Contour integration


In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane.
Contour integration is closely related to the calculus of residues, a method of complex analysis.
One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods.
Contour integration methods include
One method can be used, or a combination of these methods, or various limiting processes, for the purpose of finding these integrals or sums.

Curves in the complex plane

In complex analysis a contour is a type of curve in the complex plane. In contour integration, contours provide a precise definition of the curves on which an integral may be suitably defined. A curve in the complex plane is defined as a continuous function from a closed interval of the real line to the complex plane:.
This definition of a curve coincides with the intuitive notion of a curve, but includes a parametrization by a continuous function from a closed interval. This more precise definition allows us to consider what properties a curve must have for it to be useful for integration. In the following subsections we narrow down the set of curves that we can integrate to include only those that can be built up out of a finite number of continuous curves that can be given a direction. Moreover, we will restrict the "pieces" from crossing over themselves, and we require that each piece have a finite continuous derivative. These requirements correspond to requiring that we consider only curves that can be traced, such as by a pen, in a sequence of even, steady strokes, which only stop to start a new piece of the curve, all without picking up the pen.

Directed smooth curves

Contours are often defined in terms of directed smooth curves. These provide a precise definition of a "piece" of a smooth curve, of which a contour is made.
A smooth curve is a curve with a non-vanishing, continuous derivative such that each point is traversed only once, with the possible exception of a curve such that the endpoints match. In the case where the endpoints match the curve is called closed, and the function is required to be one-to-one everywhere else and the derivative must be continuous at the identified point. A smooth curve that is not closed is often referred to as a smooth arc.
The parametrization of a curve provides a natural ordering of points on the curve: comes before if. This leads to the notion of a directed smooth curve. It is most useful to consider curves independent of the specific parametrization. This can be done by considering equivalence classes of smooth curves with the same direction. A directed smooth curve can then be defined as an ordered set of points in the complex plane that is the image of some smooth curve in their natural order. Note that not all orderings of the points are the natural ordering of a smooth curve. In fact, a given smooth curve has only two such orderings. Also, a single closed curve can have any point as its endpoint, while a smooth arc has only two choices for its endpoints.

Contours

Contours are the class of curves on which we define contour integration. A contour is a directed curve which is made up of a finite sequence of directed smooth curves whose endpoints are matched to give a single direction. This requires that the sequence of curves be such that the terminal point of coincides with the initial point of,. This includes all directed smooth curves. Also, a single point in the complex plane is considered a contour. The symbol + is often used to denote the piecing of curves together to form a new curve. Thus we could write a contour that is made up of curves as

Contour integrals

The contour integral of a complex function is a generalization of the integral for real-valued functions. For continuous functions in the complex plane, the contour integral can be defined in analogy to the line integral by first defining the integral along a directed smooth curve in terms of an integral over a real valued parameter. A more general definition can be given in terms of partitions of the contour in analogy with the partition of an interval and the Riemann integral. In both cases the integral over a contour is defined as the sum of the integrals over the directed smooth curves that make up the contour.

For continuous functions

To define the contour integral in this way one must first consider the integral, over a real variable, of a complex-valued function. Let be a complex-valued function of a real variable,. The real and imaginary parts of are often denoted as and, respectively, so that
Then the integral of the complex-valued function over the interval is given by
Let be a continuous function on the directed smooth curve. Let be any parametrization of that is consistent with its order. Then the integral along is denoted
and is given by
This definition is well defined. That is, the result is independent of the parametrization chosen. In the case where the real integral on the right side does not exist the integral along is said not to exist.

As a generalization of the Riemann integral

The generalization of the Riemann integral to functions of a complex variable is done in complete analogy to its definition for functions from the real numbers. The partition of a directed smooth curve is defined as a finite, ordered set of points on. The integral over the curve is the limit of finite sums of function values, taken at the points on the partition, in the limit that the maximum distance between any two successive points on the partition, also known as the mesh, goes to zero.

Direct methods

Direct methods involve the calculation of the integral by means of methods similar to those in calculating line integrals in several-variable calculus. This means that we use the following method:
A fundamental result in complex analysis is that the contour integral of is, where the path of the contour is taken to be the unit circle traversed counterclockwise. In the case of the unit circle there is a direct method to evaluate the integral
In evaluating this integral, use the unit circle as a contour, parametrized by, with, then and
which is the value of the integral.

