Contour integral using residue theorem

[dykuma]

Homework Statement

Find the solution of the following integral

Homework Equations

The Attempt at a Solution

I applied the above relations getting that

Then I was able to factor the function inside the integral getting that

From here I should be able to get a solution by simply finding the poles of the function that are inside the bounds of this integral (in this case the unit circle), using that poles to find a residue, and then summing the residue's and multiplying them by 2πi.

In this case, I see four poles that are "inside" the unit circle: z= -1, z = -1/3 , z = 0, and z = 1. z= - 3 is outside of the unit circle.

However, when I do this, I get a solution of -8π/9, when I know that the solution should be 2π/9. In fact, the residue of 0 comes back as undefined, and -1 , 1 as zero.

Clearly I have done something wrong, though I am not sure what exactly.

 

[Delta2]

I am rusty in complex analysis and contour integration but what do you mean when you say that the residue of 0 comes back undefined? Maybe you do a mistake on calculating the residue of 0? Can you show us your work on how you calculate all the residues?

 

[aheight]

Are you sure there are 4 poles inside the unit circle? What's a pole? Also when you make the substitution $z=e^{it}$ the integral becomes a closed contour integral (with a circle through it). The substitutions in these are very messy and when you suspect something is wrong, first thing you should do is double, triple check those substitutions.

 

[lurflurf]

z= -1, z = 1
are not poles they have residue zero
z=0 is a pole of order 2
$$\operatorname{Res}=\lim_{z\rightarrow0}\dfrac{\operatorname{d}}{\operatorname{dx}}z^2\operatorname{f}(z)$$

 

[dykuma]

Thanks everyone! I forgot about that sneaky aspect of poles being at a higher order.