how to do inverse laplace of $(s^2+1)/s^4$?

Question

how to do the inverse laplace of $(s^2+1)/s^4$? the answer is $(t^3/6)+t$ but I do not know how to derive it.

Answer 1

Recall that

$$\mathscr{L}\left(u\right)(s)=\int_0^{\infty}u(t)e^{-st}dt=\frac{1}{s}$$

where $u$ is the unit step function.

We also know that

$$\frac{d^n}{ds^n}\left(\mathscr{L}\left(f\right)(s)\right)=\int_0^{\infty}(-1)^nt^nf(t)e^{-st}dt$$

Thus, if $f(t) =u(t)$, then

$$\frac{d^n}{ds^n}\left(\mathscr{L}\left(u\right)(s)\right)=\int_0^{\infty}(-1)^nt^nu(t)e^{-st}dt=\frac{d^n}{ds^n}s^{-1}=(-1)^nn!s^{-(n+1)}$$

Thus,

$$\mathscr{L^{-1}}\left(s^{-(n+1)}\right)(t)=\frac{1}{n!}t^n$$

Here we have $F(s) = s^{-2}+s^{-4}$ and thus $f(t)=t+\frac16 t^3$ as expected.