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In the mathematical field ofcomplex analysis,contour integration is a method of evaluating certainintegrals along paths in thecomplex plane.^[1][2][3]

Contour integration is closely related to thecalculus of residues,^[4] a method ofcomplex 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. It also has various applications in physics.^[5]

Contour integration methods include:

• direct integration of acomplex-valued function along a curve in the complex plane
• application of theCauchy integral formula
• application of theresidue 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.

Incomplex analysis, acontour is a type of curve in thecomplex plane. In contour integration, contours provide a precise definition of thecurves on which an integral may be suitably defined. Acurve in the complex plane is defined as acontinuous function from aclosed interval of thereal line to the complex plane:${\displaystyle z:[a,b]\to \mathbb{C}}$.

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.^[6]

Contours are often defined in terms of directed smooth curves.^[6] These provide a precise definition of a "piece" of a smooth curve, of which a contour is made.

Asmooth curve is a curve${\displaystyle z:[a,b]\to \mathbb{C}}$ with a non-vanishing, continuous derivative such that each point is traversed only once (z is one-to-one), with the possible exception of a curve such that the endpoints match (${\displaystyle z(a)=z(b)}$). 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 (${\displaystyle z^{\prime }(a)=z^{\prime }(b)}$). A smooth curve that is not closed is often referred to as a smooth arc.^[6]

Theparametrization of a curve provides a natural ordering of points on the curve:${\displaystyle z(x)}$ comes before${\displaystyle z(y)}$ if. This leads to the notion of adirected smooth curve. It is most useful to consider curves independent of the specific parametrization. This can be done by consideringequivalence classes of smooth curves with the same direction. Adirected 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 are the class of curves on which we define contour integration. Acontour 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${\displaystyle {\gamma }_1,\dots ,{\gamma }_n}$ be such that the terminal point of${\displaystyle {\gamma }_i}$ coincides with the initial point of${\displaystyle {\gamma }_{i+1}}$ for all${\displaystyle i}$ such that . This includes all directed smooth curves. Also, a single point in the complex plane is considered a contour. The symbol${\displaystyle +}$ is often used to denote the piecing of curves together to form a new curve. Thus we could write a contour${\displaystyle \mathrm{\Gamma }}$ that is made up of${\displaystyle n}$ curves as$${\displaystyle \mathrm{\Gamma }={\gamma }_1+{\gamma }_2+\cdots +{\gamma }_n.}$$

Thecontour integral of acomplex function${\displaystyle f:\mathbb{C}\to \mathbb{C}}$ is a generalization of the integral for real-valued functions. Forcontinuous functions in thecomplex plane, the contour integral can be defined in analogy to theline 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 thepartition of an interval and theRiemann 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.

To define the contour integral in this way one must first consider the integral, over a real variable, of a complex-valued function. Let${\displaystyle f:\mathbb{R}\to \mathbb{C}}$ be a complex-valued function of a real variable,${\displaystyle t}$. The real and imaginary parts of${\displaystyle f}$ are often denoted as${\displaystyle u(t)}$ and${\displaystyle v(t)}$, respectively, so that$${\displaystyle f(t)=u(t)+iv(t).}$$Then the integral of the complex-valued function${\displaystyle f}$ over the interval${\displaystyle [a,b]}$ is given by

Now, to define the contour integral, let${\displaystyle f:\mathbb{C}\to \mathbb{C}}$ be acontinuous function on thedirected smooth curve${\displaystyle \gamma }$. Let${\displaystyle z:[a,b]\to \mathbb{C}}$ be any parametrization of${\displaystyle \gamma }$ that is consistent with its order (direction). Then the integral along${\displaystyle \gamma }$ is denoted$${\displaystyle {\int }_{\gamma }f(z)dz}$$and is given by^[6]$${\displaystyle {\int }_{\gamma }f(z)dz:={\int }_a^{b}f(z(t))z^{\prime }(t)dt.}$$

This definition is well defined. That is, the result is independent of the parametrization chosen.^[6] In the case where the real integral on the right side does not exist the integral along${\displaystyle \gamma }$ is said not to exist.

