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:

Earlier, I evaluated this last integral using partial fractions, separating into the cases , , and . Now, I’ll calculate this same integral using contour integration. (See Wikipedia and Mathworld for more details.)

It turns out that can be rewritten as

,

where is the contour in the complex plane shown above (graphic courtesy of Mathworld). That’s because

To show that the limit of the last integral is equal to 0, I use the parameterization , so that :

.

The above limit is equal to zero because the numerator grows like while the denominator grows like . (This can be more laboriously established using L’Hopital’s rule).

Therefore, I have shown that

,

and this contour integral can be computed using residues.

I’ll continue with this fifth evaluation of the integral, starting with the case , in tomorrow’s post.

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