# 8.5: Complex Integration

- Page ID
- 90968

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We have introduced functions of a complex variable. We also established when functions are differentiable as complex functions, or holomorphic. In this chapter we will turn to integration in the complex plane. We will learn how to compute complex path integrals, or contour integrals. We will see that contour integral methods are also useful in the computation of some of the real integrals that we will face when exploring Fourier transforms in the next chapter.

## Complex Path Integrals

In this section we will investigate the computation of complex path integrals. Given two points in the complex plane, connected by a path \(\Gamma\) as shown in Figure \(\PageIndex{1}\), we would like to define the integral of \(f(z)\) along \(\Gamma\),

\[\int_{\Gamma} f(z) d z\nonumber \]

A natural procedure would be to work in real variables, by writing

\[\int_{\Gamma} f(z) d z=\int_{\Gamma}[u(x, y)+i v(x, y)](d x+i d y),\nonumber \]

since \(z=x+i y\) and \(d z=d x+i d y\).

In order to carry out the integration, we then have to find a parametrization of the path and use methods from a multivariate calculus class. Namely, let \(u\) and \(v\) be continuous in domain \(D\), and \(\Gamma\) a piecewise smooth curve in D. Let \((x(t), y(t))\) be a parametrization of \(\Gamma\) for \(t_{0} \leq t \leq t_{1}\) and \(f(z)=\) \(u(x, y)+i v(x, y)\) for \(z=x+i y\). Then

\[\int_{\Gamma} f(z) d z=\int_{t_{0}}^{t_{1}}[u(x(t), y(t))+i v(x(t), y(t))]\left(\frac{d x}{d t}+i \frac{d y}{d t}\right) d t .\label{eq:1} \]

Here we have used

\[d z=d x+i d y=\left(\frac{d x}{d t}+i \frac{d y}{d t}\right) d t .\nonumber \]

Furthermore, a set \(D\) is called a domain if it is both open and connected.

Before continuing, we first define open and connected. A set \(D\) is connected if and only if for all \(z_{1}\), and \(z_{2}\) in \(D\) there exists a piecewise smooth curve connecting \(z_{1}\) to \(z_{2}\) and lying in \(D\). Otherwise it is called disconnected. Examples are shown in Figure \(\PageIndex{2}\).

A set \(D\) is open if and only if for all \(z_{0}\) in \(D\) there exists an open disk \(\left|z-z_{0}\right|<\rho\) in \(D\). In Figure \(\PageIndex{3}\) we show a region with two disks.

For all points on the interior of the region one can find at least one disk contained entirely in the region. The closer one is to the boundary, the smaller the radii of such disks. However, for a point on the boundary, every such disk would contain points inside and outside the disk. Thus, an open set in the complex plane would not contain any of its boundary points.

We now have a prescription for computing path integrals. Let’s see how this works with a couple of examples.

Evaluate \(\int_{C} z^{2} d z\), where \(C=\) the arc of the unit circle in the first quadrant as shown in Figure \(\PageIndex{4}\).

###### Solution

There are two ways we could carry out the parametrization. First, we note that the standard parametrization of the unit circle is

\[(x(\theta), y(\theta))=(\cos \theta, \sin \theta), \quad 0 \leq \theta \leq 2 \pi .\nonumber \]

For a quarter circle in the first quadrant, \(0 \leq \theta \leq \frac{\pi}{2}\), we let \(z=\cos \theta+i \sin \theta\). Therefore, \(d z=(-\sin \theta+i \cos \theta) d \theta\) and the path integral becomes

\[\int_{\mathrm{C}} z^{2} d z=\int_{0}^{\frac{\pi}{2}}(\cos \theta+i \sin \theta)^{2}(-\sin \theta+i \cos \theta) d \theta .\nonumber \]

We can expand the integrand and integrate, having to perform some trigonometric integrations.

\[\int_{0}^{\frac{\pi}{2}}\left[\sin ^{3} \theta-3 \cos ^{2} \theta \sin \theta+i\left(\cos ^{3} \theta-3 \cos \theta \sin ^{2} \theta\right)\right] d \theta \text {. }\nonumber \]

The reader should work out these trigonometric integrations and confirm the result. For example, you can use

\[\left.\sin ^{3} \theta=\sin \theta\left(1-\cos ^{2} \theta\right)\right)\nonumber \]

to write the real part of the integrand as

\[\sin \theta-4 \cos ^{2} \theta \sin \theta \text {. }\nonumber \]

The resulting antiderivative becomes

\[-\cos \theta+\frac{4}{3} \cos ^{3} \theta \text {. }\nonumber \]

The imaginary integrand can be integrated in a similar fashion.

While this integral is doable, there is a simpler procedure. We first note that \(z=e^{i \theta}\) on \(\mathrm{C} . S_{0}, d z=i e^{i \theta} d \theta\). The integration then becomes

\[\begin{align} \int_{C} z^{2} d z &=\int_{0}^{\frac{\pi}{2}}\left(e^{i \theta}\right)^{2} i e^{i \theta} d \theta\nonumber \\ &=i \int_{0}^{\frac{\pi}{2}} e^{3 i \theta} d \theta\nonumber \\ &=\left.\frac{i e^{3 i \theta}}{3 i}\right|_{0} ^{\pi / 2}\nonumber \\ &=-\frac{1+i}{3}\label{eq:2} \end{align} \]

Evaluate \(\int_{\Gamma} z d z\), for the path \(\Gamma=\gamma_{1} \cup \gamma_{2}\) shown in Figure \(\PageIndex{5}\).

###### Solution

In this problem we have a path that is a piecewise smooth curve. We can compute the path integral by computing the values along the two segments of the path and adding the results. Let the two segments be called \(\gamma_1\) and \(\gamma_2\) as shown in Figure \(\PageIndex{5}\) and parametrize each path separately.

Over \(\gamma_{1}\) we note that \(y=0\). Thus, \(z=x\) for \(x \in[0,1]\). It is natural to take \(x\) as the parameter. So, we let \(d z=d x\) to find

\[\int_{\gamma_{1}} z d z=\int_{0}^{1} x d x=\frac{1}{2} .\nonumber \]

For path \(\gamma_{2}\) we have that \(z=1+\) iy for \(y \in[0,1]\) and \(d z=i d y\). Inserting this parametrization into the integral, the integral becomes

\[\int_{\gamma_{2}} z d z=\int_{0}^{1}(1+i y) i d y=i-\frac{1}{2} .\nonumber \]

Combining the results for the paths \(\gamma_{1}\) and \(\gamma_{2}\), we have \(\int_{\Gamma} z d z=\frac{1}{2}+\left(i-\frac{1}{2}\right)=\) \(i\).

