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CHAPTER 14
Complex Integration
Change
We now discuss the two main integration methods (indefinite integration and integration
by the use of the representation of the path) directly after the definition of the integral,
postponing the proof of the first of these methods until Cauchy’s integral formula is
available in Sec. 14.2. This order of the material seems desirable from a practical point
of view.
Comment
The introduction to the chapter mentions two reasons for the importance of complex
integration. Another practical reason is the extensive use of complex integral
representations in the higher theory of special functions; see for instance, Ref. [GR10]
listed in App. 1.
SECTION 14.1. Line Integral in the Complex Plane, page 637
Purpose.
To discuss the definition, existence, and general properties of complex line
integrals. Complex integration is rich in methods, some of them very elegant. In this
section we discuss the first two methods, integration by the use of path and (under suitable
assumptions given in Theorem 1!) by indefinite integration.
Main Content, Important Concepts
Definition of the complex line integral
Existence
Basic properties
Indefinite integration (Theorem 1)
Integration by the use of path (Theorem 2)
Integral of 1/
z
around the unit circle (basic!)
ML
-inequality (13) (needed often in our work)
Comment on Content
Indefinite integration will be justified in Sec. 14.2, after we have obtained Cauchy’s
integral theorem. We discuss this method here for two reasons: (i) to get going a little
faster and, more importantly, (ii) to answer the students’ natural question suggested by
calculus, that is, whether the method works and under what condition—that it does not
work unconditionally can be seen from Example 7!
SOLUTIONS TO PROBLEM SET 14.1, page 645
2.
Vertical straight segment from 5
1
6
i
to 5
2
6
i
4.
Circle, center 1
1
i
, radius 1, oriented clockwise, touching the axes
6.
Semicircle, center 3
1
4
i
, radius 5, passing through the origin
8.
Portion of the parabola
y
5
2(
x
2
1)
2
from
2
1
1
8
i
to 3
1
8
i
, apex at
x
5
1
10.
z
(
t
)
5
1
1
i
1
(3
2
3
i
)
t
(0
%
t
%
1)
12.
z
(
t
)
5
a
1
ib
1
[
c
2
a
1
i
(
d
2
b
)]
t
(0
%
t
%
1)
254
im14.qxd
9/21/05
12:37 PM
Page 254

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*Sign up*14.
z
(
t
)
5
a
cos
t
1
ib
sin
t
(0
%
t
%
p
)
16.
z
(
t
)
5
2
2
3
i
1
4
e
it
(0
%
t
%
2
) counterclockwise. For clockwise orientation the
exponent is
2
it
.
18.
Ellipse,
z
(
t
)
5
1
1
3 cos
t
1
(
2
2
1
2 sin
t
)
i
(0
%
t
%
2
)
20.
z
(
t
)
5
t
1
it
2
(0
%
t
%
1),
dz
5
(1
1
2
it
)
dt
,R
e
z
(
t
)
5
t
, so that
E
1
0
t
(1
1
2
it
)
dt
5
1
_
2
1
2
_
3
i
.
This differs from Prob. 19. The integrand is not analytic!

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