Module

for

Improper Integrals of Rational Functions

8.3  Improper Integrals of Rational Functions

The Cauchy principal value of an improper integral.

In calculus we studied improper integrals, and now we will extend this concept.  First, we shall assume that

is a continuous function, and is defined on the semi-infinite interval  .  Also, we assume that

for any real number ,  satisfying  ,  the integral    is properly defined.

is the area over the interval  .

Next, we can extend the integral to the entire interval  ,  and write   ,

and this can be accomplished if we can take the limit when goes to infinity.

Recall from calculus that the improper integral of    over  [)  is defined by

(8-7)               ,

provided that the limit exists.

is the area over the interval   [).

If    is defined for all real , then the integral of    over    is defined by

(8-7)               ,

provided both limits exist.

is the area over the interval  .

If the two limits in Equation (8-7) exists, then we can obtain the value of the integral by taking a single limit,

(8-8)               .

is the area over the interval .

For a wide class of functions the limit on the right side of Equation (8-8) exists, but the right side of

Equation (8-7) doesn't exist. It is our goal in this section to explain how the definition in Equation  (8-8)

leads to the construction of a certain a contour and then we will show how to use the power of complex

numbers and the residue calculus to evaluate the improper integral   .

Definition 8.2  (Cauchy Principal Value - P. V.). Let  be a continuous real valued function for all .

The Cauchy principal value () of the integral      is defined by

,

provided the limit exists.

Example 8.13.      does not exist,     and     .

Solution.

If we attempt to use Equation (8-7) then we obtain

and the last computation   ""   is undefined .

Thus, the improper integral      does not exist.

If we use Equation (8-8) then we obtain

This computation is well defined and is known as the Cauchy principal value () of   .

Therefore,

.

Explore Solution 8.13.

Extra Example 1.       and also    .

Solution.

If we attempt to use Equation (8-7) then we obtain

and we say that      diverges to  "."

If computations with the extended real numbers are permitted, then

.

If we attempt to use Equation (8-8) then we obtain

and we still say that      diverges to  "."

If computations with the extended real numbers are permitted, then

.

Explore Extra Solution 1.

Example 8.14.  Find the Cauchy principal value of   .

A portion of the area under the curve   ,   over the interval  .

Solution.

Explore Solution 8.14.

If , where  and  are polynomials, then  is called a rational function. Techniques

in calculus were developed to integrate rational functions.  We now show how the Cauchy's Residue Theorem

can be used to obtain the Cauchy Principal Value, () of the integral of    over  .

How to evaluate an improper integral with a contour integral.

The improper integral  involves a real function and the contour integral  involves

a complex function. How can they be related? It's really simple (no pun intended), we just used our standard

"trick" and replaced the variable with the variable .  But something strange is happening here, integration

over  has been replaced with integration along the contour . What is the connection between

and      ?

We must learn how to place the real integral in its proper complex variable context, and this is accomplished

by forming a very special contour in the complex plane.  The secret is the construction of the contour

.

Here we have

which is a line segment in the real -axis,

and

is a semi-circle of radius in the upper half-plane.

Theorem 8.3 (Contour Integration for Rational Functions). Let  where  and

are polynomials, of degree  ,  respectively.   If      for all real and   ,   then

the Cauchy Principal Value (or ) of the integral is

.

where    are the poles of    that lie in the upper half-plane, as shown in Figure 8.4.

Figure 8.4  The poles    of    that lie in the upper half-plane.

Remark.  The residues at the poles in the lower-half plane are not used in the computation.

Proof.

Example 8.15.  Use the residue calculus to compute   .

A portion of the area under the curve , over the interval .

Solution.

We write the integrand as  .

We see that    has simple poles at the points    and  ,  and that the

poles at    and  ,  are the only singularities of    in the upper half-plane.

The contour  consisting of the semi-circle  and the interval .

The points    lie in the upper half-plane.

Computing the residues, we obtain

Similarly,

Using Theorem 8.3, we conclude that

Explore Solution 8.15.

Example 8.16.  Use the residue calculus to compute   .

A portion of the area under the curve   ,   over the interval  .

Solution.

The integrand  has a poles of order  at the points ,

and the poles a    is the only singularity of    in the upper half-plane.

The contour  consisting of the semi-circle  and the interval .

The point    lies in the upper half-plane.

Computing the residue at   , we get

Using Theorem 8.3, we conclude that

Explore Solution 8.16.

The Next Module is

Improper Integrals Involving Trigonometric Functions

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