Evaluate \(\int_{\gamma_{3}} z d z\), where \(\gamma_{3}\), is the path shown in Figure \(\PageIndex{6}\).

###### Solution

In this case we take a path from \(z=0\) to \(z=1+i\) along a different path than in the last example. Let \(\gamma_{3}=\left\{(x, y) \mid y=x^{2}, x \in[0,1]\right\}=\left\{z \mid z=x+i x^{2}, x \in\right.\) \([0,1]\}\). Then, \(d z=(1+2 i x) d x\).

The integral becomes

\[\begin{align} \int_{\gamma_{3}} z d z &=\int_{0}^{1}\left(x+i x^{2}\right)(1+2 i x) d x\nonumber \\ &=\int_{0}^{1}\left(x+3 i x^{2}-2 x^{3}\right) d x=\nonumber \\ &=\left[\frac{1}{2} x^{2}+i x^{3}-\frac{1}{2} x^{4}\right]_{0}^{1}=i\label{eq:3} \end{align} \]

In the last case we found the same answer as we had obtained in Example \(\PageIndex{3}\). But we should not take this as a general rule for all complex path integrals. In fact, it is not true that integrating over different paths always yields the same results. However, when this is true, then we refer to this property as path independence. In particular, the integral \(\int f(z) d z\) is path independent if

\[\int_{\Gamma_{1}} f(z) d z=\int_{\Gamma_{2}} f(z) d z\nonumber \]

for all paths from \(z_{1}\) to \(z_{2}\) as shown in Figure \(\PageIndex{7}\).

We can show that if \(\int f(z) d z\) is path independent, then the integral of \(f(z)\) over all closed loops is zero,

\[\int_{\text {closed loops }} f(z) d z=0 .\nonumber \]

A common notation for integrating over closed loops is \(\oint_{C} f(z) d z\). But first we have to define what we mean by a closed loop. A simple closed contour is a path satisfying

- The end point is the same as the beginning point. (This makes the loop closed.)
- The are no self-intersections. (This makes the loop simple.)

A loop in the shape of a figure eight is closed, but it is not simple.

Now, consider an integral over the closed loop \(C\) shown in Figure \(\PageIndex{8}\). We pick two points on the loop breaking it into two contours, \(C_{1}\) and \(C_{2}\). Then we make use of the path independence by defining \(C_{2}^{-}\)to be the path along \(C_{2}\) but in the opposite direction. Then,

\[\begin{align} \oint_{C} f(z) d z &=\int_{C_{1}} f(z) d z+\int_{C_{2}} f(z) d z\nonumber \\ &=\int_{C_{1}} f(z) d z-\int_{C_{2}^{-}} f(z) d z\label{eq:4} \end{align} \]

Assuming that the integrals from point 1 to point 2 are path independent, then the integrals over \(C_{1}\) and \(C_{2}^{-}\)are equal. Therefore, we have \(\oint_{C} f(z) d z=0.\)

\(\oint_C f(z)dz=0\) if the integral is path independent.

Consider the integral \(\oint_{C} z d z\) for \(C\) the closed contour shown in Figure \(\PageIndex{6}\) starting at \(z=0\) following path \(\gamma_{1}\), then \(\gamma_{2}\) and returning to \(z=0\). Based on the earlier examples and the fact that going backwards on \(\gamma_{3}\) introduces a negative sign, we have

\[\oint_{C} z d z=\int_{\gamma_{1}} z d z+\int_{\gamma_{2}} z d z-\int_{\gamma_{3}} z d z=\frac{1}{2}+\left(i-\frac{1}{2}\right)-i=0 .\nonumber \]

## Cauchy’s Theorem

Next we want to investigate if we can determine that integrals over simple closed contours vanish without doing all the work of parametrizing the contour. First, we need to establish the direction about which we traverse the contour. We can define the orientation of a curve by referring to the normal of the curve.

Recall that the normal is a perpendicular to the curve. There are two such perpendiculars. The above normal points outward and the other normal points towards the interior of a closed curve. We will define a positively oriented contour as one that is traversed with the outward normal pointing to the right. As one follows loops, the interior would then be on the left.

A curve with parametrization \((x(t), y(t))\) has a normal \(\left(n_{x}, n_{y}\right)=\left(-\frac{d x}{d t}, \frac{d y}{d t}\right) .\)

We now consider \(\oint_{C}(u+i v) d z\) over a simple closed contour. This can be written in terms of two real integrals in the \(x y\)-plane.

\[\begin{align} \oint_{C}(u+i v) d z &=\int_{C}(u+i v)(d x+i d y)\nonumber \\ &=\int_{C} u d x-v d y+i \int_{C} v d x+u d y .\label{eq:5} \end{align} \]

These integrals in the plane can be evaluated using Green’s Theorem in the Plane. Recall this theorem from your last semester of calculus:

Let \(P(x, y)\) and \(Q(x, y)\) be continuously differentiable functions on and inside the simple closed curve \(\mathrm{C}\) as shown in Figure \(\PageIndex{9}\). Denoting the enclosed region S, we have

\[\int_{C} P d x+Q d y=\iint_{S}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\label{eq:6} \]

Green’s Theorem in the Plane is one of the major integral theorems of vector calculus. It was discovered by George Green ( \(1793-1841\) ) and published in 1828, about four years before he entered Cambridge as an undergraduate.

Using Green’s Theorem to rewrite the first integral in \(\eqref{eq:5}\), we have

\[\int_{C} u d x-v d y=\iint_{S}\left(\frac{-\partial v}{\partial x}-\frac{\partial u}{\partial y}\right) d x d y\nonumber \]

If \(u\) and \(v\) satisfy the Cauchy-Riemann equations (8.4.6), then the integrand in the double integral vanishes. Therefore,

\[\int_{C} u d x-v d y=0 .\nonumber \]

In a similar fashion, one can show that

\[\int_{C} v d x+u d y=0 .\nonumber \]

We have thus proven the following theorem:

If \(u\) and \(v\) satisfy the Cauchy-Riemann equations (8.4.6) inside and on the simple closed contour \(C\), then

\[\oint_{C}(u+i v) d z=0 .\label{eq:7} \]

**Corollary **\(\oint_{C} f(z) d z=0\) when \(f\) is differentiable in domain \(D\) with \(C \subset D\).

Either one of these is referred to as **Cauchy’s Theorem**.