The generalization of theRiemann 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${\displaystyle \gamma }$ is defined as a finite, ordered set of points on${\displaystyle \gamma }$. 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 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

  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.

A fundamental result in complex analysis is that the contour integral of⁠1/z⁠ is2πi, where the path of the contour is taken to be theunit circle traversed counterclockwise (or any positively orientedJordan curve about 0). In the case of the unit circle there is a direct method to evaluate the integral$${\displaystyle {\oint }_C\frac{1}{z}dz.}$$

In evaluating this integral, use the unit circle|z| = 1 as a contour, parametrized byz(t) =e^it, witht ∈ [0, 2π], then⁠dz/dt⁠ =ie^it and$${\displaystyle {\oint }_C\frac{1}{z}dz={\int }_0^{2\pi }\frac{1}{e^{it}}i{e}^{it}dt=i{\int }_0^{2\pi }1dt=it{|}_0^{2\pi }=\left(2\pi -0\right)i=2\pi i}$$

which is the value of the integral. This result only applies to the case in which z is raised to power of -1. If the power is not equal to -1, then the result will always be zero.

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 theCauchy integral formula orresidue 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 theCauchy integral formula orresidue theorem is possible

• application ofCauchy's integral theorem

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

• application of theCauchy integral formula orresidue theorem

  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

  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 itR), and the integral that crosses the complex plane (call itI). 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 integralI can be shown to be zero, or if the real-valued integral that is sought is improper, then if we demonstrate that the integralI as described above tends to 0, the integral alongR will tend to the integral around the contourR +I.

• conclusion

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

Consider the integral$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{\left(x^2+1\right)}^2}dx,}$$

To evaluate this integral, we look at the complex-valued function$${\displaystyle f(z)=\frac{1}{{\left(z^2+1\right)}^2}}$$

which hassingularities ati and−i. We choose a contour that will enclose the real-valued integral, here a semicircle with boundary diameter on the real line (going from, say,−a toa) will be convenient. Call this contourC.

There are two ways of proceeding, using theCauchy integral formula or by the method of residues:

Note that:$${\displaystyle {\oint }_{C}f(z)dz={\int }_{-a}^{a}f(z)dz+{\int }_{\text{Arc}}f(z)dz}$$thus$${\displaystyle {\int }_{-a}^{a}f(z)dz={\oint }_{C}f(z)dz-{\int }_{\text{Arc}}f(z)dz}$$

Furthermore, observe that$${\displaystyle f(z)=\frac{1}{{\left(z^2+1\right)}^2}=\frac{1}{(z+i{)}^2(z-i{)}^2}.}$$

Since the only singularity in the contour is the one at i, then we can write$${\displaystyle f(z)=\frac{\frac{1}{(z+i{)}^2}}{(z-i{)}^2},}$$

which puts the function in the form for direct application of the formula. Then, by using Cauchy's integral formula,$${\displaystyle {\oint }_{C}f(z)dz={\oint }_C\frac{\frac{1}{(z+i{)}^2}}{(z-i{)}^2}dz=2\pi i{\frac{d}{dz}\frac{1}{(z+i{)}^2}|}_{z=i}=2\pi i{\left[\frac{-2}{(z+i{)}^3}\right]}_{z=i}=\frac{\pi }{2}}$$

We take the first derivative, in the above steps, because the pole is a second-order pole. That is,(z −i) is taken to the second power, so we employ the first derivative off(z). If it were(z −i) taken to the third power, we would use the second derivative and divide by2!, etc. The case of(z −i) to the first power corresponds to a zero order derivative—justf(z) itself.

