Previously in this series, I have used two different techniques to show that
where this last integral is taken over the complex plane on the unit circle, a closed contour oriented counterclockwise. Also, and (and is a certain angle that is now irrelevant at this point in the calculation).
This contour integral looks more complicated; however, it’s an amazing fact that integrals over closed contours can be easily evaluated by only looking at the poles of the integrand. For this integral, that means finding the values of where the denominator is equal to 0, and then determining which of those values lie inside of the closed contour. In this case, that means finding which root(s) of the denominator lie inside the unit circle in the complex plane.
To begin, we use the quadratic formula to find the roots of the denominator:
So we have the two roots and . Earlier in this series, I showed that as long as , and so the denominator has two distinct real roots. So the integral may be rewritten as