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
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
Combining with , we
can obtain two useful substitutions
(8-5) and .
Using the substitutions for ,
Expression (8-3) transforms the real
a contour integral
Here we have used the identity
and the new complex integrand is
Remark. If involves then
it is o. k. to use the substitutions
Two graphical visualizations for the "area under the curve."
real integral can
be visualized in
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,
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.
Substitutions (8-4) and
(8-5), we transform the integral
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
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
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
We can solve for to obtain the substitutions
Using the identity for along
with Substitutions (8-4) and
(8-5), we rewrite the integral
The singularities of lying
located at the points and .
We use Theorem 8.2 to calculate the residues.
Therefore, we conclude that
Explore Solution 8.12.
Tutorial Exercises for Section 8.2. Trigonometric Integrals
Cauchy Principal Value
The Next Module is
Improper Integrals of Rational Functions
Return to the Complex Analysis Modules
Return to the Complex Analysis Project
This material is coordinated with our book Complex Analysis for Mathematics and Engineering.
(c) 2012 John H. Mathews, Russell W. Howell