Inverse Laplace Transform with e^{a s}

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SUMMARY

The discussion focuses on finding the Inverse Laplace Transform of the function $F(s) = \frac{d}{ds}\left(\frac{1-e^{5s}}{s}\right)$. Participants clarify that the correct formulation should be $F(s) = \frac{d}{ds}\left(\frac{1-e^{-5s}}{s}\right)$ to ensure convergence. The resulting inverse transform is identified as $f(t) = t(H(t-5) - 1)$ for $t > 0$. The conversation emphasizes the importance of proper exponent signs in Laplace transforms for accurate results.

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  • Familiarity with the Heaviside step function, $H(t)$
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  • Basic skills in solving differential equations
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Ricardo Jesus
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Homework Statement


How can I take the Inverse Laplace Transform of $F(s) = \frac{d}{ds}\left(\frac{1-e^{5s}}{s}\right)$?
I have tried going with inverse of the derivative and convolution (even tried evaluating the derivative and go from there) but although I can get to some results none of them seems to make sense to me, as for example $f(t) = t(H_{-5} - 1)$, though I don't think this is even correct. Any help is greatly appreciated.

Homework Equations


$L{f} = F(s)$
$L{f*g} = F(s)G(s)
 
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Should that be
$$F(s) = \frac{d}{ds}\left(\frac{1-e^{-5s}}{s}\right)$$
?
In that case ##t(H(t-5) - 1) ## should be correct (for t>0). As written above (without the minus in the exponent) the inverse expression does not converge. What is causing you uncertainty?
 

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