Chapter 6 Complex Integration
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
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
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)
Hence, by Definition (6-1),
Explore Solution 6.1.
Example 6.2. Show that .
Solution. We use the method suggested by Definitions
We can evaluate each of the integrals via integration by
parts. For example,
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,
If we divide the interval into and and
integrate f(t) over these
subintervals by using (6-1), then we
Similarly, if denotes
a complex constant, then
If the limits of integration are reversed, then
The integral of the product f(t)g(t)
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
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
and . Since , equation (6-2) takes on the familiar form
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):
Example 6.4. Use Equation (6-8) to show that .
Solution. We seek a function F
with the property that . We
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.
Exercises for Section 6.1. Complex Integrals
The Next Module is
Contours and Contour Integrals
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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.
(c) 2012 John H. Mathews, Russell W. Howell