Applications of integral theorems

Applications of integral theorems are also often used to evaluate the contour integral along a contour, which means that the real-valued integral is calculated simultaneously along with calculating the contour integral.
Integral theorems such as the Cauchy integral formula or residue theorem are generally used in the following method:
Consider the integral
To evaluate this integral, we look at the complex-valued function
which has singularities at and. We choose a contour that will enclose the real-valued integral, here a semicircle with boundary diameter on the real line will be convenient. Call this contour.
There are two ways of proceeding, using the Cauchy integral formula or by the method of residues:

Using the Cauchy integral formula

Note that:
thus
Furthermore, observe that
Since the only singularity in the contour is the one at , then we can write
which puts the function in the form for direct application of the formula. Then, by using Cauchy's integral formula,
We take the first derivative, in the above steps, because the pole is a second-order pole. That is, is taken to the second power, so we employ the first derivative of. If it were taken to the third power, we would use the second derivative and divide by 2!, etc. The case of to the first power corresponds to a zero order derivative—just itself.
We need to show that the integral over the arc of the semicircle tends to zero as, using the estimation lemma
where is an upper bound on along the arc and the length of the arc. Now,
So

Using the method of residues

Consider the Laurent series of about, the only singularity we need to consider. We then have
It is clear by inspection that the residue is, so, by the residue theorem, we have
Thus we get the same result as before.

Contour note

As an aside, a question can arise whether we do not take the semicircle to include the other singularity, enclosing. To have the integral along the real axis moving in the correct direction, the contour must travel clockwise, i.e., in a negative direction, reversing the sign of the integral overall.
This does not affect the use of the method of residues by series.

Example 2 – Cauchy distribution

The integral
resists the techniques of elementary calculus. We will evaluate it by expressing it as a limit of contour integrals along the contour that goes along the real line from to and then counterclockwise along a semicircle centered at 0 from to. Take to be greater than 1, so that the imaginary unit is enclosed within the curve. The contour integral is
Since is an entire function, this function has singularities only where the denominator is zero. Since, that happens only where or. Only one of those points is in the region bounded by this contour. The residue of at is
According to the residue theorem, then, we have
The contour may be split into a "straight" part and a curved arc, so that
and thus
It can be shown that if then
Therefore, if then
A similar argument with an arc that winds around rather than shows that if then
and finally we have this:

Example 3 – trigonometric integrals

Certain substitutions can be made to integrals involving trigonometric functions, so the integral is transformed into a rational function of a complex variable and then the above methods can be used in order to evaluate the integral.
As an example, consider
We seek to make a substitution of. Now, recall
and
Taking to be the unit circle, we substitute to get:
The singularities to be considered are at Let be a small circle about and be a small circle about Then we arrive at the following:

Example 3a – trigonometric integrals, the general procedure

The above method may be applied to all integrals of the type
where and are polynomials, i.e. a rational function in trigonometric terms is being integrated. Note that the bounds of integration may as well be and −, as in the previous example, or any other pair of endpoints 2 apart.
The trick is to use the substitution where and hence
This substitution maps the interval to the unit circle. Furthermore,
and
so that a rational function in results from the substitution, and the integral becomes
which is in turn computed by summing the residues of inside the unit circle.
The image at right illustrates this for
which we now compute. The first step is to recognize that
The substitution yields
The poles of this function are at and. Of these, and are outside the unit circle, whereas and are inside the unit circle. The corresponding residues are both equal to, so that the value of the integral is

