Methods of contour integration

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In complex analysis, the evaluation of integrals of real-valued functions along intervals on the real line, is not readily found with certain integrands and methods involving only real variables. Complex analysis methods described below give means of calculating these real-valued integrals by means of contour integrals in the complex plane.

These methods include

• direct integration of a complex-valued function along a curve in the complex plane (a contour)
• application of the Cauchy integral formula
• application of the residue theorem

One method can be used, or a combination of these methods, or various limiting processes, for the purpose of finding these integrals or sums.

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:

• parametrizing the contour

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

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

direct evaluation

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

Example

A fundamental result in complex analysis is that the integral around the contour C which is the unit circle (or any Jordan curve about 0) of z⁻¹ is 2πi. Let us evaluate the integral

[\oint_C \,dz.]

In evaluating this integral, we use the unit circle |z| = 1 as our contour, which we can parametrize by γ(t) = e^it, with t ∈ [0, 2π]. Observe that γ'(t) = ie^it. Now, substituting this for z, we have

[\oint_C \,dz = \int_0^ } ie^\,dt = i\int_0^ e^e^\,dt = i\int_0^ 1 \,dt]

[= \left.t\right]_0^ i=(2\pi-0)i = 2\pi i]

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:

• a specific contour is chosen:

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 the Cauchy-Goursat theorem

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

application of the Cauchy integral formula or residue theorem

Application of these integral formula 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

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 R), and the integral that crosses the complex plane (call it I). 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

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

conclusion

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

Example (I)

Consider

[\int_^ \,dx.]

To evaluate this integral, we look at the complex-valued function

[f(z)=]

which has singularities at i and −i. However, we will want to choose a contour that will enclose the real-valued integral, so we choose a semicircle as the one to the left, which we will let expand as to contain the whole real axis (a will tend to infinity). Call this contour C.

Now, there are two ways of proceeding, using the Cauchy integral formula or by the method of residues.

Using the Cauchy integral formula

Observe that

[f(z)==.]

Since the only singularity in the contour is the one at i, then we can write

[f(z)=+\int_}=\pi e^,]

and thus

[\int_^a =\pi e^-\int_}.]

It can be shown that if t > 0 then

[\int_} \over z^2+1}\,dz\rightarrow 0\ \mbox\ a\rightarrow\infty.]

Therefore if t > 0 then

[\int_^\infty \over z^2+1}\,dz=\pi e^.]

A similar argument with an arc that winds around −i rather than i shows that if t < 0 then

[\int_^\infty \over z^2+1}\,dz=\pi e^t,]

and finally we have

[\int_^\infty \over z^2+1}\,dz=\pi e^.\quad\square]

(If t = 0 then the integral yields immediately to real-valued calculus methods and its value is π.)

Example (III) – 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

[ \int_^ )^2} \,dt. ]

We seek to make a substitution of z = e^it.

Now, recall

[ \cos = \left(e^+e^\right) = \left(z+\right)]

and

[ = iz,\ dt = . ]

Taking C to be the unit circle, we substitute to get:

[ \oint_C (z+))^2} \,]

[ = \oint_C (z+)^2} \,dz = \oint_C z(z+)^2}\,dz = -i \oint_C z(z^2+2+)} \,dz]

[ = -i \oint_C (z^3+2z+)} \,dz = -i \oint_C z^3+z+} \,dz]

[ = -i \oint_C }\,dz = -4i \oint_C }\,dz ]

[ = -4i \oint_C \,dz. ]

We use the Cauchy integral formula. Factorize the denominator:

[ = -4i \oint_C \,dz = -4i \oint_C \,dz ]

[ = -4i \oint_C i)\left(z-\sqrti\right)\left(z+\sqrti\right)\left(z-\sqrti\right)}\,dz ]

[ = -i \oint_C i)(z-\sqrti)\left(z+\sqrti\right)\left(z-\sqrti\right)}\,dz. ]

The singularities then to be considered are at 3^−1/2i, −3^−1/2i. We can now reduce the integral:

[ = -i \oint_ i)(z-\sqrti)\left(z+\sqrti\right)}\, \over \left(z-\sqrti\right)}\,dz + -i \oint_ i)(z-\sqrti)\left(z-\sqrti\right)}\, \over \left(z+\sqrti\right)} ]

where C₁ is a small circle about 3^−1/2i, and C₂ is a small circle about −3^−1/2i. We can now apply the formula:

[ = -i \left( 2\pi i \left.\left(i)(z-\sqrti)(z+\sqrti)}\right)\right|_i} \right.]

:[ \left. + 2\pi i \left.\left(i)(z-\sqrti)(z-\sqrti)}\right)\right|_i} \right)]

[ = -i \left( 2\pi i \left( i \over (\sqrti+\sqrti)(\sqrti-\sqrti)(\sqrti+\sqrti)} \right) \right.]

: [\left. +2\pi i \left( i \over (-\sqrti+\sqrti)(-\sqrti-\sqrti)(-\sqrti-\sqrti } \right)\right)]

[ = -i \left( 2\pi i \left( i \over (}i)(-}i)(}i)} \right) +2 \pi i \left( i \over (}i)(-}i)(-}i)} \right) \right)]

[ = -i \left( 2\pi i \left( i \over i(})(})(})} \right) +2\pi i \left( i \over -i(})(})(})} \right) \right) ]

[ = -i \left( 2\pi i \left( \over (})(})(})} \right) +2\pi i \left( \over (})(})(})} \right) \right) ]

[ = -i \left( 2\pi i \left( \,\over } } \right) +2\pi i \left( \,\over } } \right) \right) ]

[ = -i \left(2 \pi i \left(\right) + 2 \pi i \left(\right)\right) = -i \left(\pi i \left(+\right)\right) = \left(\right)\pi = \pi.\quad\square]

Example (IV) – branch cuts

Consider

[\int_0^\infty \over x^2+6x+8}\,dx.]

We can begin by formulating the complex integral

[\int_C \over z^2+6z+8}\,dz=I.]

We can use the Cauchy integral formula or residue theorem again to obtain the relevant residues. However, the important thing to note is that z^1/2=e^{1/2 Log(z)}, so z^1/2 has a branch cut. This affects our choice of the contour C. Normally the logarithm branch cut is defined as the negative real axis, however, this makes the calculation of the integral slightly more complex, 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 ε 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.

Let γ be the small circle of radius ε, Γ the larger, with radius r, then

[\int_C \over z^2+6z+8}\,dz=\int_\epsilon^R \over z^2+6z+8}\,dz+\int_\Gamma \over z^2+6z+8}\,dz+\int_R^\epsilon \over z^2+6z+8}\,dz+\int_\gamma \over z^2+6z+8}\,dz.]

Since z^1/2=e^{1/2 Log(z)}, along the contour below the branch cut, we have gained 2π in argument along Γ, so

[\int_R^\epsilon \over z^2+6z+8}\,dz=\int_R^\epsilon \over z^2+6z+8}\,dz = \pi\left(1-}\right).\quad\square]

See also

• Cauchy integral formula for an example of the application of this integral theorem
• Residue (complex analysis)

External links

• [A collection of examples]

 

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