Module

for

Trigonometric Integrals via Contour Integrals

Evaluation of real integrals via complex analysis.

Needless to say, when complex analysis methods are used to evaluate real integrals, something is going to get

messy, and Section 8.2 and Section 8.3 are no exceptions.  So prepare yourself for doing the computations and

learning some new theorems.

In Section 8.2, the technique involves making special substitutions for the trigonometric functions and obtaining

a complex function that is the quotient of two polynomials.  The difficulties start with the algebraic skills necessary

to form the polynomials.  And then the difficulty is finding the roots of the polynomial in the denominator.  It is

a good idea to review the quadratic formula and factoring a polynomial before studying Section 8.2.

Some textbooks study the material in Section 8.3 about rational integrals before studying the material in Section 8.2.

The techniques will require careful attention to the details need to verify that the limit of a certain complex integral

taken around a semi-circle in the upper half plane will vanish as the radius increases.  The skills in Section 8.3

involve understanding proofs involving epsilon and limits tending to zero in the complex plane.

It is a toss-up to decide which section to study at this point in the book. Your background should help you decide.

8.2  Trigonometric Integrals

As indicated at the beginning of this chapter, our goal is to evaluate certain definite real integrals with the aid

of the Residue Theorem.  We can do this is by interpreting the definite integral as the parametric form of an

integral of an analytic function along a simple closed contour.

Suppose we wish to evaluate a real integral of the form

(8-3)               ,

where    is a function of the two real variables  .  We can transform the real integral into a

complex integral ,  where the contour of integration is the unit circle ,

and the parametrization of the circle    is given by

(8-3)               ,      for     .

The interval   for   .                     The contour    for   .

Figure 8.2 a.  Transforming integration over a real interval  to complex integration around the contour .

Use the equation   ,   and obtain the following symbolic differentials.

(8-5)               ,     and
(8-4)
(8-5)               .

Combining        with    ,    we can obtain two useful substitutions

(8-5)                    and     .

Using the substitutions for    and    in Expression (8-3) transforms the real integral into

a contour integral

,
or

.

Here we have used the identity

,

and the new complex integrand is

.

Remark.  If      involves    then it is o. k. to use the substitutions

and     .

Two graphical visualizations for the "area under the curve."

The real integral      can be visualized in   space

as the area under the curve      over the interval   .

The complex integral      can be visualized in    space as

the area under the parametric curve    over the contour

Area under the curve    over  ,     Area under the curve  ,

.                                                                     .

Figure 8.2 b.  Comparison of the    and    visualizations for the "area under the curve."

In the optional explorations for Examples 8.11 - 8.12 we show the surfaces over the unit disk  ,

in the complex plane, and we will be able to visualize the poles of the complex function    that lie

inside the unit circle  .

Theorem (Contour Integration for Trigonometric Integrals). Use the above substitutions to construct .

Then is analytic inside and on the unit circle , except at the poles  that lie inside

the unit circle  .  We can compute the integral by using the Cauchy Residue Theorem,

(8-6)               ,
(8-6)               and

(8-6)               ,

where      are the poles of     that lie inside  .

Figure 8.2 c.  The poles inside   that are used to compute  .

Example 8.10.  Evaluate    by using complex analysis and the Cauchy Residue Theorem.

Solution.  Using Substitutions (8-4) and (8-5), we transform the integral to

where   .

The singularities of    are poles located at the points where    or equivalently,

where  .  Using the quadratic formula, we find that the singular points satisfy

which implies that  or .  Hence the only singularities

that lie inside the unit circle are simple poles corresponding to the solutions of  , which are the two

points    and  .   Now we use Theorem 8.2 and L'Hôpital's rule to calculate the

residues at the points and .

Since    and  ,  the residues are given by

.

We now use Equation (8-6) to compute the value of the integral:

Explore Solutions 8.10 and 8.11.

Example 8.11.  Evaluate      by using a computer algebra system.

Solution.  We can obtain antiderivative(s) of    by using software such as or .

One of the possibilities that have been obtained is

.

Since  and  are not defined, the computations for both  and  are indeterminate.

The graph    shown below in Figure 8.3 reveals another problem.  The integrand

is a continuous function for all  ,  but the function  has a discontinuity at .  On first glance this

appears to be a violation of the fundamental theorem of calculus, which asserts that the integral of a continuous

function must be differentiable and hence continuous.  But the problem is that    is not an antiderivative

of    for all values of    in the interval  .   Oddly, it is the antiderivative at all points

except  ,  ,  and  ,  and you can easily verify that it's derivative   equals

wherever    is defined.

Figure 8.3   Graph of   .

The integration algorithm used by computer algebra systems (the Risch-Norman algorithm) gives us

the antiderivative

,

and we must take great care in using this formula. We get the proper value of the integral by using

on the open subintervals    and    where it is continuous, and by taking the appropriate

limits at  .

Explore Solutions 8.10 and 8.11.

Example 8.12.  Evaluate    by using complex analysis and the Cauchy Residue Theorem.

Solution.  For values of that lie on the unit circle  ,  we have

and

We can solve for    to obtain the substitutions

and     .

Using the identity for    along with Substitutions (8-4) and (8-5), we rewrite the integral as

where  .

The singularities of    lying inside    are poles located at the points      and   .

We use Theorem 8.2 to calculate the residues.

and

Therefore, we conclude that

Explore Solution 8.12.

Library Research Experience for Undergraduates

The Next Module is

Improper Integrals of Rational Functions

(c) 2012 John H. Mathews, Russell W. Howell