2Contour integration
IB Complex Analysis
2.1 Basic properties of complex integration
We start by considering functions
f
: [
a, b
]
→ C
. We say such a function is
Riemann integrable if
Re
(
f
) and
Im
(
f
) are individually, and the integral is
defined to be
Z
b
a
f(t) dt =
Z
b
a
Re(f(t)) dt + i
Z
b
a
Im(f(t)) dt.
While Riemann integrability is a technical condition to check, we know that all
continuous functions are integrable, and this will apply in most cases we care
about in this course. After all, this is not a course on exotic functions.
We start from some less interesting facts, and slowly develop and prove some
really amazing results.
Lemma. Suppose f : [a, b] → C is continuous (and hence integrable). Then
Z
b
a
f(t) dt
≤ (b − a) sup
t
|f(t)|
with equality if and only if f is constant.
Proof. We let
θ = arg
Z
b
a
f(t) dt
!
,
and
M = sup
t
|f(t)|.
Then we have
Z
b
a
f(t) dt
=
Z
b
a
e
−iθ
f(t) dt
=
Z
b
a
Re(e
−iθ
f(t)) dt
≤ (b − a)M,
with equality if and only if
|f
(
t
)
|
=
M
and
arg f
(
t
) =
θ
for all
t
, i.e.
f
is
constant.
Integrating functions of the form
f
: [
a, b
]
→ C
is easy. What we really care
about is integrating a genuine complex function
f
:
U ⊆ C → C
. However,
we cannot just “integrate” such a function. There is no given one-dimensional
domain we can integrate along. Instead, we have to make some up ourselves.
We have to define some paths in the complex plane, and integrate along them.
Definition
(Path)
.
A path in
C
is a continuous function
γ
: [
a, b
]
→ C
, where
a, b ∈ R.
For general paths, we just require continuity, and do not impose any conditions
about, say, differentiability.
Unfortunately, the world is full of weird paths. There are even paths that fill
up the whole of the unit square. So we might want to look at some nicer paths.
Definition
(Simple path)
.
A path
γ
: [
a, b
]
→ C
is simple if
γ
(
t
1
) =
γ
(
t
2
) only
if t
1
= t
2
or {t
1
, t
2
} = {a, b}.
In other words, it either does not intersect itself, or only intersects itself at
the end points.
Definition (Closed path). A path γ : [a, b] → C is closed if γ(a) = γ(b).
Definition
(Contour)
.
A contour is a simple closed path which is piecewise
C
1
,
i.e. piecewise continuously differentiable.
For example, it can look something like this:
Most of the time, we are just interested in integration along contours. However,
it is also important to understand integration along just simple
C
1
smooth paths,
since we might want to break our contour up into different segments. Later, we
will move on to consider more general closed piecewise
C
1
paths, where we can
loop around a point many many times.
We can now define what it means to integrate along a smooth path.
Definition
(Complex integration)
.
If
γ
: [
a, b
]
→ U ⊆ C
is
C
1
-smooth and
f : U → C is continuous, then we define the integral of f along γ as
Z
γ
f(z) dz =
Z
b
a
f(γ(t))γ
0
(t) dt.
By summing over subdomains, the definition extends to piecewise
C
1
-smooth
paths, and in particular contours.
We have the following elementary properties:
(i)
The definition is insensitive to reparametrization. Let
φ
: [
a
0
, b
0
]
→
[
a, b
]
be
C
1
such that
φ
(
a
0
) =
a, φ
(
b
0
) =
b
. If
γ
is a
C
1
path and
δ
=
γ ◦φ
, then
Z
γ
f(z) dz =
Z
δ
f(z) dz.
This is just the regular change of variables formula, since
Z
b
0
a
0
f(γ(φ(t)))γ
0
(φ(t))φ
0
(t) dt =
Z
b
a
f(γ(u))γ
0
(u) du
if we let u = φ(t).
(ii) If a < u < b, then
Z
γ
f(z) dz =
Z
γ|
[a,u]
f(z) dz +
Z
γ|
[u,b]
f(z) dz.
These together tells us the integral depends only on the path itself, not how we
look at the path or how we cut up the path into pieces.
We also have the following easy properties:
(iii) If −γ is γ with reversed orientation, then
Z
−γ
f(z) dz = −
Z
γ
f(z) dz.
(iv) If we set for γ : [a, b] → C the length
length(γ) =
Z
b
a
|γ
0
(t)| dt,
then
Z
γ
f(z) dz
≤ length(γ) sup
t
|f(γ(t))|.
Example.
Take
U
=
C
∗
, and let
f
(
z
) =
z
n
for
n ∈ Z
. We pick
φ
: [0
,
2
π
]
→ U
that sends θ 7→ e
iθ
. Then
Z
φ
f(z) dz =
(
2πi n = −1
0 otherwise
To show this, we have
Z
φ
f(z) dz =
Z
2π
0
e
inθ
ie
iθ
dθ
= i
Z
2π
0
e
i(n+1)θ
dθ.
If
n
=
−
1, then the integrand is constantly 1, and hence gives 2
πi
. Otherwise, the
integrand is a non-trivial exponential which is made of trigonometric functions,
and when integrated over 2π gives zero.
Example. Take γ to be the contour
γ
1
γ
2
R−R
iR
We parametrize the path in segments by
γ
1
: [−R, R] → C γ
2
: [0, 1] → C
t 7→ t t 7→ Re
iπt
Consider the function f(z) = z
2
. Then the integral is
Z
γ
f(z) dz =
Z
R
−R
t
2
dt +
Z
1
0
R
2
e
2πit
iπRe
iπt
dt
=
2
3
R
3
+ R
3
iπ
Z
1
0
e
3πit
dt
=
2
3
R
3
+ R
3
iπ
e
3πit
3πi
1
0
= 0
We worked this out explicitly, but we have just wasted our time, since this is
just an instance of the fundamental theorem of calculus!
Definition
(Antiderivative)
.
Let
U ⊆ C
and
f
:
U → C
be continuous. An
antiderivative of
f
is a holomorphic function
F
:
U → C
such that
F
0
(
z
) =
f
(
z
).
Then the fundamental theorem of calculus tells us:
Theorem
(Fundamental theorem of calculus)
.
Let
f
:
U → C
be continuous
with antiderivative F . If γ : [a, b] → U is piecewise C
1
-smooth, then
Z
γ
f(z) dz = F (γ(b)) − F (γ(a)).
In particular, the integral depends only on the end points, and not the path
itself. Moreover, if γ is closed, then the integral vanishes.
Proof. We have
Z
γ
f(z) dz =
Z
b
a
f(γ(t))γ
0
(t) dt =
Z
b
a
(F ◦γ)
0
(t) dt.
Then the result follows from the usual fundamental theorem of calculus, applied
to the real and imaginary parts separately.
Example.
This allows us to understand the first example we had. We had the
function f(z) = z
n
integrated along the path φ(t) = e
it
(for 0 ≤ t ≤ 2π).
If n 6= −1, then
f =
d
dt
z
n+1
n + 1
.
So
f
has a well-defined antiderivative, and the integral vanishes. On the other
hand, if n = −1, then
f(z) =
d
dz
(log z),
where
log
can only be defined on a slit plane. It is not defined on the whole unit
circle. So we cannot apply the fundamental theorem of calculus.
Reversing the argument around, since
R
φ
z
−1
d
z
does not vanish, this implies
there is not a continuous branch of log on any set U containing the unit circle.