Inverse Laplace without Partial Fractions

Question

How do I find Inverse Laplace of $s^3/(s^4+4a^4)$ without using Partial fractions. I solved it using Partial Fractions but I wonder if there is some way solving it using properties of Laplace Transformation.

Answer 1

Not sure if this helps, but since $\mathcal{L}^{-1}\left(\frac{1}{s^n}\right)=\frac{x^{n-1}}{(n-1)!}$ for any $n>0$, the decomposition

$$ \frac{s^3}{s^4+4a^4} = \frac{1}{s}\cdot\frac{1}{1+\frac{4a^4}{s^4}} = \sum_{n\geq 0}\frac{4^n a^{4n} (-1)^n}{s^{4n+1}} \tag{1}$$ grants $$ \mathcal{L}^{-1}\left(\frac{s^3}{s^4+4a^4}\right)(x)=\sum_{n\geq 0}\frac{4^n a^{4n} (-1)^n}{(4n)!} x^{4n} \tag{2}$$ where the RHS can be computed by applying a discrete Fourier transform to the Maclaurin series of $e^x$.
This is pretty much the same as performing a partial fraction decomposition of the original function, and the outcome, of course, still is $\color{red}{\cos(ax)\cosh(ax)}$.