Module

for

Complex Integrals

Chapter 6  Complex Integration

Overview

Of the two main topics studied in calculus - differentiation and integration - we have so far only studied derivatives of complex functions. We now turn to the problem of integrating complex functions. The theory you will learn is elegant, powerful, and a useful tool for physicists and engineers. It also connects widely with other branches of mathematics. For example, even though the ideas presented here belong to the general area of mathematics known as analysis, you will see as an application of them one of the simplest proofs of the fundamental theorem of algebra.

6.1  Complex Integrals

In Section 3.1 we saw how the derivative of a complex function is defined.  We now turn our attention to the problem of integrating complex functions.  We will find that integrals of analytic functions are well behaved and that many properties from calculus carry over to the complex case.

We introduce the integral of a complex function by defining the integral of a complex-valued function of a real variable

Definition 6.1 (Definite Integral of a Complex Integrand).  Let    where u(t) and v(t) are real-valued functions of the real variable t for   .  Then

(6-1)            .

We generally evaluate integrals of this type by finding the antiderivatives of u(t) and v(t) and evaluating the definite integrals on the right side of Equation (6-1).  That is, if    and  ,  we have

(6-2)            .

Example 6.1.  Show that  .

Solution.  We write the integrand in terms of its real and imaginary parts, i.e.,   .  Here,    and  .  The integrals of u(t) and v(t) are

,   and

.

Hence, by Definition (6-1),

Explore Solution 6.1.

Example 6.2.  Show that  .

Solution.  We use the method suggested by Definitions (6-1) and (6-2).

We can evaluate each of the integrals via integration by parts.  For example,

Adding    to both sides of this equation and then dividing by 2 gives  .  Likewise,  .  Therefore,

.

Explore Solution 6.2.

Complex integrals have properties that are similar to those of real integrals.  We now trace through several commonalities.  Let    and    be continuous on  .

Using Definition (6-1), we can easily show that the integral of their sum is the sum of their integrals, that is

(6-3)            .

If we divide the interval    into    and    and integrate f(t) over these subintervals by using (6-1), then we get

(6-4)            .

Similarly, if    denotes a complex constant, then

(6-5)            .

If the limits of integration are reversed, then

(6-6)            .

The integral of the product f(t)g(t) becomes

(6-7)

Example 6.3.  Let us verify property (6-5).  We start by writing

Using Definition (6-1), we write the left side of Equation (6-5) as

which is equivalent to

Therefore,  .

Explore Solution 6.3.

It is worthwhile to point out the similarity between equation (6-2) and its counterpart in calculus.  Suppose that U and V are differentiable on
and  .  Since ,  equation (6-2) takes on the familiar form

(6-8)            .

where .  We can view Equation (6-8) as an extension of the fundamental theorem of calculus.  In Section 6.4 we show how to generalize this extension to analytic functions of a complex variable.  For now, we simply note an important case of Equation (6-8):

(6-9)            .

Example 6.4.  Use Equation (6-8) to show that  .

Solution.  We seek a function F with the property that   .  We note that   satisfies this requirement, so

which is the same result we obtained in Example 6.2, but with a lot less work.

Explore Solution 6.4.

Remark 6.1  Example 6.4 illustrates the potential computational advantage we have when we lift our sights to the complex domain.  Using ordinary calculus techniques to evaluate , for example, required a lengthy integration by parts procedure (Example 6.2).  When we recognize this expression as the real part of , however, the solution comes quickly.  This is just one of the many reasons why good physicists and engineers, in addition to mathematicians, benefit from a thorough working knowledge of complex analysis.

Extra Example 1.  Show that  .

Explore Solution for Extra Example 1.

The Next Module is

Contours and Contour Integrals

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