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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?

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  • $\begingroup$ Do you remember the convolution theorem? $\endgroup$ Oct 9, 2013 at 15:10
  • $\begingroup$ @DanielFischer, Convolution Theorem? I'm lost, I've never learn about that before. Could you please show how to use it? $\endgroup$
    – alasd
    Oct 9, 2013 at 15:15
  • $\begingroup$ $$\mathcal{L}[f\ast g](s) = \mathcal{L}[f](s)\cdot \mathcal{L}[g](s)$$ So here, you get $\sin \ast \sin \ast \sin$. Of course, you can also compute the inverse Laplace transform via the residue theorem, if that's more familiar. $\endgroup$ Oct 9, 2013 at 15:18

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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.

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  • $\begingroup$ thanks a lot, how about using convolution theorem? So I can compare it. $\endgroup$
    – alasd
    Oct 9, 2013 at 16:02
  • $\begingroup$ Well, what you want to do is first set up a first convolution $$\int_0^t dt' \, \sin{t'} \, \sin{(t-t')} = \frac12 (\sin{t} - t \cos{t})$$ Then set up a convolution of this result with another sine, viz: $$\frac12 \int_0^t dt' \, (\sin{t'} - t' \cos{t'})\, \sin{(t-t')}$$ This should return the result that I posted. $\endgroup$
    – Ron Gordon
    Oct 9, 2013 at 16:07
  • $\begingroup$ Ron Gordon, I'm a bit confused how you calculate the derivative. Could you please elaborate it? $\endgroup$
    – alasd
    Oct 9, 2013 at 17:29
  • $\begingroup$ @alasd: use the quotient rule twice. The first derivative brings $$\frac{(s\pm i)^3 t e^{s t} - 3 (s \pm i)^2 e^{s t}}{(s \pm i)^6}$$ Simplify, and take the derivative of the result. $\endgroup$
    – Ron Gordon
    Oct 9, 2013 at 17:33

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