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I am trying to understand Laplace transformations. Could someone tell me from where the fraction (1/-s), in red on the first line is originated?

http://imgur.com/bTqjLGl

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$\color{red}{\left(\dfrac{1}{-s}\right)}$ comes because you do $dt\color{red}{(-s)}$

you mistaken $1$ for $s$ You change one $1$ for and $s$ accidentally

edit other way: since $\left(\dfrac{e^{-st}}{-s}\right)'=e^{-st}$ then $\mathcal{L\{1\}}=\displaystyle\lim_{b\to\infty} \left( \dfrac{e^{-sb}}{-s}-\dfrac{e^{-s.0}}{-s} \right)$

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  • $\begingroup$ Why do you need to add this? $\endgroup$ – privetDruzia Oct 24 '15 at 11:29
  • $\begingroup$ If you see your resolución, you can solve the integral by differents ways, but the indication in your paper shows that you solve the integral by change of variable $t\mapsto t'(-s):=-st$ so $dt=(-sd)t'=-sdt $ Is clear now? $\endgroup$ – Luis Felipe Oct 24 '15 at 11:32
  • $\begingroup$ No, it is not clear from why you need to add the .(-s) at the end and why you add the fraction in the beginning. I see the logic more or less but don't understand the reason $\endgroup$ – privetDruzia Oct 24 '15 at 11:50
  • $\begingroup$ @privetDruzia and now? $\endgroup$ – Luis Felipe Oct 24 '15 at 11:58
  • $\begingroup$ Thanks, clearer. But what is the reason of the second line? Why do you bring the s down? Is there a reason or is it just part of the algorithm? $\endgroup$ – privetDruzia Oct 24 '15 at 12:06
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the neutral factor $\frac{-s}{-s}$ has simply be introduced, then one part kept in the integral and the other out...

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    $\begingroup$ Why is it introduced? $\endgroup$ – privetDruzia Oct 24 '15 at 11:29

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