Evaluate \(\oint_{|z-1|=3} z^{4} d z\).

###### Solution

Since \(f(z)=z^{4}\) is differentiable inside the circle \(|z-1|=3\), this integral vanishes.

We can use Cauchy’s Theorem to show that we can deform one contour into another, perhaps simpler, contour.

One can deform contours into simpler ones.

If \(f(z)\) is holomorphic between two simple closed contours, \(C\) and \(C^{\prime}\), then \(\oint_{C} f(z) d z=\oint_{C^{\prime}} f(z) d z\).

**Proof**

We consider the two curves \(C\) and \(C^{\prime}\) as shown in Figure \(\PageIndex{10}\). Connecting the two contours with contours \(\Gamma_{1}\) and \(\Gamma_{2}\) (as shown in the figure), \(C\) is seen to split into contours \(C_{1}\) and \(C_{2}\) and \(C^{\prime}\) into contours \(C_{1}^{\prime}\) and \(C_{2}^{\prime}\). Note that \(f(z)\) is differentiable inside the newly formed regions between the curves. Also, the boundaries of these regions are now simple closed curves. Therefore, Cauchy’s Theorem tells us that the integrals of \(f(z)\) over these regions are zero.

Noting that integrations over contours opposite to the positive orientation are the negative of integrals that are positively oriented, we have from Cauchy’s Theorem that

\[\int_{C_{1}} f(z) d z+\int_{\Gamma_{1}} f(z) d z-\int_{C_{1}^{\prime}} f(z) d z+\int_{\Gamma_{2}} f(z) d z=0\nonumber \]

and

\[\int_{C_{2}} f(z) d z-\int_{\Gamma_{2}} f(z) d z-\int_{C_{2}^{\prime}} f(z) d z-\int_{\Gamma_{1}} f(z) d z=0 .\nonumber \]

In the first integral we have traversed the contours in the following order: \(C_{1}, \Gamma_{1}, C_{1}^{\prime}\) backwards, and \(\Gamma_{2}\). The second integral denotes the integration over the lower region, but going backwards over all contours except for \(C_{2}\).

Combining these results by adding the two equations above, we have

\[\int_{C_{1}} f(z) d z+\int_{C_{2}} f(z) d z-\int_{C_{1}^{\prime}} f(z) d z-\int_{C_{2}^{\prime}} f(z) d z=0 .\nonumber \]

Noting that \(C=C_{1}+C_{2}\) and \(C^{\prime}=C_{1}^{\prime}+C_{2}^{\prime}\), we have

\[\oint_{C} f(z) d z=\oint_{C^{\prime}} f(z) d z,\nonumber \]

as was to be proven.

Compute \(\oint_{R} \frac{d z}{z}\) for \(R\) the rectangle \([-2,2] \times[-2 i, 2 i]\).

###### Solution

We can compute this integral by looking at four separate integrals over the sides of the rectangle in the complex plane. One simply parametrizes each line segment, perform the integration and sum the four separate results. From the last theorem, we can instead integrate over a simpler contour by deforming the rectangle into a circle as long as \(f(z)=\frac{1}{z}\) is differentiable in the region bounded by the rectangle and the circle. So, using the unit circle, as shown in Figure \(\PageIndex{11}\), the integration might be easier to perform.

More specifically, the last theorem tells us that

\[\oint_{R} \frac{d z}{z}=\oint_{|z|=1} \frac{d z}{z}\nonumber \]

The latter integral can be computed using the parametrization \(z=e^{i \theta}\) for \(\theta \in\) \([0,2 \pi]\). Thus,

\[\begin{align} \oint_{|z|=1} \frac{d z}{z} &=\int_{0}^{2 \pi} \frac{i e^{i \theta} d \theta}{e^{i \theta}}\nonumber \\ &=i \int_{0}^{2 \pi} d \theta=2 \pi i .\label{eq:8} \end{align} \]

Therefore, we have found that \(\oint_{R} \frac{d z}{z}=2 \pi i\) by deforming the original simple closed contour.

For fun, let’s do this the long way to see how much effort was saved. We will label the contour as shown in Figure \(\PageIndex{12}\). The lower segment, \(\gamma_{4}\) of the square can be simple parametrized by noting that along this segment \(z=x-2 i\) for \(x \in[-2,2]\). Then, we have

\[\begin{align} \oint_{\gamma_{4}} \frac{d z}{z} &=\int_{-2}^{2} \frac{d x}{x-2 i}\nonumber \\ &=\ln |x-2 i|_{-2}^{2}\nonumber \\ &=\left(\ln (2 \sqrt{2})-\frac{\pi i}{4}\right)-\left(\ln (2 \sqrt{2})-\frac{3 \pi i}{4}\right)\nonumber \\ &=\frac{\pi i}{2} .\label{eq:9} \end{align} \]

We note that the arguments of the logarithms are determined from the angles made by the diagonals provided in Figure \(\PageIndex{12}\).

Similarly, the integral along the top segment, \(z=x+2 i, x \in[-2,2]\), is computed as

\[\begin{align} \oint_{\gamma_{2}} \frac{d z}{z} &=\int_{2}^{-2} \frac{d x}{x+2 i}\nonumber \\ &=\ln |x+2 i|_{2}^{-2}\nonumber \\ &=\left(\ln (2 \sqrt{2})+\frac{3 \pi i}{4}\right)-\left(\ln (2 \sqrt{2})+\frac{\pi i}{4}\right)\nonumber \\ &=\frac{\pi i}{2} .\label{eq:10} \end{align} \]

The integral over the right side, \(z=2+i y, y \in[-2,2]\), is

\[\begin{align} \oint_{\gamma_{1}} \frac{d z}{z} &=\int_{-2}^{2} \frac{i d y}{2+i y}\nonumber \\ &=\ln |2+i y|_{-2}^{2}\nonumber \\ &=\left(\ln (2 \sqrt{2})+\frac{\pi i}{4}\right)-\left(\ln (2 \sqrt{2})-\frac{\pi i}{4}\right)\nonumber \\ &=\frac{\pi i}{2} .\label{eq:11} \end{align} \]

Finally, the integral over the left side, \(z=-2+i y, y \in[-2,2]\), is

\[\begin{align} \oint_{\gamma_{3}} \frac{d z}{z} &=\int_{2}^{-2} \frac{i d y}{-2+i y}\nonumber \\ &=\ln |-2+i y|_{-2}^{2}\nonumber \\ &=\left(\ln (2 \sqrt{2})+\frac{5 \pi i}{4}\right)-\left(\ln (2 \sqrt{2})+\frac{3 \pi i}{4}\right)\nonumber \\ &=\frac{\pi i}{2} .\label{eq:12} \end{align} \]

Therefore, we have that

\[\begin{align} \oint_{R} \frac{d z}{z}& =\int_{\gamma_{1}} \frac{d z}{z}+\int_{\gamma_{2}} \frac{d z}{z}+\int_{\gamma_{3}} \frac{d z}{z}+\int_{\gamma_{4}} \frac{d z}{z}\nonumber \\ &=\frac{\pi i}{2}+\frac{\pi i}{2}+\frac{\pi i}{2}+\frac{\pi i}{2}\nonumber \\ &=4\left(\frac{\pi i}{2}\right)=2 \pi i .\label{eq:13} \end{align} \]

This gives the same answer we had found using a simple contour deformation.