We need to show that the integral over the arc of the semicircle tends to zero asa → ∞, using theestimation lemma$${\displaystyle \left|{\int }_{\text{Arc}}f(z)dz\right|\le ML}$$

whereM is an upper bound on|f(z)| along the arc andL the length of the arc. Now,$${\displaystyle \left|{\int }_{\text{Arc}}f(z)dz\right|\le \frac{a\pi }{{\left(a^2-1\right)}^2}\to 0\text{ as }a\to \mathrm{\infty }.}$$So$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{\left(x^2+1\right)}^2}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(z)dz=\underset{a\to +\mathrm{\infty }}{lim}{\int }_{-a}^{a}f(z)dz=\frac{\pi }{2}.◻}$$

Consider theLaurent series off(z) abouti, the only singularity we need to consider. We then have$${\displaystyle f(z)=\frac{-1}{4(z-i{)}^2}+\frac{-i}{4(z-i)}+\frac{3}{16}+\frac{i}{8}(z-i)+\frac{-5}{64}(z-i{)}^2+\cdots }$$

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

It is clear by inspection that the residue is−⁠i/4⁠, so, by theresidue theorem, we have$${\displaystyle {\oint }_{C}f(z)dz={\oint }_C\frac{1}{{\left(z^2+1\right)}^2}dz=2\pi i{\mathrm{Res}}_{z=i}f(z)=2\pi i\left(-\frac{i}{4}\right)=\frac{\pi }{2}◻}$$

Thus we get the same result as before.

As an aside, a question can arise whether we do not take the semicircle to include theother singularity, enclosing−i. 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.

The integral$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{itx}}{x^2+1}dx}$$

(which arises inprobability theory as a scalar multiple of thecharacteristic function of theCauchy distribution) resists the techniques of elementarycalculus. We will evaluate it by expressing it as a limit of contour integrals along the contourC that goes along thereal line from−a toa and then counterclockwise along a semicircle centered at 0 froma to−a. Takea to be greater than 1, so that theimaginary uniti is enclosed within the curve. The contour integral is$${\displaystyle {\int }_C\frac{e^{itz}}{z^2+1}dz.}$$

Sincee^itz is anentire function (having nosingularities at any point in the complex plane), this function has singularities only where the denominatorz² + 1 is zero. Sincez² + 1 = (z +i)(z −i), that happens only wherez =i orz = −i. Only one of those points is in the region bounded by this contour. Theresidue off(z) atz =i is$${\displaystyle \underset{z\to i}{lim}(z-i)f(z)=\underset{z\to i}{lim}(z-i)\frac{e^{itz}}{z^2+1}=\underset{z\to i}{lim}(z-i)\frac{e^{itz}}{(z-i)(z+i)}=\underset{z\to i}{lim}\frac{e^{itz}}{z+i}=\frac{e^{-t}}{2i}.}$$

According to theresidue theorem, then, we have$${\displaystyle {\int }_{C}f(z)dz=2\pi i{\mathrm{Res}}_{z=i}f(z)=2\pi i\frac{e^{-t}}{2i}=\pi e^{-t}.}$$

The contourC may be split into a "straight" part and a curved arc, so that$${\displaystyle {\int }_{\text{straight}}+{\int }_{\text{arc}}=\pi e^{-t},}$$and thus$${\displaystyle {\int }_{-a}^a=\pi e^{-t}-{\int }_{\text{arc}}.}$$

According toJordan's lemma,ift > 0 then$${\displaystyle {\int }_{\text{arc}}\frac{e^{itz}}{z^2+1}dz\to 0\text{ as }a\to \mathrm{\infty }.}$$

Therefore,ift > 0 then$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{itx}}{x^2+1}dx=\pi e^{-t}.}$$

A similar argument with an arc that winds around−i rather thani shows thatift < 0 then$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{itx}}{x^2+1}dx=\pi e^t,}$$and finally we have this:$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{itx}}{x^2+1}dx=\pi e^{-|t|}.}$$

(Ift = 0 then the integral yields immediately to real-valued calculus methods and its value isπ.)

