Is the residue at ##2i## incorrect?

[kq6up]

Homework Statement

Show that $\int _{ 0 }^{ \infty }{ \frac{x^2dx}{(x^2+9)(x^2+4)^2} } =\frac{\pi}{200}$.

Homework Equations

$Res=\frac{1}{n!}\frac{d^n}{dz^n}[f(z)(z-z_0)^{n+1}]$ where the order of the pole is $n+1$.

The Attempt at a Solution

Integreading of a semicircle contour one finds poles at $2i$ and $3i$. The second residue $3i$ was correct (I checked with mathematica). I used the $Res=\frac{g(z)}{h^{\prime}(z)}$ because it is first order, and I can just include the squared term on bottom into $g(z)$. The first residue $2i$ is not as easy, and I need to use the equation above. $Res=\frac{1}{1!}\frac{d}{dz}[\frac{z^2(z^2+4)^2}{(z^2+4)^2(z^2+9)}]$. This simplifies to $\frac{d}{dz}[\frac{z^2}{z^2+9}]$. Which in turn simplifies to $\frac{2z}{z^2+9}-\frac{2z^3}{(z^2+9)^2}$. Evaluating at $2i$ yields $\frac{36i}{25}$. This is wrong. Mathematica yields: $-\frac{13i}{200}$. Calculating this with the other term I got $\frac{1}{2}2\pi i (\frac{3i}{50}-\frac{13i}{200})=\frac{\pi}{200}$. So the Mathematica term is correct, but my $\frac{36i}{25}$ is wrong. I am not sure what I did wrong here.

Any suggestions?

Thanks,.
Chris

 

[mfb]

$(z-z_0)^2$ is not what you used in your numerator at $Res=\frac{1}{1!}\frac{d}{dz}[\frac{z^2(z^2+4)^2}{(z^2+4)^2(z^2+9)}]$

 

[kq6up]

It should be cancelling the term that goes to zero in the denominator, no?

Chris

 

[RUber]

Is $z_0=2i$? Or is it the negative?
I don't think you can do both at the same time.

 

[kq6up]

Good point.

Chris

 

[mfb]

Exactly.
x^2 + 4 = (x+2i)(x-2i). Those are independent terms.

 

[kq6up]

Thanks, I fixed it and it worked finally.

Chris