Solving integral using residue theorem

[arpitm08]

Homework Statement

Find the value of ∫ x^2/(((x^2 + 4)^2)(x^2+9)) dx. With the limits from 0 to infinity.

Homework Equations

The Attempt at a Solution

I know that this is an even function so we can calculate the value of the integral from -∞ to ∞ and just divide by 2. The upper half plane singularities are 3i and 2i and 2i. I calculated the residue for 3i, but couldn't figure out a way to calculate it for 2i. I tried to take the derivate of the denominator, but if you plug in 2i into that, you would get 0. I tried other ways, but they only resulted in negative answers. Is there a way that I'm not thinking of to find the residue at 2i? Thanks for any help.

 

[Mute]

The pole at $z = 2i$ is not a simple pole, so you have to use the residue formula for nth order poles. If a function f(z) has a pole of order n at z₀, then

$$Res(f,z_0) = \lim_{z\rightarrow z_0} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}\left[(z-z_0)^n f(z) \right]$$

 

[arpitm08]

The pole at $z = 2i$ is not a simple pole, so you have to use the residue formula for nth order poles. If a function f(z) has a pole of order n at z₀, then

$$Res(f,z_0) = \lim_{z\rightarrow z_0} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}\left[(z-z_0)^n f(z) \right]$$

It's a 2nd order pole. I took the derivative of h(x) = f(x)*(x-2i)^2. Then h(2i)=-13 * i / 200. Is that right?

 

[Dick]

It's a 2nd order pole. I took the derivative of h(x) = f(x)*(x-2i)^2. Then h(2i)=-13 * i / 200. Is that right?

Yes, it is. Well done!