Mathematics: Set up an integral for finding the laplace transform of f(t)=t−6. f(s)=l{f(t)}=∫ba e^(-s*t)(t-6)dt equation editorequation editor (formulas) where a= 0 equation editorequation editor and b= inf equation editorequation editor . find the antiderivative (with constant term 0) corresponding to the previous part. equation editorequation editor evaluate appropriate limits to compute the laplace transform of f(t): f(s)=l{f(t)}= equation editorequation editor where does the laplace transform you found exist?

Based on the other present answer the solution should be: .19%...

Set up an integral for finding the laplace transform

Ответ:

jetblackcap

09.01.2022

Mathematics

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Step-by-step explanation:

By definition the Laplace transform of f(t) = t-6 is

$\bf F(s)=L[f(t)](s)=\int_{0}^{\infty}e^{-st}(t-6)dt$

To find an anti-derivative, let's consider the indefinite integral

$\bf \int e^{-st}(t-6)dt$

and use integration by parts.

If we set

$\bf u=(t-6)\\dv=e^{-st}dt$

then

$\bf du=dt\\v=\int e^{-st}dt=\frac{e^{-st}}{-s}$

hence

$\bf \int e^{-st}(t-6)dt=uv-\int vdu=(t-6)\frac{e^{-st}}{-s}-\int \frac{e^{-st}}{-s}dt=\\=(t-6)\frac{e^{-st}}{-s}+\frac{1}{s}\int e^{-st}dt=(t-6)\frac{e^{-st}}{-s}+\frac{1}{s}\frac{e^{-st}}{(-s)}=\\=(t-6)\frac{e^{-st}}{-s}-\frac{e^{-st}}{s^2}=\frac{e^{-st}}{s}(6-t-\frac{1}{s})$

To find the Laplace transform we evaluate the improper integral as

$\bf \lim_{n \rightarrow \infty}\int_{0}^{n}e^{-st}(t-6)dt=\lim_{n \rightarrow \infty}\left[\frac{e^{-st}}{s}(6-t-\frac{1}{s}) \right]_0^{n}=\\=\lim_{n \rightarrow \infty}\left[\frac{e^{-sn}}{s}(6-n-\frac{1}{s}) -\frac{1}{s}(6-\frac{1}{s})\right]$

But

$\bf \lim_{n \rightarrow \infty}\left[\frac{e^{-sn}}{s}(6-n-\frac{1}{s})\right]=0\;(s0)$

and does not exist for s<0.

So, the Laplace transform only exists for s>0 and equals

$\bf \boxed{F(s)=-\frac{1}{s}(6-\frac{1}{s})=6(\frac{1}{s^2}-\frac{1}{s})}$ for s>0

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