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I have this problem to solve, but I couldn't get the solution:

$F(s) = \frac{s-2}{s^{2}-1}$

$s>1$

$\int_{0}^{+\infty}\left ( \frac{f(t)}{e^{2t}} \right )dt$

$L\left [ e^{-2t}f(t)) \right ]=F\left ( s-a \right )=\frac{s}{(s+2)^{2}-1}$

and

$L\left [ \int_{0}^{\infty}\left ( \frac{f(t))}{e^{2t}} \right ) \right ] = \frac{F(s))}{s}=\frac{1}{(s+2)^{2}-1}$

so

$\int_{0}^{\infty}\left ( \frac{f(t))}{e^{2t}} \right ) = L^{-1}\left [ \frac{1}{(s+2)^{2}-1} \right ]$

$\int_{0}^{\infty}\left ( \frac{f(t))}{e^{2t}} \right ) = \frac{1}{2}\left ( e^{-t} + e^{-3t} \right )$

I am struggling to solve this with Laplace properties. I have to evaluate the integral without explicitly calculate f(t). F(s) is the Laplace transform of f(t). I suppose that my resolution is wrong since the superior limit is infinity. The solution must be achieved without performing f(t) at the beginning. The method must use F(s) and not f(t) directly on the integral. Can anyone help me?

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    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Nov 17, 2021 at 12:04
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    $\begingroup$ The integal is just $F(2)$. $\endgroup$ Commented Nov 17, 2021 at 12:12
  • $\begingroup$ If you are writing $F(s)$ for the Laplace transform of $f$ then $F(2)$ is equal to that integral by definition. $\endgroup$ Commented Nov 17, 2021 at 12:15
  • $\begingroup$ @Kavi Rama Murthy - How? $\endgroup$
    – Pedro R.
    Commented Nov 17, 2021 at 12:16
  • $\begingroup$ Just use partial fractions and already proved Laplace transforms. Please edit your post and make it easier to understand what exactly you are asking for. $\endgroup$ Commented Nov 17, 2021 at 13:25

1 Answer 1

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$$ F(s)=\frac{s-2}{s^{2}-1},\quad{s}\gt{1}. \\ L^{-1}\left[F(s)\right]=L^{-1}\left[\frac{s-2}{s^{2}-1}\right]=L^{-1}\left[\frac{s}{s^{2}-1}\right]+L^{-1}\left[\frac{-2}{s^{2}-1}\right]= \\ =\left(\cosh(t)-2\sinh(t)\right)=\frac{e^{t}+e^{-t}-2e^{t}+2e^{-t}}{2}=\frac{-e^{t}+3e^{-t}}{2},\implies \\ \implies{f(t)}=\frac{-e^{t}+3e^{-t}}{2}. $$ $$ \int{\frac{(-e^{t}+3e^{-t})e^{-2t}}{2}}dt=\left(\int{\frac{-e^{-t}}{2}}dt+\int{\frac{3e^{-3t}}{2}}dt\right)=0.5e^{-t}-1.5e^{-3t}. $$

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