Find the inverse Laplace transform of...

Question

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

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

Answer 1

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 $s^r$

Answer 2

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 $a=0,b=1$.