grabrovi0u

2022-10-21

How to find the Laplace transform of the following function:
$f\left(t\right)=t{e}^{t}$
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}\left(t{e}^{t}{e}^{-st}\right)dt$
What method do I use to find the integral?

Kamden Simmons

We have:
$f\left(t\right)=t{e}^{t}$
So the Laplace transform of this function is:
$\begin{array}{rl}F\left(s\right)& =\mathcal{L}\left\{f\left(t\right)\right\}\\ & ={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}\phantom{\rule{thinmathspace}{0ex}}dt\\ & ={\int }_{0}^{\mathrm{\infty }}t{e}^{t}{e}^{-st}\phantom{\rule{thinmathspace}{0ex}}dt\\ & ={\int }_{0}^{\mathrm{\infty }}t{e}^{\left(1-s\right)t}\phantom{\rule{thinmathspace}{0ex}}dt\\ & =\underset{b\to \mathrm{\infty }}{lim}\left[{\int }_{0}^{b}t{e}^{\left(1-s\right)t}\phantom{\rule{thinmathspace}{0ex}}dt\right]\\ & =\underset{b\to \mathrm{\infty }}{lim}\left({\left[t\cdot \frac{1}{1-s}{e}^{\left(1-s\right)t}\right]}_{t=0}^{b}-{\int }_{0}^{b}\frac{1}{1-s}{e}^{\left(1-s\right)t}\phantom{\rule{thinmathspace}{0ex}}dt\right)\\ & =\frac{1}{1-s}\underset{b\to \mathrm{\infty }}{lim}\left(b{e}^{\left(1-s\right)b}-{\int }_{0}^{b}{e}^{\left(1-s\right)t}\phantom{\rule{thinmathspace}{0ex}}dt\right)\\ & =\frac{1}{s-1}\underset{b\to \mathrm{\infty }}{lim}{\int }_{0}^{b}{e}^{\left(1-s\right)t}\phantom{\rule{thinmathspace}{0ex}}dt& \left(s>1\right)\\ & =\frac{1}{s-1}\underset{b\to \mathrm{\infty }}{lim}{\left[\frac{1}{1-s}{e}^{\left(1-s\right)t}\right]}_{0}^{b}& \left(s>1\right)\\ & =\frac{1}{\left(s-1{\right)}^{2}}\underset{b\to \mathrm{\infty }}{lim}\left[1-{e}^{\left(1-s\right)b}\right]& \left(s>1\right)\\ & =\frac{1}{\left(s-1{\right)}^{2}}& \left(s>1\right)\end{array}$

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