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696 CHAPTER 20 Complex Integration

y

2 + 4i

–1 + i

x

FIGURE 20.2 δ in Example 20.2.

EXAMPLE 20.2
2
Let δ(t) = t + it for −1 ≤ t ≤ 2. Then δ is a simple, smooth curve from −1 + i to 2 + 4i,as
2
shown in Figure 20.2. The coordinate functions are x = t and y = t , and we can think of C as
the part of the parabola y = x from (−1,1) to (2,4).
2
We are now ready to define the complex integral, which we will do in two stages.

First Stage—Integral Over a Closed Interval Suppose f is a complex function, and f (x) =
u(x) + iv(x) is defined at least for a ≤ x ≤ b. Define
b b b
f (x)dx = u(x)dx + i v(x)dx.
a a a
EXAMPLE 20.3
2
Let f (x) = x − ix for 1 ≤ x ≤ 2. Then
2 2 2 3 7

2
f (x)dx = xdx − i x dx = − i.
2 3
1 1 1
Second Stage—Integral Over a Curve Now we can define the integral of a complex function
over a smooth curve in the plane. Let f be a complex function and γ a curve with γ(t) defined
for a ≤ t ≤ b. Assume that f is continuous at all points on the curve. Then we define the integral
of f over γ by
b

f (z)dz = f (γ (t))γ (t)dt.
γ a
This integral also may be formulated as
b

f (z)dz = f (z(t))z (t)dt.
γ a

Evaluate f (z)dz by replacing z by z(t) = γ(t) and integrating f (γ (t))γ (t) over [a,b],

γ
according to the first stage in the definition of the integral.
EXAMPLE 20.4
it
We will evaluate zdz if γ(t) = e for 0 ≤ t ≤ π.
γ
The graph of γ is the upper half of the unit circle oriented counterclockwise from initial
it
it
point 1 to terminal point −1. On γ , z(t) = e and z (t) = ie ,so

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October 14, 2010 15:32 THM/NEIL Page-696 27410_20_ch20_p695-714