This series was inspired by a question that my wife asked me: calculate
Originally, I multiplied the top and bottom of the integrand by
and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.
where I’ve made the assumption that . In the above derivation, is the contour in the complex plane shown below (graphic courtesy of Mathworld).
are the two poles of the final integrand that lie within this contour.
It now remains to simplify the final algebraic expression. To begin, I note
And so, at long last, I’ve completed a fifth different evaluation of .