Laplace inverse for Taylor expansion

Question

By using infinite series find Laplace inverse for |1/(S^3+1)| .... I don't know what to do after using taylor expansion.. when I use it I got polynomial of $ S $ in the nominator which I can not deal with

Answer 1

Here is how you advance.

$$ \frac{1}{1+s^3} = \frac{1}{s^3(1+\frac{1}{s^3})} = \sum_{k=0}^{\infty} \frac{(-1)^k}{s^{3k+3}}$$

Apply inverse Laplace you get

$$ \sum_{k=0}^{\infty} \frac{(-1)^k\, x^{3 k+2}}{\Gamma(3k+3)}. $$

Now you sum the above series.

Added: Here is the summation of the series using CAS

$$ \frac{1}{3}\,{{\rm e}^{1/2\,x}}\sin \left( 1/2\,\sqrt {3}x \right) \sqrt {3}-1 /3\,{{\rm e}^{1/2\,x}}\cos \left( 1/2\,\sqrt {3}x \right) +1/3\,{ {\rm e}^{-x}} .$$

Answer 2

This is far easier using the definition of the inverse transform, i.e.,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s^3+1} $$

If you know complex analysis, then you know that this integral is simply the sum of the residues of the poles of the integrand. The poles are at $s=e^{i \pi/3}$, $s=e^{i \pi}$, and $s=e^{i 5 \pi/3} = e^{-i \pi/3}$. The sum of the residues is then

$$\frac{e^{e^{i \pi/3} t}}{3 e^{i 2 \pi/3}} + \frac{e^{e^{-i \pi/3} t}}{3 e^{-i 2 \pi/3}} + \frac13 e^{-t} = \frac{1}{3} e^{t/2} \left [\sqrt{3} \sin{\left (\frac{\sqrt{3}}{2} t \right )}-\cos{\left (\frac{\sqrt{3}}{2} t \right )} \right ]+\frac13 e^{-t}$$