The converse of Cauchy’s Theorem is not true, namely \(\oint_{C} f(z) d z=0\) does not always imply that \(f(z)\) is differentiable. What we do have is Morera’s Theorem(Giacinto Morera, 1856-1909):

Theorem 8.6. Let \(f\) be continuous in a domain \(D\). Suppose that for every simple closed contour \(C\) in \(D, \oint_{C} f(z) d z=0\). Then \(f\) is differentiable in \(D\).

The proof is a bit more detailed than we need to go into here. However, this theorem is useful in the next section.

## Analytic Functions and Cauchy’s Integral Formula

In the previous section we saw that Cauchy’s Theorem was useful for computing particular integrals without having to parametrize the contours or for deforming contours into simpler contours. The integrand needs to possess certain differentiability properties. In this section, we will generalize the functions that we can integrate slightly so that we can integrate a larger family of complex functions. This will lead us to the Cauchy’s Integral Formula, which extends Cauchy’s Theorem to functions analytic in an annulus. However, first we need to explore the concept of analytic functions.

A function \(f(z)\) is analytic in domain \(D\) if for every open disk \(\left|z-z_{0}\right|<\rho\) lying in \(D, f(z)\) can be represented as a power series in \(z_{0}\). Namely,

\[f(z)=\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n} .\nonumber \]

This series converges uniformly and absolutely inside the circle of convergence, \(\left|z-z_{0}\right|<R\), with radius of convergence \(R\). [See the Appendix for a review of convergence.]

Since \(f(z)\) can be written as a uniformly convergent power series, we can integrate it term by term over any simple closed contour in \(D\) containing \(z_{0}\). In particular, we have to compute integrals like \(\oint_{C}\left(z-z_{0}\right)^{n} d z\). As we will see in the homework exercises, these integrals evaluate to zero for most \(n\). Thus, we can show that for \(f(z)\) analytic in \(D\) and on any closed contour \(C\) lying in \(D, \oint_{C} f(z) d z=0\). Also, \(f\) is a uniformly convergent sum of continuous functions, so \(f(z)\) is also continuous. Thus, by Morera’s Theorem, we have that \(f(z)\) is differentiable if it is analytic. Often terms like analytic, differentiable and holomorphic are used interchangeably, though there is a subtle distinction due to their definitions.

There are various types of complex-valued functions.

A holomorphic function is (complex) differentiable in a neighborhood of every point in its domain.

An analytic function has a convergent Taylor series expansion in a neighborhood of each point in its domain. We see here that analytic functions are holomorphic and vice versa.

If \(\mathrm{a}\) function is holomorphic throughout the complex plane, then it is called an entire function.

Finally, a function which is holomorphic on all of its domain except at a set of isolated poles (to be defined later), then it is called a meromorphic function.

As examples of series expansions about a given point, we will consider series expansions and regions of convergence for \(f(z)=\frac{1}{1+z}\).

Find the series expansion of \(f(z)=\frac{1}{1+z}\) about \(z_{0}=0\).

###### Solution

This case is simple. From Chapter 1 we recall that \(f(z)\) is the sum of a geometric series for \(|z|<1\). We have

\[f(z)=\frac{1}{1+z}=\sum_{n=0}^{\infty}(-z)^{n} .\nonumber \]

Thus, this series expansion converges inside the unit circle \((|z|<1)\) in the complex plane.

Find the series expansion of \(f(z)=\frac{1}{1+z}\) about \(z_{0}=\frac{1}{2}\).

###### Solution

We now look into an expansion about a different point. We could compute the expansion coefficients using Taylor’s formula for the coefficients. However, we can also make use of the formula for geometric series after rearranging the function. We seek an expansion in powers of \(z-\frac{1}{2}\). So, we rewrite the function in a form that has is a function of \(z-\frac{1}{2}\). Thus,

\[f(z)=\frac{1}{1+z}=\frac{1}{1+\left(z-\frac{1}{2}+\frac{1}{2}\right)}=\frac{1}{\frac{3}{2}+\left(z-\frac{1}{2}\right)} .\nonumber \]

This is not quite in the form we need. It would be nice if the denominator were of the form of one plus something. [Note: This is similar to what we had seen in Example 11.7.7.] We can get the denominator into such a form by factoring out the \(\frac{3}{2}\). Then we would have

\[f(z)=\frac{2}{3} \frac{1}{1+\frac{2}{3}\left(z-\frac{1}{2}\right)} .\nonumber \]

The second factor now has the form \(\frac{1}{1-r}\), which would be the sum of a geometric series with first term \(a=1\) and ratio \(r=-\frac{2}{3}\left(z-\frac{1}{2}\right)\) provided that \(|r|<1\). Therefore, we have found that

\[f(z)=\frac{2}{3} \sum_{n=0}^{\infty}\left[-\frac{2}{3}\left(z-\frac{1}{2}\right)\right]^{n}\nonumber \]

for

\[\left|-\frac{2}{3}\left(z-\frac{1}{2}\right)\right|<1 \text {. }\nonumber \]

This convergence interval can be rewritten as

\[\left|z-\frac{1}{2}\right|<\frac{3}{2},\nonumber \]

which is a circle centered at \(z=\frac{1}{2}\) with radius \(\frac{3}{2}\).

In Figure \(\PageIndex{13}\) we show the regions of convergence for the power series expansions of \(f(z)=\frac{1}{1+z}\) about \(z=0\) and \(z=\frac{1}{2}\). We note that the first expansion gives that \(f(z)\) is at least analytic inside the region \(|z|<1\). The second expansion shows that \(f(z)\) is analytic in a larger region, \(\left|z-\frac{1}{2}\right|<\frac{3}{2}\). We will see later that there are expansions which converge outside of these regions and that some yield expansions involving negative powers of \(z-z_{0}\).

