Inverse Laplace Transform Problem

Question

I have this problem $\frac{1}{(s^2+1)^3}$. I have to find its Inverse Laplace Tranformation. I already try using partial fraction but it didn't work because I found it will back to the problem form.

Any other way for solutions?

Answer 1

You may use the residue theorem. The ILT is

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

where $c \gt 0$. Consider a contour integral in the complex $s$ plane:

$$\frac{1}{i 2 \pi} \oint_C ds \frac{e^{s t}}{(s^2+1)^3}$$

where $C$ includes the line $[c-i R,c+i R]$ and a semicircle of radius $R$ that closes to the left when $t \gt 0$ and to the right when $t \lt 0$. We take the limit as $R \to \infty$. When $t \gt 0$, the contour integral - and therefore the ILT, is then equal to the sum of the residues of the integrand at the poles $z=\pm i$. In this case, each pole is a triple pole, so that the residues are given by

$$\sum_{\pm} \frac12 \left [\frac{d^2}{ds^2} \frac{e^{s t}}{(s \pm i)^3} \right ]_{z=\pm i} $$

The derivative term produces

$$\frac{d^2}{ds^2} \frac{e^{s t}}{(s \pm i)^3} = \frac{(s\pm i)^2 t^2 - 6 (s\pm i) t + 12}{(s \pm i)^5} e^{s t}$$

so that performing the above sum produces, as the ILT

$$\frac12 \frac{-4 t^2 - i 12 t + 12}{32 i} e^{i t} + \frac12 \frac{-4 t^2 + i 12 t+ 12}{-32 i} e^{-i t} = -\frac18 (t^2-3) \sin{t} - \frac{3}{8} t \, \cos{t}$$

When $t \lt 0$, on the other hand, there are no poles within $C$, so the ILT is zero there, as is expected.