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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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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. – Thomas Andrews Jan 18 '13 at 18:20
I edited the message @ThomasAndrews, – Vivh Jan 18 '13 at 18:29
Can you invert $1/\sqrt{s}$ and $1/(1+\sqrt{s})$? – leonbloy Jan 18 '13 at 18:31
yes I can @leonbloy, thank you. Alright it seems good now. This means that $$ \mathcal{L}^{-1} \left\{\cfrac{1}{\sqrt{s}+s}\right \} = \mathcal{L}^{-1}\left\{\cfrac{1}{\sqrt{s}} \right\} * \mathcal{L}^{-1}\left\{\cfrac{1}{{1+\sqrt{s}}}\right\} $$? – Vivh Jan 18 '13 at 18:38
@Vivh: Yes. we have to use convolution. – Babak S. Jan 18 '13 at 18:49
up vote 3 down vote accepted

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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Thank you @Tim . – Vivh Jan 18 '13 at 21:47

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 $a=0,b=1$.

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Thank you @Babak Sorouh – Vivh Jan 18 '13 at 21:47
Great "hint"! Miss you! +1 – amWhy Feb 14 '13 at 0:08

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