We now present the main theorem of this section:

Let \(f(z)\) be analytic in \(\left|z-z_{0}\right|<\rho\) and let \(C\) be the boundary (circle) of this disk. Then,

\[f\left(z_{0}\right)=\frac{1}{2 \pi i} \oint_{C} \frac{f(z)}{z-z_{0}} d z .\label{eq:14} \]

**Proof**

In order to prove this, we first make use of the analyticity of \(f(z)\). We insert the power series expansion of \(f(z)\) about \(z_{0}\) into the integrand. Then we have

\[\begin{align} \frac{f(z)}{z-z_{0}} &=\frac{1}{z-z_{0}}\left[\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n}\right]\nonumber \\ &=\frac{1}{z-z_{0}}\left[c_{0}+c_{1}\left(z-z_{0}\right)+c_{2}\left(z-z_{0}\right)^{2}+\ldots\right]\nonumber \\ &=\frac{c_{0}}{z-z_{0}}+\underbrace{c_{1}+c_{2}\left(z-z_{0}\right)+\ldots}_{\text {analytic function }}\label{eq:15} \end{align} \]

As noted the integrand can be written as

\[\frac{f(z)}{z-z_{0}}=\frac{c_{0}}{z-z_{0}}+h(z),\nonumber \]

where \(h(z)\) is an analytic function, since \(h(z)\) is representable as a series expansion about \(z_{0}\). We have already shown that analytic functions are differentiable, so by Cauchy’s Theorem \(\oint_{C} h(z) d z=0\).

Noting also that \(c_{0}=f\left(z_{0}\right)\) is the first term of a Taylor series expansion about \(z=z_{0}\), we have

\[\oint_{C} \frac{f(z)}{z-z_{0}} d z=\oint_{C}\left[\frac{c_{0}}{z-z_{0}}+h(z)\right] d z=f\left(z_{0}\right) \oint_{C} \frac{1}{z-z_{0}} d z .\nonumber \]

We need only compute the integral \(\oint_{C} \frac{1}{z-z_{0}} d z\) to finish the proof of Cauchy’s Integral Formula. This is done by parametrizing the circle, \(\left|z-z_{0}\right|=\rho\), as shown in Figure \(\PageIndex{14}\). This is simply done by letting

\[z-z_{0}=\rho e^{i \theta} .\nonumber \]

(Note that this has the right complex modulus since \(\left|e^{i \theta}\right|=1\). Then \(d z=\) \(i \rho e^{i \theta} d \theta\). Using this parametrization, we have

\[\oint_{C} \frac{d z}{z-z_{0}}=\int_{0}^{2 \pi} \frac{i \rho e^{i \theta} d \theta}{\rho e^{i \theta}}=i \int_{0}^{2 \pi} d \theta=2 \pi i .\nonumber \]

Therefore,

\[\oint_{C} \frac{f(z)}{z-z_{0}} d z=f\left(z_{0}\right) \oint_{C} \frac{1}{z-z_{0}} d z=2 \pi i f\left(z_{0}\right),\nonumber \]

as was to be shown.

Compute \(\oint_{|z|=4} \frac{\cos z}{z^{2}-6 z+5} d z\).

###### Solution

In order to apply the Cauchy Integral Formula, we need to factor the denominator, \(z^{2}-6 z+5=(z-1)(z-5)\). We next locate the zeros of the denominator. In Figure \(\PageIndex{15}\) we show the contour and the points \(z=1\) and \(z=5\). The only point inside the region bounded by the contour is \(z=1\). Therefore, we can apply the Cauchy Integral Formula for \(f(z)=\frac{\cos z}{z-5}\) to the integral

\[\int_{|z|=4} \frac{\cos z}{(z-1)(z-5)} d z=\int_{|z|=4} \frac{f(z)}{(z-1)} d z=2 \pi i f(1) \text {. }\nonumber \]

Therefore, we have

\[\int_{|z|=4} \frac{\cos z}{(z-1)(z-5)} d z=-\frac{\pi i \cos (1)}{2} \text {. }\nonumber \]

We have shown that \(f\left(z_{0}\right)\) has an integral representation for \(f(z)\) analytic in \(\left|z-z_{0}\right|<\rho\). In fact, all derivatives of an analytic function have an integral representation. This is given by

\[f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz.\label{eq:16} \]

This can be proven following a derivation similar to that for the Cauchy Integral Formula. Inserting the Taylor series expansion for \(f(z)\) into the integral on the right hand side, we have

\[\begin{align} \oint_{C} \frac{f(z)}{\left(z-z_{0}\right)^{n+1}} d z &=\sum_{m=0}^{\infty} c_{m} \oint_{C} \frac{\left(z-z_{0}\right)^{m}}{\left(z-z_{0}\right)^{n+1}} d z\nonumber \\ &=\sum_{m=0}^{\infty} c_{m} \oint_{C} \frac{d z}{\left(z-z_{0}\right)^{n-m+1}}\label{eq:17} \end{align} \]

Picking \(k=n-m\), the integrals in the sum can be computed by using the following result:

\[\oint_{C} \frac{d z}{\left(z-z_{0}\right)^{k+1}}=\left\{\begin{array}{cc} 0, & k \neq 0 \\ 2 \pi i, & k=0 . \end{array}\right.\label{eq:18} \]

The proof is left for the exercises.

The only nonvanishing integrals, \(\oint_{C} \frac{d z}{\left(z-z_{0}\right)^{n-m+1}}\), occur when \(k=n-m=\) 0 , or \(m=n\). Therefore, the series of integrals collapses to one term and we have

\[\oint_{C} \frac{f(z)}{\left(z-z_{0}\right)^{n+1}} d z=2 \pi i c_{n} .\nonumber \]

We finish the proof by recalling that the coefficients of the Taylor series expansion for \(f(z)\) are given by

\[c_{n}=\frac{f^{(n)}\left(z_{0}\right)}{n !} .\nonumber \]

Then,

\[\oint_{C} \frac{f(z)}{\left(z-z_{0}\right)^{n+1}} d z=\frac{2 \pi i}{n !} f^{(n)}\left(z_{0}\right)\nonumber \]

and the result follows.

## Laurent Series

Until this point we have only talked about series whose terms have nonnegative powers of \(z-z_{0}\). It is possible to have series representations in which there are negative powers. In the last section we investigated expansions of \(f(z)=\frac{1}{1+z}\) about \(z=0\) and \(z=\frac{1}{2}\). The regions of convergence for each series was shown in Figure \(\PageIndex{13}\). Let us reconsider each of these expansions, but for values of \(z\) outside the region of convergence previously found.