Example 4 – branch cuts

Consider the real integral
We can begin by formulating the complex integral
We can use the Cauchy integral formula or residue theorem again to obtain the relevant residues. However, the important thing to note is that, so has a branch cut. This affects our choice of the contour. Normally the logarithm branch cut is defined as the negative real axis, however, this makes the calculation of the integral slightly more complicated, so we define it to be the positive real axis.
Then, we use the so-called keyhole contour, which consists of a small circle about the origin of radius say, extending to a line segment parallel and close to the positive real axis but not touching it, to an almost full circle, returning to a line segment parallel, close, and below the positive real axis in the negative sense, returning to the small circle in the middle.
Note that and are inside the big circle. These are the two remaining poles, derivable by factoring the denominator of the integrand. The branch point at was avoided by detouring around the origin.
Let be the small circle of radius, the larger, with radius, then
It can be shown that the integrals over and both tend to zero as and, by an estimation argument above, that leaves two terms. Now since, on the contour outside the branch cut, we have gained 2 in argument along. So
Therefore:
By using the residue theorem or the Cauchy integral formula one obtains

Example 5 – the square of the logarithm

This section treats a type of integral of which
is an example.
To calculate this integral, one uses the function
and the branch of the logarithm corresponding to.
We will calculate the integral of along the keyhole contour shown at right. As it turns out this integral is a multiple of the initial integral that we wish to calculate and by the Cauchy residue theorem we have
Let be the radius of the large circle, and the radius of the small one. We will denote the upper line by, and the lower line by. As before we take the limit when and. The contributions from the two circles vanish. For example, one has the following upper bound with the lemma:
In order to compute the contributions of and we set on and on, with :
which gives

Example 6 – logarithms and the residue at infinity

We seek to evaluate
This requires a close study of
We will construct so that it has a branch cut on, shown in red in the diagram. To do this, we choose two branches of the logarithm, setting
and
The cut of is therefore and the cut of is. It is easy to see that the cut of the product of the two, i.e., is, because is actually continuous across. This is because when and we approach the cut from above, has the value
When we approach from below, has the value
But
so that we have continuity across the cut. This is illustrated in the diagram, where the two black oriented circles are labelled with the corresponding value of the argument of the logarithm used in and.
We will use the contour shown in green in the diagram. To do this we must compute the value of along the line segments just above and just below the cut.
Let , where. Along the upper segment, we find that has the value
and along the lower segment,
It follows that the integral of along the upper segment is in the limit, and along the lower segment,.
If we can show that the integrals along the two green circles vanish in the limit, then we also have the value of, by the Cauchy residue theorem. Let the radius of the green circles be, where and, and apply the inequality. For the circle on the left, we find
Similarly, for the circle on the right, we have
Now using the Cauchy residue theorem, we have
where the minus sign is due to the clockwise direction around the residues. Using the branch of the logarithm from before, clearly
The pole is shown in blue in the diagram. The value simplifies to
We use the following formula for the residue at infinity:
Substituting, we find
and
where we have used the fact that for the second branch of the logarithm. Next we apply the binomial expansion, obtaining
The conclusion is that
Finally, it follows that the value of is
which yields

Evaluation with residue theorem

Using the residue theorem, we can evaluate closed contour integrals. The following are examples on evaluating contour integrals with the residue theorem.
Using the residue theorem, let's evaluate this contour integral.
As a refresher, the residue theorem states
where is the residue of.
has only one pole,. From that, we can determine the residue of to be
Thus, using the residue theorem, we can determine:

Multivariable Contour Integrals

To solve multivariable contour integrals, we must use the divergence theorem. For right now, let be interchangeable with. These will both serve as the divergence of the vector field denoted as . This theorem states:
In addition, we also need to evaluate where is an alternate notation of. The Divergence of any dimension can be described as

Example 1

Let the vector field and be bounded by the following
The corresponding double contour integral would be set up as such:
We now evaluate. While we're at it, let's set up the corresponding triple integral:

Example 2

For example, let the vector field, and is the fourth dimension. Let this vector field be bounded by the following:
To evaluate this, we must utilize the divergence theorem as stated before, and we must evaluate. For right now, let
Thus, we can evaluate a contour integral of the fourth dimension.

Integral representation

An integral representation of a function is an expression of the function involving a contour integral. Various integral representations are known for many special functions. Integral representations can be important for theoretical reasons, e.g. giving analytic continuation or functional equations, or sometimes for numerical evaluations.
For example, the original definition of the Riemann zeta function via a Dirichlet series,
is valid only for. But
where the integration is done over the Hankel contour, is valid for all complex s not equal to 1.