Certain substitutions can be made to integrals involvingtrigonometric 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$${\displaystyle {\int }_{-\pi }^{\pi }\frac{1}{1+3(\cos t{)}^2}dt.}$$

We seek to make a substitution ofz =e^it. Now, recall$${\displaystyle \cos t=\frac{1}{2}\left(e^{it}+e^{-it}\right)=\frac{1}{2}\left(z+\frac{1}{z}\right)}$$and$${\displaystyle \frac{dz}{dt}=iz,dt=\frac{dz}{iz}.}$$

TakingC to be the unit circle, we substitute to get:

The singularities to be considered are at${\displaystyle \frac{\pm i}{\sqrt{3}}.}$ LetC₁ be a small circle about${\displaystyle \frac{i}{\sqrt{3}},}$ andC₂ be a small circle about${\displaystyle \frac{-i}{\sqrt{3}}.}$ Then we arrive at the following:

The above method may be applied to all integrals of the type$${\displaystyle {\int }_0^{2\pi }\frac{P(\sin (t),\sin (2t),\dots ,\cos (t),\cos (2t),\dots )}{Q(\sin (t),\sin (2t),\dots ,\cos (t),\cos (2t),\dots )}dt}$$

whereP andQ are polynomials, i.e. a rational function in trigonometric terms is being integrated. Note that the bounds of integration may as well beπ and −π, as in the previous example, or any other pair of endpoints 2π apart.

The trick is to use the substitutionz =e^it wheredz =ie^it dt and hence$${\displaystyle \frac{1}{iz}dz=dt.}$$

This substitution maps the interval[0, 2π] to the unit circle. Furthermore,$${\displaystyle \sin (kt)=\frac{e^{ikt}-e^{-ikt}}{2i}=\frac{z^k-z^{-k}}{2i}}$$and$${\displaystyle \cos (kt)=\frac{e^{ikt}+e^{-ikt}}{2}=\frac{z^k+z^{-k}}{2}}$$so that a rational functionf(z) inz results from the substitution, and the integral becomes$${\displaystyle {\oint }_{|z|=1}f(z)\frac{1}{iz}dz}$$which is in turn computed by summing the residues off(z)⁠1/iz⁠ inside the unit circle.

The image at right illustrates this for$${\displaystyle I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+(\sin t{)}^2}dt,}$$which we now compute. The first step is to recognize that$${\displaystyle I=\frac{1}{4}{\int }_0^{2\pi }\frac{1}{1+(\sin t{)}^2}dt.}$$

The substitution yields$${\displaystyle \frac{1}{4}{\oint }_{|z|=1}\frac{4iz}{z^4-6z^2+1}dz={\oint }_{|z|=1}\frac{iz}{z^4-6z^2+1}dz.}$$

The poles of this function are at1 ±√2 and−1 ±√2. Of these,1 +√2 and−1 −√2 are outside the unit circle (shown in red, not to scale), whereas1 −√2 and−1 +√2 are inside the unit circle (shown in blue). The corresponding residues are both equal to−⁠i√2/16⁠, so that the value of the integral is$${\displaystyle I=2\pi i2\left(-\frac{\sqrt{2}}{16}i\right)=\pi \frac{\sqrt{2}}{4}.}$$

Consider the real integral$${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{x^2+6x+8}dx.}$$

We can begin by formulating the complex integral$${\displaystyle {\int }_C\frac{\sqrt{z}}{z^2+6z+8}dz=I.}$$

We can use the Cauchy integral formula or residue theorem again to obtain the relevant residues. However, the important thing to note is thatz^1/2 =e^(Logz)/2, soz^1/2 has abranch cut. This affects our choice of the contourC. 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-calledkeyhole contour, which consists of a small circle about the origin of radiusε 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 thatz = −2 andz = −4 are inside the big circle. These are the two remaining poles, derivable by factoring the denominator of the integrand. The branch point atz = 0 was avoided by detouring around the origin.

