tau }=1-\left(1+t\right){e}^{-t}$
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}{e}^{-st}& ={\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}t\phantom{\rule{thinmathspace}{0ex}}{e}^{-st}-{\int }_{0}^{\mathrm{\infty }}dt\phantom{\rule{thinmathspace}{0ex}}\left(t+{t}^{2}\right){e}^{-\left(s+1\right)t}\\ & =\frac{1}{{s}^{2}}-\frac{1}{\left(s+1{\right)}^{2}}-\frac{2}{\left(s+1{\right)}^{3}}\end{array}$