Chapter Four
Integration
4.1. Introduction. If Lð : D ¸ð C is simply a function on a real interval D =ð ßðJð, KðÄ…ð , then the
Kð
integral Xð LðÅ»ðtÞðdt is, of course, simply an ordered pair of everyday 3rd grade calculus
Jð
integrals:
Kð Kð Kð
LðÅ»ðtÞðdt =ð xÅ»ðtÞðdt +ð i yÅ»ðtÞðdt,
Xð Xð Xð
Jð Jð Jð
where LðÅ»ðtÞð =ð xÅ»ðtÞð +ð iyÅ»ðtÞð. Thus, for example,
1
4 i
Å»ðt2 +ð 1Þð +ð it3 dt =ð +ð .
Xðßð Ä…ð
3 4
0
Nothing really new here. The excitement begins when we consider the idea of an integral
of an honest-to-goodness complex function f : D ¸ð C, where D is a subset of the complex
plane. Let s define the integral of such things; it is pretty much a straight-forward extension
to two dimensions of what we did in one dimension back in Mrs. Turner s class.
Suppose f is a complex-valued function on a subset of the complex plane and suppose a
and b are complex numbers in the domain of f. In one dimension, there is just one way to
get from one number to the other; here we must also specify a path from a to b. Let C be a
path from a to b, and we must also require that C be a subset of the domain of f.
4.1
(Note we do not even require that a ®ð b; but in case a =ð b, we must specify an orientation
for the closed path C.) Next, let P be a partition of the curve; that is, P =ð áðz0, z1, z2, uð , znâð
is a finite subset of C, such that a =ð z0, b =ð zn, and such that zj comes immediately after
zj?ð1 as we travel along C from a to b.
A Riemann sum associated with the partition P is just what it is in the real case:
n
SÅ»ðPÞð =ð fÅ»ðzDðÞðAðzj,
>ð
j
j=ð1
where zDð is a point on the arc between zj?ð1 and zj , and Aðzj =ð zj ?ð zj?ð1. (Note that for a
j
given partition P, there are many SÅ»ðPÞð depending on how the points zDð are chosen.) If
j
there is a number L so that given any Pð >ð 0, there is a partition PPð of C such that
SÅ»ðPÞð ?ð L <ð Pð
| |
whenever P Ńð PPð, then f is said to be integrable on C and the number L is called the
integral of f on C. This number L is usually written Xð fÅ»ðzÞðdz.
C
Some properties of integrals are more or less evident from looking at Riemann sums:
cfÅ»ðzÞðdz =ð c fÅ»ðzÞðdz
Xð Xð
C C
for any complex constant c.
4.2
Å»ðfÅ»ðzÞð +ð gÅ»ðzÞðÞðdz =ð fÅ»ðzÞðdz +ð gÅ»ðzÞðdz
Xð Xð Xð
C C C
4.2 Evaluating integrals. Now, how on Earth do we ever find such an integral? Let
Lð : ßðJð, KðÄ…ð ¸ð C be a complex description of the curve C. We partition C by partitioning the
interval ßðJð, KðÄ…ð in the usual way: Jð =ð t0 <ð t1 <ð t2 <ðuð <ð tn =ð Kð. Then
áða =ð LðÅ»ðJðÞð, LðÅ»ðt1Þð, LðÅ»ðt2Þð, uð , LðÅ»ðKðÞð =ð bâð is partition of C. (Recall we assume that LðvðÅ»ðtÞð ®ð 0
for a complex description of a curve C.) A corresponding Riemann sum looks like
n
SÅ»ðPÞð =ð fÅ»ðLðÅ»ðtDðÞðÞðÅ»ðLðÅ»ðtjÞð ?ð LðÅ»ðtj?ð1ÞðÞð.
>ð
j
j=ð1
We have chosen the points zDð =ð LðÅ»ðtDðÞð, where tj?ð1 ²ð tDð ²ð tj. Next, multiply each term in the
j j j
sum by 1 in disguise:
n
LðÅ»ðtjÞð ?ð LðÅ»ðtj?ð1Þð
SÅ»ðPÞð =ð fÅ»ðLðÅ»ðtDðÞðÞðÅ»ð
>ð
j
tj ?ð tj?ð1 ÞðÅ»ðtj ?ð tj?ð1Þð.
j=ð1
I hope it is now reasonably convincing that in the limit , we have
Kð
fÅ»ðzÞðdz =ð fÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞðdt.
