Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to find a limit in a function and need to calculate the following inverse Laplace transform:

$$ \mathcal{L}^{-1}\left\{\cfrac{1}{\sqrt{s}+s}\right\} $$

share|improve this question
    
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. –  B. S. Jan 18 '13 at 18:49

2 Answers 2

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 $s^r$

share|improve this answer
    
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 $a=0,b=1$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.