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.

Since is independent of , I can substitute any convenient value of that I want without changing the value of . As shown in previous posts, substituting yields the following simplification:

,

where I’ve made the assumption that . In the above derivation, is the contour in the complex plane shown below (graphic courtesy of Mathworld).

Also,

and

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

.

Similarly,

.

Therefore,

.

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