Letγ be the small circle of radiusε,Γ the larger, with radiusR, then$${\displaystyle {\int }_C={\int }_{\epsilon }^R+{\int }_{\mathrm{\Gamma }}+{\int }_R^{\epsilon }+{\int }_{\gamma }.}$$

It can be shown that the integrals overΓ andγ both tend to zero asε → 0 andR → ∞, by an estimation argument above, that leaves two terms. Now sincez^1/2 =e^(Logz)/2, on the contour outside the branch cut, we have gained 2π in argument alongγ. (ByEuler's identity,e^iπ represents theunit vector, which therefore hasπ as its log. Thisπ is what is meant by the argument ofz. The coefficient of⁠1/2⁠ forces us to use 2π.) So

Therefore:$${\displaystyle {\int }_C\frac{\sqrt{z}}{z^2+6z+8}dz=2{\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{x^2+6x+8}dx.}$$

By using the residue theorem or the Cauchy integral formula (first employing the partial fractions method to derive a sum of two simple contour integrals) one obtains$${\displaystyle \pi i\left(\frac{i}{\sqrt{2}}-i\right)={\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{x^2+6x+8}dx=\pi \left(1-\frac{1}{\sqrt{2}}\right).◻}$$

This section treats a type of integral of which$${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\log x}{{\left(1+x^2\right)}^2}dx}$$is an example.

To calculate this integral, one uses the function$${\displaystyle f(z)={\left(\frac{\log z}{1+z^2}\right)}^2}$$and the branch of the logarithm corresponding to−π < argz ≤ π.

We will calculate the integral off(z) 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

LetR be the radius of the large circle, andr the radius of the small one. We will denote the upper line byM, and the lower line byN. As before we take the limit whenR → ∞ andr → 0. The contributions from the two circles vanish. For example, one has the following upper bound with theML lemma:$${\displaystyle \left|{\int }_{R}f(z)dz\right|\le 2\pi R\frac{(\log R{)}^2+{\pi }^2}{{\left(R^2-1\right)}^2}\to 0.}$$

In order to compute the contributions ofM andN we setz = −x +iε onM andz = −x −iε onN, with0 <x < ∞:

which gives$${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\log x}{{\left(1+x^2\right)}^2}dx=-\frac{\pi }{4}.}$$

We seek to evaluate$${\displaystyle I={\int }_0^3\frac{x^{\frac{3}{4}}(3-x{)}^{\frac{1}{4}}}{5-x}dx.}$$

This requires a close study of$${\displaystyle f(z)=z^{\frac{3}{4}}(3-z{)}^{\frac{1}{4}}.}$$

We will constructf(z) so that it has a branch cut on[0, 3], shown in red in the diagram. To do this, we choose two branches of the logarithm, settingand

The cut ofz^3⁄4 is therefore(−∞, 0] and the cut of(3 −z)^1/4 is(−∞, 3]. It is easy to see that the cut of the product of the two, i.e.f(z), is[0, 3], becausef(z) is actually continuous across(−∞, 0). This is because whenz = −r < 0 and we approach the cut from above,f(z) has the value$${\displaystyle r^{\frac{3}{4}}{e}^{\frac{3}{4}\pi i}(3+r{)}^{\frac{1}{4}}{e}^{\frac{2}{4}\pi i}=r^{\frac{3}{4}}(3+r{)}^{\frac{1}{4}}{e}^{\frac{5}{4}\pi i}.}$$

When we approach from below,f(z) has the value$${\displaystyle r^{\frac{3}{4}}{e}^{-\frac{3}{4}\pi i}(3+r{)}^{\frac{1}{4}}{e}^{\frac{0}{4}\pi i}=r^{\frac{3}{4}}(3+r{)}^{\frac{1}{4}}{e}^{-\frac{3}{4}\pi i}.}$$

But$${\displaystyle e^{-\frac{3}{4}\pi i}=e^{\frac{5}{4}\pi i},}$$

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 inz^3⁄4 and(3 −z)^1/4.