Xð Xð
C
Jð
(We are, of course, assuming that the derivative Lðvð exists.)
Example
We shall find the integral of fÅ»ðzÞð =ð Å»ðx2 +ð yÞð +ð iÅ»ðxyÞð from a =ð 0 to b =ð 1 +ð i along three
different paths, or contours, as some call them.
First, let C1 be the part of the parabola y =ð x2 connecting the two points. A complex
description of C1 is Lð1Å»ðtÞð =ð t +ð it2, 0 ²ð t ²ð 1:
4.3
1
0.8
0.6
0.4
0.2
0
0.2 0.4 0.6 0.8 1
x
Now, Lðvð Å»ðtÞð =ð 1 +ð 2ti, and fÅ»ð Lð1Å»ðtÞðÞð =ð Å»ðt2 +ð t2Þð +ð itt2 =ð 2t2 +ð it3. Hence,
1
1
fÅ»ðzÞðdz =ð fÅ»ð Lð1Å»ðtÞðÞðLðvð Å»ðtÞðdt
Xð Xð
1
C1 0
1
=ð Å»ð2t2 +ð it3ÞðÅ»ð1 +ð 2tiÞðdt
Xð
0
1
=ð Å»ð2t2 ?ð 2t4 +ð 5t3iÞðdt
Xð
0
4 5
=ð +ð i
15 4
Next, let s integrate along the straight line segment C2 joining 0 and 1 +ð i.
1
0.8
0.6
0.4
0.2
0 0.2 0.4 0.6 0.8 1
x
Here we have Lð2Å»ðtÞð =ð t +ð it, 0 ²ð t ²ð 1. Thus, Lðvð Å»ðtÞð =ð 1 +ð i, and our integral looks like
2
4.4
1
fÅ»ðzÞðdz =ð fÅ»ð Lð2Å»ðtÞðÞðLðvð Å»ðtÞðdt
Xð Xð
2
C2 0
1
=ð Å»ðt2 +ð tÞð +ð it2 Å»ð1 +ð iÞðdt
Xðßð Ä…ð
0
1
=ð ßðt +ð i t +ð 2t2 Ä…ðdt
Xð Å»ð Þð
0
1 7
=ð +ð i
2 6
Finally, let s integrate along C3, the path consisting of the line segment from 0 to 1
together with the segment from 1 to 1 +ð i.
1
0.8
0.6
0.4
0.2
0
0.2 0.4 0.6 0.8 1
We shall do this in two parts: C31, the line from 0 to 1 ; and C32, the line from 1 to 1 +ð i.
Then we have
fÅ»ðzÞðdz =ð fÅ»ðzÞðdz +ð fÅ»ðzÞðdz.
Xð Xð Xð
C3 C31 C32
For C31 we have LðÅ»ðtÞð =ð t, 0 ²ð t ²ð 1. Hence,
1
1
fÅ»ðzÞðdz =ð dt =ð .
Xð Xð
3
C31 0
For C32 we have LðÅ»ðtÞð =ð 1 +ð it, 0 ²ð t ²ð 1. Hence,
1
1 5
fÅ»ðzÞðdz =ð Å»ð1 +ð t +ð itÞðitdt =ð ?ð +ð i.
Xð Xð
3 6
C32 0
4.5
Thus,
fÅ»ðzÞðdz =ð fÅ»ðzÞðdz +ð fÅ»ðzÞðdz
Xð Xð Xð
C3 C31 C32
5
=ð i.
6
Suppose there is a number M so that fÅ»ðzÞð ²ð M for all zOðC. Then
| |
Kð
fÅ»ðzÞðdz =ð fÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞðdt
Xð Xð
C
Jð
Kð
²ð fÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞð dt
|
Xð |
Jð
Kð
²ð M LðvðÅ»ðtÞð dt =ð ML,
| |
Xð
Jð
Kð
where L =ð Xð LðvðÅ»ðtÞð dt is the length of C.
| |
Jð
Exercises
1. Evaluate the integral Xð zdz, where C is the parabola y =ð x2 from 0 to 1 +ð i.
C
1
2. Evaluate Xð dz, where C is the circle of radius 2 centered at 0 oriented
z
C
counterclockwise.
