That depends what you're integrating. For example, suppose:
and I let , then I'd get:
where the sum is over the residues of the enclosed poles.
You screwed up the quadratic equation.
- Wolfram|Alpha -Step 1
- Wolfram|Alpha -Step 2...validattion.
The zeros are . Now, which ones are in the unit circle? Isn't it the ones? So then we have:
.
I'll do one of the residues:
Edit: Also, forgot to mention, if it's a simple pole and the derivative of the denominator is not zero there then:
May be easier to calculate these residues this way.