We will use the contour shown in green in the diagram. To do this we must compute the value off(z) along the line segments just above and just below the cut.

Letz =r (in the limit, i.e. as the two green circles shrink to radius zero), where0 ≤r ≤ 3. Along the upper segment, we find thatf(z) has the value$${\displaystyle r^{\frac{3}{4}}{e}^{\frac{0}{4}\pi i}(3-r{)}^{\frac{1}{4}}{e}^{\frac{2}{4}\pi i}=i{r}^{\frac{3}{4}}(3-r{)}^{\frac{1}{4}}}$$and along the lower segment,$${\displaystyle r^{\frac{3}{4}}{e}^{\frac{0}{4}\pi i}(3-r{)}^{\frac{1}{4}}{e}^{\frac{0}{4}\pi i}=r^{\frac{3}{4}}(3-r{)}^{\frac{1}{4}}.}$$

It follows that the integral of⁠f(z)/5 −z⁠ along the upper segment is−iI in the limit, and along the lower segment,I.

If we can show that the integrals along the two green circles vanish in the limit, then we also have the value ofI, by theCauchy residue theorem. Let the radius of the green circles beρ, whereρ < 0.001 andρ → 0, and apply theML inequality. For the circleC_L on the left, we find$${\displaystyle \left|{\int }_{C_{\mathrm{L}}}\frac{f(z)}{5-z}dz\right|\le 2\pi \rho \frac{{\rho }^{\frac{3}{4}}{3.001}^{\frac{1}{4}}}{4.999}\in \mathcal{O}\left({\rho }^{\frac{7}{4}}\right)\to 0.}$$

Similarly, for the circleC_R on the right, we have$${\displaystyle \left|{\int }_{C_{\mathrm{R}}}\frac{f(z)}{5-z}dz\right|\le 2\pi \rho \frac{{3.001}^{\frac{3}{4}}{\rho }^{\frac{1}{4}}}{1.999}\in \mathcal{O}\left({\rho }^{\frac{5}{4}}\right)\to 0.}$$

Now using theCauchy residue theorem, we have$${\displaystyle (-i+1)I=-2\pi i\left({\mathrm{Res}}_{z=5}\frac{f(z)}{5-z}+{\mathrm{Res}}_{z=\mathrm{\infty }}\frac{f(z)}{5-z}\right).}$$where the minus sign is due to the clockwise direction around the residues. Using the branch of the logarithm from before, clearly$${\displaystyle {\mathrm{Res}}_{z=5}\frac{f(z)}{5-z}=-5^{\frac{3}{4}}{e}^{\frac{1}{4}\log (-2)}.}$$

The pole is shown in blue in the diagram. The value simplifies to$${\displaystyle -5^{\frac{3}{4}}{e}^{\frac{1}{4}(\log 2+\pi i)}=-e^{\frac{1}{4}\pi i}{5}^{\frac{3}{4}}{2}^{\frac{1}{4}}.}$$

We use the following formula for the residue at infinity:$${\displaystyle {\mathrm{Res}}_{z=\mathrm{\infty }}h(z)={\mathrm{Res}}_{z=0}\left(-\frac{1}{z^2}h\left(\frac{1}{z}\right)\right).}$$

Substituting, we find$${\displaystyle \frac{1}{5-\frac{1}{z}}=-z\left(1+5z+5^2{z}^2+5^3{z}^3+\cdots \right)}$$and$${\displaystyle {\left(\frac{1}{z^3}\left(3-\frac{1}{z}\right)\right)}^{\frac{1}{4}}=\frac{1}{z}(3z-1{)}^{\frac{1}{4}}=\frac{1}{z}{e}^{\frac{1}{4}\pi i}(1-3z{)}^{\frac{1}{4}},}$$where we have used the fact that−1 =e^πi for the second branch of the logarithm. Next we apply the binomial expansion, obtaining$${\displaystyle \frac{1}{z}{e}^{\frac{1}{4}\pi i}\left(1-(\genfrac{}{}{0ex}{}{1/4}{1})3z+(\genfrac{}{}{0ex}{}{1/4}{2})3^2{z}^2-(\genfrac{}{}{0ex}{}{1/4}{3})3^3{z}^3+\cdots \right).}$$

