Complex Integration
The magic and power of calculus ultimately rests on the amazing fact that differentiation and integration are mutually inverse operations. And, just as complex functions enjoy remarkable differentiability properties not shared by their real counterparts, so the sublime beauty of complex integration goes far beyond its real progenitor.
Contour integral
Consider a contour
The right hand side of (
Now let us write the integrand
Example 1: Let's evaluate

Here we have that
Example 2: Now let's evaluate

In this case
Numerical evaluation of complex integrals
Exploration 1
Use the following applet to explore numerically the integral
- Line segments.
- Semicircles.
- Circles, positively and negatively oriented.
You can also change the domain coloring plotting option. Drag the points around and observe carefully what happens. Then solve Exercise 1 below.
The arrows on the contours indicate direction.
Exercise 1: Use definition (
- Line segment. That is,
with - Right-hand semicircle. That is,
with - Left-hand semicircle. That is,
with
Use the applet to confirm your results.
What conclusions (if any) can you draw about the function
Exploration 2
Now use the applet below to explore numerically the integrals
z^2+z
or 1/z^2
from the list at the left-top corner.
Then solve Exercises 2 and 3.
Exercise 2: Consider the integral
is any contour from to is the circle with center and radius ; positively or negatively oriented. In this cases selectCircle ↺
orCircle ↻
.
What conclusions (if any) can you draw about the value of
Exercise 3: Now considering integral
f(z)=1/z^2
.
Then analyze the values of
is any contour from to What happens when you selectLine Segment
in the applet? What happens when you selectSemicircles
? is the circle with center and radius ; positively or negatively oriented. In this case selectCircle ↺
orCircle ↻
. What happens if is inside or outside the circle? What happens if lies on the contour, e.g. when and ?
What conclusions (if any) can you draw about the value of
Properties
All the following familiar properties of integrals can be proved directly
from the definition given in (
From definition (
Now consider the contour defined for the integral (
The contour
If we use the substitution
Hence, we obtain the following property
Consider now a path
Exercise 4:
Use the properties for integrals of functions
Finally we introduce an inequality involving contour integrals that is
extremely important in various applications. Consider
If
Note that since
The property (
Example 3:
Let
Note that if
Thus
Example 4:
We can also use the ML-inequality to find the value of
the integral
Note that when
Since the length of
The term on the right tends to zero as
Antiderivatives
While the value of a contour integral of a function
As a consequence of the previous Theorem 1, for closed curves we have
If the function
Considering Figure 6, we have
In addition, we have the following sufficient condition for the existence of an antiderivate:
Using properties of the integral we have
Since
From (
Now, since
Therefore, we have proved that