\(f(z)=\frac{1}{1+z}\) for \(|z|>1\).

###### Solution

As before, we make use of the geometric series. Since \(|z|>1\), we instead rewrite the function as

\[f(z)=\frac{1}{1+z}=\frac{1}{z} \frac{1}{1+\frac{1}{z}} .\nonumber \]

We now have the function in a form of the sum of a geometric series with first term \(a=1\) and ratio \(r=-\frac{1}{2}\). We note that \(|z|>1\) implies that \(|r|<1\). Thus, we have the geometric series

\[f(z)=\frac{1}{z} \sum_{n=0}^{\infty}\left(-\frac{1}{z}\right)^{n} .\nonumber \]

This can be re-indexed\(^{1}\) as

\[f(z)=\sum_{n=0}^{\infty}(-1)^{n} z^{-n-1}=\sum_{j=1}^{\infty}(-1)^{j-1} z^{-j} .\nonumber \]

Note that this series, which converges outside the unit circle, \(|z|>1\), has negative powers of \(z\).

Re-indexing a series is often useful in the geometric series series manipulations. In this case, we have the series

\[\sum_{n=0}^{\infty}(-1)^{n} z^{-n-1}=z^{-1}-z^{-2}+z^{-3}+\ldots\nonumber \]

The index is \(n\). You can see that the index does not appear when the sum is expanded showing the terms. The summation index is sometimes refered to as a dummy index for this reason. Reindexing allows one to rewrite the shorthand summation notation while capturing the same terms. In this example, the exponents are \(-n-1\). We can simplify the notation by letting \(-n-1=-j\), or \(j=n+1\). Noting that \(j=1\) when \(n=0\), we get the sum \(\sum_{j=1}^{\infty}(-1)^{j-1} z^{-j}\).

\(f(z)=\frac{1}{1+z}\) for \(\left|z-\frac{1}{2}\right|>\frac{3}{2}\).

###### Solution

As before, we express this in a form in which we can use a geometric series expansion. We seek powers of \(z-\frac{1}{2}\). So, we add and subtract \(\frac{1}{2}\) to the \(z\) to obtain:

\[f(z)=\frac{1}{1+z}=\frac{1}{1+\left(z-\frac{1}{2}+\frac{1}{2}\right)}=\frac{1}{\frac{3}{2}+\left(z-\frac{1}{2}\right)} .\nonumber \]

Instead of factoring out the \(\frac{3}{2}\) as we had done in Example \(\PageIndex{8}\), we factor out the \(\left(z-\frac{1}{2}\right)\) term. Then, we obtain

\[f(z)=\frac{1}{1+z}=\frac{1}{\left(z-\frac{1}{2}\right)} \frac{1}{\left[1+\frac{3}{2}\left(z-\frac{1}{2}\right)^{-1}\right]} .\nonumber \]

Now we identify \(a=1\) and \(r=-\frac{3}{2}\left(z-\frac{1}{2}\right)^{-1}\). This leads to the series

\[\begin{align} f(z) &=\frac{1}{z-\frac{1}{2}} \sum_{n=0}^{\infty}\left(-\frac{3}{2}\left(z-\frac{1}{2}\right)^{-1}\right)^{n}\nonumber \\ &=\sum_{n=0}^{\infty}\left(-\frac{3}{2}\right)^{n}\left(z-\frac{1}{2}\right)^{-n-1} .\label{eq:19} \end{align} \]

This converges for \(\left|z-\frac{1}{2}\right|>\frac{3}{2}\) and can also be re-indexed to verify that this series involves negative powers of \(z-\frac{1}{2}\).

This leads to the following theorem:

Let \(f(z)\) be analytic in an annulus, \(R_{1}<\left|z-z_{0}\right|<R_{2}\), with \(\mathrm{C}\) a positively oriented simple closed curve around \(z_{0}\) and inside the annulus as shown in Figure \(\PageIndex{16}\). Then,

\[f(z)=\sum_{j=0}^{\infty} a_{j}\left(z-z_{0}\right)^{j}+\sum_{j=1}^{\infty} b_{j}\left(z-z_{0}\right)^{-j},\nonumber \]

with

\[a_{j}=\frac{1}{2 \pi i} \oint_{C} \frac{f(z)}{\left(z-z_{0}\right)^{j+1}} d z,\nonumber \]

and

\[b_{j}=\frac{1}{2 \pi i} \oint_{C} \frac{f(z)}{\left(z-z_{0}\right)^{-j+1}} d z .\nonumber \]

The above series can be written in the more compact form

\[f(z)=\sum_{j=-\infty}^{\infty} c_{j}\left(z-z_{0}\right)^{j} .\nonumber \]

Such a series expansion is called a Laurent series expansion named after its discoverer Pierre Alphonse Laurent (1813-1854).

Expand \(f(z)=\frac{1}{(1-z)(2+z)}\) in the annulus \(1<|z|<2\).

###### Solution

Using partial fractions, we can write this as

\[f(z)=\frac{1}{3}\left[\frac{1}{1-z}+\frac{1}{2+z}\right] .\nonumber \]

We can expand the first fraction, \(\frac{1}{1-z}\), as an analytic function in the region \(|z|>1\) and the second fraction, \(\frac{1}{2+z}\), as an analytic function in \(|z|<2\). This is done as follows. First, we write

\[\frac{1}{2+z}=\frac{1}{2\left[1-\left(-\frac{z}{2}\right)\right]}=\frac{1}{2} \sum_{n=0}^{\infty}\left(-\frac{z}{2}\right)^{n} .\nonumber \]

Then, we write

\[\frac{1}{1-z}=-\frac{1}{z\left[1-\frac{1}{z}\right]}=-\frac{1}{z} \sum_{n=0}^{\infty} \frac{1}{z^{n}} .\nonumber \]

Therefore, in the common region, \(1<|z|<2\), we have that

\[\begin{align} \frac{1}{(1-z)(2+z)} &=\frac{1}{3}\left[\frac{1}{2} \sum_{n=0}^{\infty}\left(-\frac{z}{2}\right)^{n}-\sum_{n=0}^{\infty} \frac{1}{z^{n+1}}\right]\nonumber \\ &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{6\left(2^{n}\right)} z^{n}+\sum_{n=1}^{\infty} \frac{(-1)}{3} z^{-n}\label{eq:20} \end{align} \]

We note that this is not a Taylor series expansion due to the existence of terms with negative powers in the second sum.