The conclusion is that$${\displaystyle {\mathrm{Res}}_{z=\mathrm{\infty }}\frac{f(z)}{5-z}=e^{\frac{1}{4}\pi i}\left(5-\frac{3}{4}\right)=e^{\frac{1}{4}\pi i}\frac{17}{4}.}$$

Finally, it follows that the value ofI is$${\displaystyle I=2\pi i\frac{e^{\frac{1}{4}\pi i}}{-1+i}\left(\frac{17}{4}-5^{\frac{3}{4}}{2}^{\frac{1}{4}}\right)=2\pi 2^{-\frac{1}{2}}\left(\frac{17}{4}-5^{\frac{3}{4}}{2}^{\frac{1}{4}}\right)}$$which yields$${\displaystyle I=\frac{\pi }{2\sqrt{2}}\left(17-5^{\frac{3}{4}}{2}^{\frac{9}{4}}\right)=\frac{\pi }{2\sqrt{2}}\left(17-{40}^{\frac{3}{4}}\right).}$$

Using theresidue theorem, we can evaluate closed contour integrals. The following are examples on evaluating contour integrals with the residue theorem.

Using the residue theorem, let us evaluate this contour integral.$${\displaystyle {\oint }_C\frac{e^z}{z^3}dz}$$

Recall that the residue theorem states$${\displaystyle {\oint }_{C}f(z)dz=2\pi i\cdot \sum \mathrm{Res}(f,a_k)}$$

where${\displaystyle \mathrm{Res}}$ is the residue of${\displaystyle f(z)}$, and the${\displaystyle a_k}$ are the singularities of${\displaystyle f(z)}$ lying inside the contour${\displaystyle C}$ (with none of them lying directly on${\displaystyle C}$).

${\displaystyle f(z)}$ has only one pole,${\displaystyle 0}$. From that, we determine that theresidue of${\displaystyle f(z)}$ to be${\displaystyle \frac{1}{2}}$

Thus, using theresidue theorem, we can determine:$${\displaystyle {\oint }_C\frac{e^z}{z^3}dz=\pi i.}$$

To solve multivariable contour integrals (i.e.surface integrals, complexvolume integrals, and higher orderintegrals), we must use thedivergence theorem. For now, let${\displaystyle \mathrm{\nabla }\cdot }$ be interchangeable with${\displaystyle \mathrm{div}}$. These will both serve as the divergence of thevector field denoted as${\displaystyle \mathbf{F}}$. This theorem states:$${\displaystyle \underset{n}{\underbrace{\int \cdots {\int }_U}}\mathrm{div}(\mathbf{F})dV=\underset{n-1}{\underbrace{\oint \cdots {\oint }_{\mathrm{\partial }U}}}\mathbf{F}\cdot \mathbf{n}dS}$$

In addition, we also need to evaluate${\displaystyle \mathrm{\nabla }\cdot \mathbf{F}}$ where${\displaystyle \mathrm{\nabla }\cdot \mathbf{F}}$ is an alternate notation of${\displaystyle \mathrm{div}(\mathbf{F})}$. Thedivergence of any dimension can be described as

Let thevector field${\displaystyle \mathbf{F}=\sin (2x){\mathbf{e}}_x+\sin (2y){\mathbf{e}}_y+\sin (2z){\mathbf{e}}_z}$ and be bounded by the following$${\displaystyle 0\le x\le 10\le y\le 3-1\le z\le 4}$$