I'm trying to find a limit in a function and need to calculate the following inverse Laplace transform:
$$ \mathcal{L}^{-1}\left\{\cfrac{1}{\sqrt{s}+s}\right\} $$
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$\begingroup$ When you say "need to inverse," do you mean "compute the inverse Laplace transform?" "Inverse" is not a verb, "invert" is the verb, but I don't think you mean "invert," either. $\endgroup$– Thomas AndrewsCommented Jan 18, 2013 at 18:20
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$\begingroup$ I edited the message @ThomasAndrews, $\endgroup$– VivhCommented Jan 18, 2013 at 18:29
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$\begingroup$ Can you invert $1/\sqrt{s}$ and $1/(1+\sqrt{s})$? $\endgroup$– leonbloyCommented Jan 18, 2013 at 18:31
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$\begingroup$ yes I can @leonbloy, thank you. Alright it seems good now. This means that $$ \mathcal{L}^{-1} \left\{\cfrac{1}{\sqrt{s}+s}\right \} = \mathcal{L}^{-1}\left\{\cfrac{1}{\sqrt{s}} \right\} * \mathcal{L}^{-1}\left\{\cfrac{1}{{1+\sqrt{s}}}\right\} $$? $\endgroup$– VivhCommented Jan 18, 2013 at 18:38
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$\begingroup$ @Vivh: Yes. we have to use convolution. $\endgroup$– MikasaCommented Jan 18, 2013 at 18:49
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2 Answers
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First, notice that $\dfrac{1}{\sqrt{s}+s}=\dfrac{1}{\sqrt{s}}-\dfrac{1}{1+\sqrt{s}}$ (using partial fraction decomposition).
Using the linearity of the Laplace transform, we can inverse transform each of these separately. The final result in the time domain is: $$\dfrac{1}{\sqrt{\pi t}} -\left(\dfrac{1}{\sqrt{\pi t}}-e^t\cdot\text{erfc}(\sqrt{t})\right)=e^t\cdot\text{erfc}(\sqrt{t})$$
Where "erfc" is the complementary error function.
Edit: in general if $p(s)$ is a posynomial, and $r$ is a real number that evenly divides every exponent, then we should be able to decompose $1/p(s)$ into a sum of fractions with constant numerators and the denominators each linear in $s^r$
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Note that we have: $$\mathcal{L}\left(\text{e}^{ab}\text{e}^{b^{2}t}\text{erfc}\left(b\sqrt{t}+\frac{a}{2\sqrt{2}}\right)\right)=\frac{\text{e}^{-a\sqrt{s}}}{\sqrt{s}(\sqrt{s}+b)}$$ Now set $a=0,b=1$.
