Inverse Laplace of $\frac{s^3}{2+s^3}$ How I can find the Inverse Laplace of $\displaystyle \frac{s^3}{2+s^3}$
Thanks
 A: Write
$$\frac{s^3}{2+s^3} = \frac{2+s^3 -2}{2+s^3} = 1 - 2 \frac{1}{s^3 +2},$$
then break $s^3+2$ into factors and apply partial fractions decomposition. It should be simple to invert then.
A: Use the expansion 
$$
\frac{s^3}{2+s^3}=\frac1{1+2s^{-3}}=\sum\limits_{k\geqslant0}(-2)^ks^{-3k},
$$ 
and the identity
$$
\int_0^\infty x^{n}\mathrm e^{-sx}\mathrm dx=\frac{n!}{s^{n+1}},
$$ 
for every $n\geqslant0$, to deduce that
$$
\frac{s^3}{2+s^3}-1=\sum_{k\geqslant1}\frac{(-2)^k}{(3k-1)!}\int_0^\infty x^{3k-1}\mathrm e^{-sx}\mathrm dx=\int_0^\infty u(x)\mathrm e^{-sx}\mathrm dx,
$$
where
$$
u(x)=\sum\limits_{k\geqslant1}(-2)^k\frac{x^{3k-1}}{(3k-1)!}=-2^{1/3}\sum\limits_{k\geqslant1}\frac{z^{3k-1}}{(3k-1)!},
$$
with
$$
z=-2^{1/3}x.
$$
Now, if $\mathrm j=\mathrm e^{\mathrm i2\pi/3}=-\frac12+\mathrm i\frac{\sqrt3}2$, then $1+\mathrm j^{n+1}+\mathrm j^{2n+2}=0$ unless $n+1$ is a multiple of $3$ hence
$$
3u(x)=-2^{1/3}\sum\limits_{n\geqslant0}\frac1{n!}z^{n}(1+\mathrm j^n+\mathrm j^{2n})=-2^{1/3}(\mathrm e^{z}+\mathrm j\mathrm e^{\mathrm jz}+\mathrm j^2\mathrm e^{\mathrm j^2z}).
$$
Finally, $\mathrm j=-\frac12+\mathrm i\frac{\sqrt3}2$ and $\mathrm j^2=-\frac12-\mathrm i\frac{\sqrt3}2$ hence
$$
\mathrm j\mathrm e^{\mathrm jz}+\mathrm j^2\mathrm e^{\mathrm j^2z}=-\mathrm e^{-z/2}(\cos(\sqrt3z/2)+\sqrt3\sin(\sqrt3z/2)),
$$
hence $s^3/(2+s^3)$ is the Laplace transform of the measure $\delta_0(x)+u(x)\mathrm dx$ where
$$
u(x)=\frac{2^{1/3}}3\left(\mathrm e^{-z/2}(\cos(\sqrt3z/2)+\sqrt3\sin(\sqrt3z/2))-\mathrm e^{z}\right).
$$
