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Wikipedia / 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 used to study complex-valued functions that are holomorphic in a region.

Contour integration is closely related to the calculus of residues, a method of complex analysis. The power of contour integration comes from the fact that the integrals of holmorphic functions are invariant under deforming the contour, provided the deformation does not cross a singularity or branch cut. Thus the value of a contour integral between fixed endpoints is not governed by the precise shape of the contour, but by its winding around the singularities of the integrand.

One use of contour integrals is the evaluation of certain integrals of functions over the real line. Regarding the real line as a contour, and deforming it into the complex plane often leads to simpler integrals than those which can be found by using only real variable methods. In modern language, the integral of a holomorphic or meromorphic function is a pairing between a cohomology class of differential forms and a homology class of cycles in the domain of the function. It also has various applications to physics.

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.

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 (non-vanishing) 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 stop only 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 with a non-vanishing, continuous derivative such that each point is traversed only once (). A smooth curve that is not closed is often referred to as a smooth arc.

The parametrization. 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 (according to the parametrization). 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

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

be a continuous function on the directed smooth curve and is given by

This definition is well defined. That is, the result is independent of the parametrization chosen. is said not to exist.

As a generalization of the Riemann integral

The generalization of the Riemann integral. 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 (in the two-dimensional complex plane), also known as the mesh, goes to zero.

Direct methods

Direct methods involve the calculation of the integral through methods similar to those in calculating line integrals in multivariate calculus. This means that we use the following method:

  • parametrizing the contour

    1. The contour is parametrized by a differentiable complex-valued function of real variables, or the contour is broken up into pieces and parametrized separately.

    substitution of the parametrization into the integrand

    1. Substituting the parametrization into the integrand transforms the integral into an integral of one real variable.

    direct evaluation

    1. The integral is evaluated in a method akin to a real-variable integral.

Example

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 (or any positively oriented Jordan curve about

In evaluating this integral, use the unit circle as a contour, parametrized by , with , then which is the value of the integral. This result only applies to the case in which is raised to power of . If the power is not equal to , then the result will always be zero.

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:

  • a specific contour is chosen:

    1. The contour is chosen so that the contour follows the part of the complex plane that describes the real-valued integral, and also encloses singularities of the integrand so application of the Cauchy integral formula or residue theorem is possible

    application of Cauchy's integral theorem

    1. The integral is reduced to only an integration around a small circle about each pole.

    application of the Cauchy integral formula or residue theorem

    1. Application of these integral formulae gives us a value for the integral around the whole of the contour.

    division of the contour into a contour along the real part and imaginary part

    1. The whole of the contour can be divided into the contour that follows the part of the complex plane that describes the real-valued integral as chosen before (call it ), and the integral that crosses the complex plane (call it ). The integral over the whole of the contour is the sum of the integral over each of these contours.

    demonstration that the integral that crosses the complex plane plays no part in the sum

    1. If the integral can be shown to be zero, or if the real-valued integral that is sought is improper, then if we demonstrate that the integral as described above tends to 0, the integral along will tend to the integral around the contour .

    conclusion

    1. If we can show the above step, then we can directly calculate , the real-valued integral.

Example 1

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 (going from, say, to ) 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

Since the only singularity in the contour is the one at

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 , 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

Using the method of residues

Consider the Laurent series of about

(See the sample Laurent calculation from Laurent series for the derivation of this series.)

It is clear by inspection that the residue is , so, by the residue theorem

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

(which arises in probability theory as a scalar multiple of the characteristic function of the Cauchy distribution) 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 , so that the imaginary unit

Since is an entire function (having no singularities at any point in the complex plane), 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

According to the residue theorem

The contour

According to Jordan's lemma, if then

Therefore, if then

A similar argument with an arc that winds around rather than shows that if then

(If then the integral yields immediately to real-valued calculus methods and its value is .)

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.

We seek to make a substitution of

Taking

Let and

Example 3a – trigonometric integrals, the general procedure

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

This substitution maps the interval so that a rational function in which is in turn computed by summing the residues of inside the unit circle.

The poles of this function are at and . Of these, and are outside the unit circle (shown in red, not to scale), whereas and are inside the unit circle (shown in blue). The corresponding residues are both equal to

Example 4 – branch cuts

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

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 . (By Euler's identity, represents the unit vector, which therefore has as its log. This is what is meant by the argument of . The coefficient of forces us to use 2

Example 5 – the square of the logarithm

is an example.

and the branch of the logarithm corresponding to .

We will calculate the integral of

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 ML lemma

In order to compute the contributions of and we set on and on , with

Example 6 – logarithms and the residue at infinity

We will construct so that it has a branch cut on

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,

When we approach from below,

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 (in the limit, i.e. as the two green circles shrink to radius zero), where . Along the upper segment, we find that

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 ML inequality. For the circle

Similarly, for the circle

Now using the Cauchy residue theorem

where we have used the fact that

Finally, it follows that the value of

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.

).

. From that, we determine that the residue

Thus, using the residue theorem

Multivariable contour integrals

To solve multivariable contour integrals (i.e. surface integrals, complex volume integrals, and higher order integrals), we must use the divergence theorem. These will both serve as the divergence of the vector field

. The divergence

Example 1

Let the vector field

The corresponding double contour integral would be set up as such:

Example 2

Let the vector field, and remark that there are 4 parameters in this case. Let this vector field

To evaluate this, we must utilize the divergence theorem

. We can use the same method to evaluate contour integrals for any vector field as well.

Integral representation

In complex analysis, an integral representation expresses a function as a contour integral in the complex plane. Such representations are central to the theory of holomorphic functions and are closely tied to the fundamental theorems of complex integration.

One of the most important examples is Cauchy's integral formula, which provides a way to reconstruct an analytic function from its values on a surrounding contour:

inside the contour are determined by its values along the contour.

Examples

Inverse Laplace transform

The inverse Laplace transform is defined by a complex contour integral known as the Bromwich integral

in terms of its Laplace transform.

Sinc function representation

The following integral gives a representation the sinc function:

Although this is a real integral, methods from contour integration are often used in its derivation or evaluation.

Gamma function

The Gamma function has the following integral representation:

Extensions of this definition involve contour integrals in the complex plane.

Riemann zeta function

The original definition of the Riemann zeta function via a Dirichlet series where the integration is done over the Hankel contour.

Applications

Integral representations are used to evaluate definite integrals, derive function identities, and solve differential equations. They also appear in complex asymptotic analysis, potential theory, and mathematical physics.

See also

References

Further reading