Find series representations of \(f(z)=\frac{1}{(1-z)(2+z)}\) throughout the complex plane.

###### Solution

In the last example we found series representations of \(f(z)=\frac{1}{(1-z)(2+z)}\) in the annulus \(1<|z|<2\). However, we can also find expansions which converge for other regions. We first write

\[f(z)=\frac{1}{3}\left[\frac{1}{1-z}+\frac{1}{2+z}\right] .\nonumber \]

We then expand each term separately.

The first fraction, \(\frac{1}{1-z}\), can be written as the sum of the geometric series

\[\frac{1}{1-z}=\sum_{n=0}^{\infty} z^{n}, \quad|z|<1 .\nonumber \]

This series converges inside the unit circle. We indicate this by region 1 in Figure \(\PageIndex{17}\).

In the last example, we showed that the second fraction, \(\frac{1}{2+z}\), has the series expansion

\[\frac{1}{2+z}=\frac{1}{2\left[1-\left(-\frac{z}{2}\right)\right]}=\frac{1}{2} \sum_{n=0}^{\infty}\left(-\frac{z}{2}\right)^{n} .\nonumber \]

which converges in the circle \(|z|<2\). This is labeled as region 2 in Figure \(\PageIndex{17}\).

Regions 1 and 2 intersect for \(|z|<1\), so, we can combine these two series representations to obtain

\[\frac{1}{(1-z)(2+z)}=\frac{1}{3}\left[\sum_{n=0}^{\infty} z^{n}+\frac{1}{2} \sum_{n=0}^{\infty}\left(-\frac{z}{2}\right)^{n}\right], \quad|z|<1 .\nonumber \]

In the annulus, \(1<|z|<2\), we had already seen in the last example that we needed a different expansion for the fraction \(\frac{1}{1-z}\). We looked for an expansion in powers of \(1 / z\) which would converge for large values of \(z\). We had found that

\[\frac{1}{1-z}=-\frac{1}{z\left(1-\frac{1}{z}\right)}=-\frac{1}{z} \sum_{n=0}^{\infty} \frac{1}{z^{n}}, \quad|z|>1 .\nonumber \]

This series converges in region 3 in Figure \(\PageIndex{17}\). Combining this series with the one for the second fraction, we obtain a series representation that converges in the overlap of regions 2 and 3 . Thus, in the annulus \(1<|z|<2\) we have

\[\frac{1}{(1-z)(2+z)}=\frac{1}{3}\left[\frac{1}{2} \sum_{n=0}^{\infty}\left(-\frac{z}{2}\right)^{n}-\sum_{n=0}^{\infty} \frac{1}{z^{n+1}}\right] .\nonumber \]

So far, we have series representations for \(|z|<2\). The only region not covered yet is outside this disk, \(|z|>2\). In in Figure \(\PageIndex{17}\) we see that series 3, which converges in region 3, will converge in the last section of the complex plane. We just need one more series expansion for \(1 /(2+z)\) for large \(z\). Factoring out a \(z\) in the denominator, we can write this as a geometric series with \(r=2 / z\),

\[\frac{1}{2+z}=\frac{1}{z\left[\frac{2}{z}+1\right]}=\frac{1}{z} \sum_{n=0}^{\infty}\left(-\frac{2}{z}\right)^{n} .\nonumber \]

This series converges for \(|z|>2\). Therefore, it converges in region 4 and the final series representation is

\[\frac{1}{(1-z)(2+z)}=\frac{1}{3}\left[\frac{1}{z} \sum_{n=0}^{\infty}\left(-\frac{2}{z}\right)^{n}-\sum_{n=0}^{\infty} \frac{1}{z^{n+1}}\right] .\nonumber \]

## Singularities and The Residue Theorem

In the last section we found that we could integrate functions satisfying some analyticity properties along contours without using detailed parametrizations around the contours. We can deform contours if the function is analytic in the region between the original and new contour. In this section we will extend our tools for performing contour integrals.

The integrand in the Cauchy Integral Formula was of the form \(g(z)=\) \(\frac{f(z)}{z-z_{0}}\), where \(f(z)\) is well behaved at \(z_{0}\). The point \(z=z_{0}\) is called a singularity of \(g(z)\), as \(g(z)\) is not defined there. More specifically, a singularity of \(f(z)\) is a point at which \(f(z)\) fails to be analytic.

We can also classify these singularities. Typically these are isolated singularities. As we saw from the proof of the Cauchy Integral Formula, \(g(z)=\frac{f(z)}{z-z_{0}}\) has a Laurent series expansion about \(z=z_{0}\), given by

\[g(z)=\frac{f\left(z_{0}\right)}{z-z_{0}}+f^{\prime}\left(z_{0}\right)+\frac{1}{2} f^{\prime \prime}\left(z_{0}\right)\left(z-z_{0}\right)+\ldots\nonumber \]

It is the nature of the first term that gives information about the type of singularity that \(g(z)\) has. Namely, in order to classify the singularities of \(f(z)\), we look at the principal part of the Laurent series of \(f(z)\) about \(z=z_{0}\), \(\sum_{j-1}^{\infty} b_{j}\left(z-z_{0}\right)^{-j}\), which consists of the negative powers of \(z-z_{0}\).

There are three types of singularities, removable, poles, and essential singularities. They are defined as follows:

- If \(f(z)\) is bounded near \(z_{0}\), then \(z_{0}\) is a removable singularity.
- If there are a finite number of terms in the principal part of the Laurent series of \(f(z)\) about \(z=z_{0}\), then \(z_{0}\) is called a pole.
- If there are an infinite number of terms in the principal part of the Laurent series of \(f(z)\) about \(z=z_{0}\), then \(z_{0}\) is called an essential singularity.

\(f(z)=\frac{\sin z}{z}\) has a removable singularity at \(z=0\).

###### Solution

At first it looks like there is a possible singularity at \(z=0\), since the denominator is zero at \(z=0\). However, we know from the first semester of calculus that \(\lim _{z \rightarrow 0} \frac{\sin z}{z}=1\). Furthermore, we can expand \(\sin z\) about \(z=0\) and see that

\[\frac{\sin z}{z}=\frac{1}{z}\left(z-\frac{z^{3}}{3 !}+\ldots\right)=1-\frac{z^{2}}{3 !}+\ldots\nonumber \]

Thus, there are only nonnegative powers in the series expansion. So, \(z=0\) is a removable singularity.

\(f(z)=\frac{e^{2}}{(z-1)^{n}}\) has poles at \(z=1\) for \(n\) a positive integer.

