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,
are the two distinct roots of the denominator (as long as ). In these formulas, and . (Also, 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.
Let’s now see if either of the two roots of the denominator lies inside of the unit circle in the complex plane. In other words, let’s determine if and/or .
I’ll begin with . Clearly, the numbers , , and are the lengths of three sides of a right triangle with hypotenuse . So, since the hypotenuse is the longest side,
Also, by the triangle inequality,
Combining these inequalities, we see that
and so I see that , so that does lie inside of the contour .
The second root is easier to handle:
Therefore, since lies outside of the contour, this root is not important for the purposes of computing the above contour integral.