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## Homework Statement

The problem is to find the inverse laplace of [tex]\frac{s^2-a^2}{(s^2+a^2)^2}[/tex]

I am supposed to use the residue definition of inverse laplace (given below)

The poles of F(s) are at ai and at -ai and they are both double poles.

## Homework Equations

[tex]f(t) = \sum_{j=1}^{N}Res[F(s)*e^{st};s_{j}][/tex]

Where [tex]s_{j}[/tex] are the poles of F(s)

[tex]Residue(f(z)) = \frac{1}{(m-1)!}lim_{z->z0}\frac{d^{m-1}}{dz^{m-1}}((z-z0)^{m}f(z))[/tex]

where z0 is a pole of order m of f(z)

## The Attempt at a Solution

The poles of F(s) are at ai and at -ai and they are both double poles.

If you can just help find the residue of ai I'm sure I can figure out -ai.

[tex]Residue[F(s);ai] = lim_{s->ai}[/tex][tex]\frac{d}{ds}\frac{(s-ai)^2(s^2-a^2)}{(s-ai)^2*(s+ai)^2}[/tex]

[tex]= lim_{s->ai}[/tex][tex]\frac{d}{ds}\frac{s^2-a^2}{(s+ai)^2}[/tex]

using quotient rule I get:

[tex]= lim_{s->ai}[/tex][tex]\frac{2s(s+ai)^2-2(s+ai)(s^2-a^2)}{(s+ai)^4}[/tex]

substitute ai for s

[tex]=\frac{2ai(2ai)^2-2(2ai)(-2a^2)}{(2ai)^4}[/tex]

[tex]=\frac{2ai(-4a^2)+8a^3i}{16a^4} = 0[/tex]

But it can't equal zero right? I also get zero when I try to calculate the residue at -ai.