4. Evaluate Xð fÅ»ðzÞðdz, where C is the curve y =ð x3 from ?ð1 ?ð i to 1 +ð i , and
C
1 for y <ð 0
fÅ»ðzÞð =ð .
4y for y Å‚ð 0
5. Let C be the part of the circle LðÅ»ðtÞð =ð eit in the first quadrant from a =ð 1 to b =ð i. Find as
4
small an upper bound as you can for XðCÅ»ðz2 ?ð z +ð 5Þðdz .
4.6
6. Evaluate Xð fÅ»ðzÞðdz where fÅ»ðzÞð =ð z +ð 2 z and C is the path from z =ð 0 to z =ð 1 +ð 2i
C
consisting of the line segment from 0 to 1 together with the segment from 1 to 1 +ð 2i.
4.3 Antiderivatives. Suppose D is a subset of the reals and Lð : D ¸ð C is differentiable at t.
Suppose further that g is differentiable at LðÅ»ðtÞð. Then let s see about the derivative of the
composition gÅ»ðLðÅ»ðtÞðÞð. It is, in fact, exactly what one would guess. First,
gÅ»ðLðÅ»ðtÞðÞð =ð uÅ»ðxÅ»ðtÞð, yÅ»ðtÞðÞð +ð ivÅ»ðxÅ»ðtÞð, yÅ»ðtÞðÞð,
where gÅ»ðzÞð =ð uÅ»ðx, yÞð +ð ivÅ»ðx, yÞð and LðÅ»ðtÞð =ð xÅ»ðtÞð +ð iyÅ»ðtÞð. Then,
dy dy
d /ðu dx /ðu /ðv dx /ðv
gÅ»ðLðÅ»ðtÞðÞð =ð +ð +ð i +ð .
dt /ðx dt /ðy dt /ðx dt /ðy dt
The places at which the functions on the right-hand side of the equation are evaluated are
obvious. Now, apply the Cauchy-Riemann equations:
dy dy
d /ðu dx /ðv /ðv dx /ðu
gÅ»ðLðÅ»ðtÞðÞð =ð ?ð +ð i +ð
dt /ðx dt /ðx dt /ðx dt /ðx dt
dy
/ðu /ðv dx
=ð +ð i +ð i
/ðx /ðx dt dt
=ð gvðÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞð.
The nicest result in the world!
Now, back to integrals. Let F : D ¸ð C and suppose FvðÅ»ðzÞð =ð fÅ»ðzÞð in D. Suppose moreover
that a and b are in D and that C Ðð D is a contour from a to b. Then
Kð
fÅ»ðzÞðdz =ð fÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞðdt,
Xð Xð
C
Jð
where Lð : ßðJð, KðÄ…ð ¸ð C describes C. From our introductory discussion, we know that
d
FÅ»ðLðÅ»ðtÞðÞð =ð FvðÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞð =ð fÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞð. Hence,
dt
4.7
Kð
fÅ»ðzÞðdz =ð fÅ»ðLðÅ»ðtÞðÞðLðvðÅ»ðtÞðdt
Xð Xð
C
Jð
Kð
d
=ð FÅ»ðLðÅ»ðtÞðÞðdt =ð FÅ»ðLðÅ»ðKðÞðÞð ?ð FÅ»ðLðÅ»ðJðÞðÞð
Xð
dt
Jð
=ð FÅ»ðbÞð ?ð FÅ»ðaÞð.
This is very pleasing. Note that integral depends only on the points a and b and not at all
on the path C. We say the integral is path independent. Observe that this is equivalent to
saying that the integral of f around any closed path is 0. We have thus shown that if in D
the integrand f is the derivative of a function F, then any integral Xð fÅ»ðzÞðdz for C Ðð D is path
C
independent.
Example
1 i
Let C be the curve y =ð from the point z =ð 1 +ð i to the point z =ð 3 +ð . Let s find
9
x2
z2dz.
Xð
C
1
This is easy we know that FvðÅ»ðzÞð =ð z2 , where FÅ»ðzÞð =ð z3. Thus,
3
3
1 i
z2dz =ð 1 +ð i ?ð 3 +ð
Xð Å»ð Þð3
3 9
C
260 728
=ð ?ð ?ð i
27 2187
Now, instead of assuming f has an antiderivative, let us suppose that the integral of f
between any two points in the domain is independent of path and that f is continuous.
