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 assumed , the contour in the complex plane is shown below (graphic courtesy of Mathworld),

and the positive constants and are given by

,

.

Now we have the small matter of simplifying our expression for . Actually, this isn’t a small matter because Mathematica 10.1 is not able to simplify this expression much at all:

Fortunately, humans can still do some things that computers can’t. The numbers and are chosen so that and are the roots of the denominator . In other words,

,

These relationships will be very handy for simplifying our expression for :

To complete the calculation, we recall that

,

,

so that

,

,

and hence

since and are both positive. Also,

,

so that

.

Finally,

.

Therefore,

.

### Like this:

Like Loading...

*Related*

## 1 Comment