3.1 The Fundamental Theorem of Calculus
In Example 5 we saw that

where is the contour shown in Figure 23. Our method was to write down a smooth parametrisation for each of the two line segments, replace
in the integral by these parametrisations, and then integrate.




It is, however, tempting to approach this integral as you would a corresponding real integral and write

The Fundamental Theorem of Calculus for contour integrals tells us that this method of evaluation is permissible under certain conditions. Before stating it, we need the idea of a primitive of a complex function, which is defined in a similar way to the primitive of a real function (Section 1.3).
Definition
Let and
be functions defined on a region
. Then
is a primitive of
on
if
is analytic on
and

The function is also called an antiderivative or indefinite integral of
on
.
For example, is a primitive of
on
, since
is analytic on
and
, for all
. Another primitive is
; indeed, any function of the form
, where
, is a primitive of
on
.
Exercise 10
Write down a primitive of each of the following functions
on the given region
.
Answer
We now state the Fundamental Theorem of Calculus for contour integrals, which gives us a quick way of evaluating a contour integral of a function with a primitive that we can determine. The theorem will be proved later in this section.
Theorem 7 Fundamental Theorem of Calculus
Let be a function that is continuous and has a primitive
on a region
, and let
be a contour in
with initial point
and final point
. Then

We often use the notation

Some texts write instead of
.
For an example of the use of the Fundamental Theorem of Calculus, observe that if , then
is continuous on
and has a primitive
there. Hence, for the contour
in Figure 23, we can write

Exercise 11
Use the Fundamental Theorem of Calculus to evaluate

where is the semicircular path shown in Figure 24.




Answer
Let ,
and
. Then
is continuous on
, and
is a primitive of
on
. Thus, by the Fundamental Theorem of Calculus,

The final simplification follows from the formula

with .
You have seen that

both when is the contour in Figure 23 and also when
is the line segment from
to
(see Example 2). This is not a coincidence: in fact, it is a particular case of the following important consequence of the Fundamental Theorem of Calculus.
Theorem 8 Contour Independence Theorem
Let be a function that is continuous and has a primitive
on a region
, and let
and
be contours in
with the same initial point
and the same final point
. Then

Proof
By the Fundamental Theorem of Calculus for contour integrals, the value of each of these integrals is .
The idea that a contour integral may, under suitable hypotheses, depend only on the endpoints of the contour (and not on the contour itself) has great significance.
Exercise 12
Use the Fundamental Theorem of Calculus to evaluate the following integrals.
, where
is any contour from
to
.
, where
is any contour from 2 to
.
, where
is any contour from
to 1.
, where
is any contour from 0 to
.
, where
is any contour from 0 to
lying in
.
Answer
Let
,
and
. Then
is continuous on
, and
is a primitive of
on
. Thus, by the Fundamental Theorem of Calculus,
Let
,
and
. Then
is continuous on
, and
is a primitive of
on
. Thus, by the Fundamental Theorem of Calculus,
Let
,
and
. Then
is continuous on
, and
is a primitive of
on
. Thus, by the Fundamental Theorem of Calculus,
The integrand
can be written as
which equals
by the Chain Rule. So let
,
and
. Then
is continuous on
, and
is a primitive of
on
. Thus, by the Fundamental Theorem of Calculus,
Remark: If you have a good deal of experience at differentiating and integrating real and complex functions, then you may have chosen to write down the primitive
of
straight away.
The integrand
can be written as
which equals
So let
Then
is continuous on
, and
is a primitive of
on
. Thus, by the Fundamental Theorem of Calculus,
(In this solution, note that the region
does not contain the point
, as
; thus
cannot be chosen to be a path that contains
. In particular, the real integral
does not exist.)
Next we give a version of Integration by Parts for contour integrals.
Theorem 9 Integration by Parts
Let and
be functions that are analytic on a region
, and suppose that
and
are continuous on
. Let
be a contour in
with initial point
and final point
. Then

Proof
Let and
. Then
is continuous on
, by hypothesis. Also,
has primitive
, since
is analytic on
and

by the Product Rule for differentiation. It follows from the Fundamental Theorem of Calculus that

that is,

Using the Sum Rule (Theorem 5(a)) and rearranging the resulting equation, we obtain

as required.
Example 7
Use Integration by Parts to evaluate

where is any contour from
to
.
Solution
We take ,
and
. Then
and
are analytic on
, and
and
are continuous on
.
Integrating by parts, we obtain

Exercise 13
Use Integration by Parts to evaluate the following integrals.
where
is any contour from 0 to
.
, where
is any contour from 1 to
lying in the cut plane
.
(Hint: For part (b), take and
.)
Answer
We take
,
and
.
Then
and
are analytic on
, and
and
are continuous on
.
Integrating by parts, we obtain
We take
,
and
. Then
and
are analytic on
, and
and
are continuous on
.
Integrating by parts, we obtain
The Fundamental Theorem of Calculus is a useful tool when the function being integrated has an easily determined primitive
. However, if the function
has no primitive, or if we are unable to find one, then we have to resort to the definition of an integral and use parametrisation. For example, we cannot use the Fundamental Theorem of Calculus to evaluate

along any contour, since the function has no primitive on any region.
To see why this is so, suppose that is a function that is defined on a region in the complex plane. We observe that if
is not differentiable, then
has no primitive
. This is because any differentiable complex function can be differentiated as many times as we like. Thus, if
has a primitive
, then
is differentiable with
. Hence
is also differentiable.
It follows that we cannot use the Fundamental Theorem of Calculus to evaluate integrals of non-differentiable functions such as

We conclude this section by proving the Fundamental Theorem of Calculus.
Proof
The proof of the Fundamental Theorem of Calculus is in two parts. We first prove the result in the case when is a smooth path, and then extend the proof to contours.
Let
be a smooth path. Then
by the Chain Rule. Now, if we write
as a sum of its real and imaginary parts
, then
The Fundamental Theorem of Calculus for real integrals (Theorem 2) tells us that
Hence
since
and
.
To extend the proof to a general contour
with initial point
and final point
, we argue as follows.
Let
, for smooth paths
, and let the initial and final points of
be
and
, for
. Then
By part (a),
for
(where
). Hence
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