Assume also that every point in the domain D is an interior point of D and that D is
connected. We shall see that in this case, f has an antiderivative. To do so, let z0 be any
point in D, and define the function F by
FÅ»ðzÞð =ð fÅ»ðzÞðdz,
Xð
Cz
where Cz is any path in D from z0 to z. Here is important that the integral is path
independent, otherwise FÅ»ðzÞð would not be well-defined. Note also we need the assumption
that D is connected in order to be sure there always is at least one such path.
4.8
Now, for the computation of the derivative of F:
FÅ»ðz +ðAðzÞð ?ð FÅ»ðzÞð =ð fÅ»ðsÞðds,
Xð
LAðz
where LAðz is the line segment from z to z +ðAðz.
1
Next, observe that Xð ds =ð Aðz. Thus, fÅ»ðzÞð =ð Xð fÅ»ðzÞðds, and we have
Aðz
LAðz LAðz
FÅ»ðz +ðAðzÞð ?ð FÅ»ðzÞð
1
?ð fÅ»ðzÞð =ð fÅ»ðsÞð ?ð fÅ»ðzÞð ds.
Xð Å»ð Þð
Aðz Aðz
LAðz
Now then,
1 1
fÅ»ðsÞð ?ð fÅ»ðzÞð ds ²ð Aðz max fÅ»ðsÞð ?ð fÅ»ðzÞð : sOðLAðz
Xð Å»ð Þð | | áð| | âð
Aðz Aðz
LAðz
²ð max fÅ»ðsÞð ?ð fÅ»ðzÞð : sOðLAðz .
áð| | âð
We know f is continuous at z, and so lim max fÅ»ðsÞð ?ð fÅ»ðzÞð : sOðLAðz =ð 0. Hence,
áð| | âð
Aðz¸ð0
FÅ»ðz +ðAðzÞð ?ð FÅ»ðzÞð
1
lim ?ð fÅ»ðzÞð =ð lim fÅ»ðsÞð ?ð fÅ»ðzÞð ds
Xð Å»ð Þð
Aðz Aðz
Aðz¸ð0 Aðz¸ð0
LAðz
=ð 0.
4.9
In other words, FvðÅ»ðzÞð =ð fÅ»ðzÞð, and so, just as promised, f has an antiderivative! Let s
summarize what we have shown in this section:
Suppose f : D ¸ð C is continuous, where D is connected and every point of D is an interior
point. Then f has an antiderivative if and only if the integral between any two points of D is
path independent.
Exercises
7. Suppose C is any curve from 0 to ^ð +ð 2i. Evaluate the integral
z
cos dz.
Xð
2
C
1
8. a)Let FÅ»ðzÞð =ð log z, 0 <ð arg z <ð 2^ð. Show that the derivative FvðÅ»ðzÞð =ð .
z
^ð 7^ð 1
b)Let GÅ»ðzÞð =ð log z, ?ð <ð arg z <ð . Show that the derivative GvðÅ»ðzÞð =ð .
z
4 4
c)Let C1 be a curve in the right-half plane D1 =ð áðz : Re z Å‚ð 0âð from ?ði to i that does not
pass through the origin. Find the integral
1
dz.
Xð
z
C1
d)Let C2 be a curve in the left-half plane D2 =ð áðz : Re z ²ð 0âð from ?ði to i that does not
pass through the origin. Find the integral.
1
dz.
Xð
z
C2
9. Let C be the circle of radius 1 centered at 0 with the clockwise orientation. Find
1
dz.
Xð
z
C
10. a)Let HÅ»ðzÞð =ð zc,?ð^ð <ð arg z <ð ^ð. Find the derivative HvðÅ»ðzÞð.
^ð 7^ð
b)Let KÅ»ðzÞð =ð zc, ?ð <ð arg z <ð . What is the largest subset of the plane on which
4 4
HÅ»ðzÞð =ð KÅ»ðzÞð?
c)Let C be any path from ?ð1 to 1 that lies completely in the upper half-plane. (Upper
4.10
half-plane =ð áðz : Imz Å‚ð 0âð.) Find
FÅ»ðzÞðdz,
Xð
C
where FÅ»ðzÞð =ð zi, ?ð^ð <ð arg z ²ð ^ð.
11. Suppose P is a polynomial and C is a closed curve. Explain how you know that
Xð PÅ»ðzÞðdz =ð 0.
C
4.11
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