INTRODUCTION

In real calculus when a function f possesses an elementary antiderivative, that is, a function F for which F′(x) = f(x), a definite integral can be evaluated by the Fundamental Theorem of Calculus:

(1)

Note that (x) dx depends only on the numbers a and b at the initial and terminal points of the interval of integration. In contrast, the value of a real-line integral ∫_C P dx + Q dy generally depends on the curve C. However, we saw in Section 9.9 that there exist line integrals whose value depends only on the initial point A and terminal point B of the curve C, and not on C itself. In this case we say that the line integral is independent of the path. These integrals can be evaluated by the Fundamental Theorem for Line Integrals (Theorem 9.9.1). It seems natural then to ask:

In this section we will see that the answer to both of these questions is “yes.”

Path Independence

As the next definition shows, the definition of path independence for a contour integral ∫_Cf(z) dz is essentially the same as for a real-line integral ∫_C P dx + Q dy.

At the end of the preceding section we noted that the Cauchy–Goursat theorem also holds for closed contours, not just simple closed contours, in a simply connected domain D. Now suppose, as shown in FIGURE 18.3.1, that C and C₁ are two contours in a simply connected domain D, both with initial point z₀ and terminal point z₁. Note that C and −C₁ form a closed contour. Thus, if f is analytic in D, it follows from the Cauchy–Goursat theorem that

(2)

But (2) is equivalent to

(3)

The result in (3) is also an example of the principle of deformation of contours introduced in (3) of Section 18.2. We summarize the last result as a theorem.

EXAMPLE 1 Choosing a Different Path

Evaluate ∫_C 2z dz, where C is the contour with initial point z = −1 and terminal point z = −1 + i shown in FIGURE 18.3.2.

SOLUTION

Since the function f(z) = 2z is entire, we can replace the path C by any convenient contour C₁ joining z = −1 and z = −1 + i. In particular, by choosing C₁ to be the straight line segment x = −1, 0 ≤ y ≤ 1, shown in red in Figure 18.3.2, we have z = −1 + iy, dz = i dy. Therefore,

≡

A contour integral ∫_C f(z) dz that is independent of the path C is usually written (z) dz, where z₀ and z₁ are the initial and terminal points of C. Hence in Example 1 we can write dz.

There is an easier way to evaluate the contour integral in Example 1, but before proceeding we need another definition.

For example, the function F(z) = −cos z is an antiderivative of f(z) = sin z, since F′(z) = sin z. As in real calculus, the most general antiderivative, or indefinite integral, of a function f(z) is written ∫ f(z) dz = F(z) + C, where F′(z) = f(z) and C is some complex constant.

Since an antiderivative F of a function f has a derivative at each point in a domain D, it is necessarily analytic and hence continuous in D (recall that differentiability implies continuity).

We are now in a position to prove the complex analogue of (1).

PROOF:

We will prove (4) in the case when C is a smooth curve defined by z = z(t), a ≤ t ≤ b. Using (3) of Section 18.1 and the fact that F′(z) = f(z) for each z in D, we have

≡

EXAMPLE 2 Using an Antiderivative

EXAMPLE 3 Using an Antiderivative

Evaluate ∫_C cos z dz, where C is any contour with initial point z = 0 and terminal point z = 2 + i.

SOLUTION

F(z) = sin z is an antiderivative of f(z) = cos z, since F′(z) = cos z. Therefore from (4) we have

If we desired a complex number of the form a + bi for an answer, we can use sin(2 + i) = 1.4031 − 0.4891i (see Example 1 in Section 17.7). Hence,

≡

We can draw several immediate conclusions from Theorem 18.3.2. First, observe that if the contour C is closed, then z₀ = z₁ and consequently

(5)

Next, since the value of ∫_C f(z) dz depends on only the points z₀ and z₁, this value is the same for any contour C in D connecting these points. In other words:

In addition we have the following sufficient condition for the existence of an antiderivative:

The last statement is important and deserves a proof. Assume that f is continuous, ∫_C f(z) dz is independent of the path in a domain D, and F is a function defined by F(z) = (s) ds where s denotes a complex variable, z₀ is a fixed point in D, and z represents any point in D. We wish to show that F′(z) = f(z); that is, F is an antiderivative of f in D. Now,

(8)

Because D is a domain we can choose Δz so that z + Δz is in D. Moreover, z and z + Δz can be joined by a straight segment lying in D, as shown in FIGURE 18.3.3. This is the contour we use in the last integral in (8). With z fixed, we can write^*

(9)

From (8) and (9) it follows that

Now f is continuous at the point z. This means that for any ϵ > 0 there exists a δ > 0 so that |f(s) − f(z)| < ϵ whenever |s − z| < δ. Consequently, if we choose Δ z so that |Δ z| < δ, we have

Hence, we have shown that

If f is an analytic function in a simply connected domain D, it is necessarily continuous throughout D. This fact, when put together with the results in Theorem 18.3.1 and (7), leads to a theorem that states that an analytic function possesses an analytic antiderivative.

In (9) of Section 17.6 we saw that 1/z is the derivative of Ln z. This means that under some circumstances Ln z is an antiderivative of 1/z. Care must be exercised in using this result. For example, suppose D is the entire complex plane without the origin. The function 1/z is analytic in this multiply connected domain. If C is any simple closed contour containing the origin, it does not follow from (5) that dz/z = 0. In fact, from (4) of Section 18.2 with the identification z₀ = 0, we see that

In this case, Ln z is not an antiderivative of 1/z in D, since Ln z is not analytic in D. Recall that Ln z fails to be analytic on the nonpositive real axis (the branch cut off the principal branch of the logarithm).

EXAMPLE 4 Using the Logarithmic Function

Evaluate dz, where C is the contour shown in FIGURE 18.3.4.

SOLUTION

Suppose that D is the simply connected domain defined by x = Re(z) > 0, y = Im(z) > 0. In this case, Ln z is an antiderivative of 1/z, since both these functions are analytic in D. Hence by (4),

From (7) of Section 17.6, we have

and so ≡

REMARKS

Suppose f and g are analytic in a simply connected domain D that contains the contour C. If z₀ and z₁ are the initial and terminal points of C, then the integration by parts formula is valid in D:

This can be proved in a straightforward manner using Theorem 18.3.2 on the function (d/dz)(fg). See Problems 21–24 in Exercises 18.3.

18.3 Exercises Answers to selected odd-numbered problems begin on page ANS-46.

In Problems 1 and 2, evaluate the given integral, where C is the contour given in the figure, by (a) finding an alternative path of integration and (b) using Theorem 18.3.2.

1. (4z − 1) dz

   

2. e^z dz

   

In Problems 3 and 4, evaluate the given integral along the indicated contour C.

3. 2z dz, where C is z(t) = 2t³ + i(t⁴ − 4t³ + 2), −1 ≤ t ≤ 1
4. 6z² dz, where C is z(t) = 2 cos³πt − i sin² t, 0 ≤ t ≤ 2

In Problems 5–24, use Theorem 18.3.2 to evaluate the given integral. Write each answer in the form a + ib.

5. 
6. 
7. 
8. 
9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. dz, C is the arc of the circle z = 4e^it, −π/2 ≤ t ≤ π/2
18. dz, C is the straight line segment between z = 1 + i and z = 4 + 4i
19. dz, C is any contour not passing through the origin
20. dz, C is any contour in the right half-plane Re(z) > 0
21. 
22. 
23. 
24. 

*See Problem 29 in Exercises 18.1.