###### Solution

For \(n=1\) we have \(f(z)=\frac{e^{z}}{z-1}\). This function has a singularity at \(z=1\). The series expansion is found by expanding \(e^{z}\) about \(z=1\) :

\[f(z)=\frac{e}{z-1} e^{z-1}=\frac{e}{z-1}+e+\frac{e}{2 !}(z-1)+\ldots .\nonumber \]

Note that the principal part of the Laurent series expansion about \(z=1\) only has one term, \(\frac{e}{z-1}\). Therefore, \(z=1\) is a pole. Since the leading term has an exponent of \(-1, z=1\) is called a pole of order one, or a simple pole.

For \(n=2\) we have \(f(z)=\frac{e^{z}}{(z-1)^{2}}\). The series expansion is found again by expanding \(e^{z}\) about \(z=1\) :

\[f(z)=\frac{e}{(z-1)^{2}} e^{z-1}=\frac{e}{(z-1)^{2}}+\frac{e}{z-1}+\frac{e}{2 !}+\frac{e}{3 !}(z-1)+\ldots .\nonumber \]

Note that the principal part of the Laurent series has two terms involving \((z-1)^{-2}\) and \((z-1)^{-1}\). Since the leading term has an exponent of \(-2, z=1\) is called a pole of order 2, or a double pole.

\(f(z)=e^{\frac{1}{2}}\) has an essential singularity at \(z=0\).

###### Solution

In this case we have the series expansion about \(z=0\) given by

\[f(z)=e^{\frac{1}{2}}=\sum_{n=0}^{\infty} \frac{\left(\frac{1}{z}\right)^{n}}{n !}=\sum_{n=0}^{\infty} \frac{1}{n !} z^{-n} .\nonumber \]

We see that there are an infinite number of terms in the principal part of the Laurent series. So, this function has an essential singularity at \(z=0\).

In the above examples we have seen poles of order one (a simple pole) and two (a double pole). In general, we can say that \(f(z)\) has a pole of order \(k\) at \(z_{0}\) if and only if \(\left(z-z_{0}\right)^{k} f(z)\) has a removable singularity at \(z_{0}\), but \(\left(z-z_{0}\right)^{k-1} f(z)\) for \(k>0\) does not.

Determine the order of the pole of \(f(z)=\cot z \csc z\) at \(z=0\).

###### Solution

First we rewrite \(f(z)\) in terms of sines and cosines.

\[f(z)=\cot z \csc z=\frac{\cos z}{\sin ^{2} z} .\nonumber \]

We note that the denominator vanishes at \(z=0\).

How do we know that the pole is not a simple pole? Well, we check to see if \((z-0) f(z)\) has a removable singularity at \(z=0\) :

\[\begin{align} \lim _{z \rightarrow 0}(z-0) f(z) &=\lim _{z \rightarrow 0} \frac{z \cos z}{\sin ^{2} z}\nonumber \\ &=\left(\lim _{z \rightarrow 0} \frac{z}{\sin z}\right)\left(\lim _{z \rightarrow 0} \frac{\cos z}{\sin z}\right)\nonumber \\ &=\lim _{z \rightarrow 0} \frac{\cos z}{\sin z} .\label{eq:21} \end{align} \]

We see that this limit is undefined. So, now we check to see if \((z-0)^{2} f(z)\) has a removable singularity at \(z=0\) :

\[\begin{align} \lim _{z \rightarrow 0}(z-0)^{2} f(z)&=\lim _{z \rightarrow 0} \frac{z^{2} \cos z}{\sin ^{2} z}\nonumber \\ &=\left(\lim _{z \rightarrow 0} \frac{z}{\sin z}\right)\left(\lim _{z \rightarrow 0} \frac{z \cos z}{\sin z}\right)\nonumber \\ &=\lim _{z \rightarrow 0} \frac{z}{\sin z} \cos (0)=1 .\label{eq:22}\end{align} \]

In this case, we have obtained a finite, nonzero, result. \(S_{0}, z=0\) is a pole of order 2 .

We could have also relied on series expansions. Expanding both the sine and cosine functions in a Taylor series expansion, we have

\[f(z)=\frac{\cos z}{\sin ^{2} z}=\frac{1-\frac{1}{2 !} z^{2}+\ldots}{\left(z-\frac{1}{3 !} z^{3}+\ldots\right)^{2}} .\nonumber \]

Factoring a \(z\) from the expansion in the denominator,

\[f(z)=\frac{1}{z^{2}} \frac{1-\frac{1}{2 !} z^{2}+\ldots}{\left(1-\frac{1}{3 !} z+\ldots\right)^{2}}=\frac{1}{z^{2}}\left(1+O\left(z^{2}\right)\right),\nonumber \]

we can see that the leading term will be a \(1 / z^{2}\), indicating a pole of order 2.

Integral of a function with a simple pole inside \(C\).

Residues of a function with poles of order \(k\).

We will see how knowledge of the poles of a function can aid in the computation of contour integrals. We now show that if a function, \(f(z)\), has a pole of order \(k\), then

\[\oint_{C} f(z) d z=2 \pi i \operatorname{Res}\left[f(z) ; z_{0}\right]\nonumber \]

where we have defined \(\operatorname{Res}\left[f(z) ; z_{0}\right]\) as the residue of \(f(z)\) at \(z=z_{0}\). In particular, for a pole of order \(k\) the residue is given by

\[\operatorname{Res}\left[f(z) ; z_{0}\right]=\lim _{z \rightarrow z_{0}} \frac{1}{(k-1) !} \frac{d^{k-1}}{d z^{k-1}}\left[\left(z-z_{0}\right)^{k} f(z)\right] .\label{eq:23} \]

**Proof**

Let \(\phi(z)=\left(z-z_{0}\right)^{k} f(z)\) be an analytic function. Then \(\phi(z)\) has a Taylor series expansion about \(z_{0}\). As we had seen in the last section, we can write the integral representation of any derivative of \(\phi\) as

\[\phi^{(k-1)}\left(z_{0}\right)=\frac{(k-1) !}{2 \pi i} \oint_{C} \frac{\phi(z)}{\left(z-z_{0}\right)^{k}} d z .\nonumber \]

Inserting the definition of \(\phi(z)\), we then have

\[\phi^{(k-1)}\left(z_{0}\right)=\frac{(k-1) !}{2 \pi i} \oint_{C} f(z) d z .\nonumber \]

Solving for the integral, we have the result

\[\begin{align} \oint_{C} f(z) d z &=\frac{2 \pi i}{(k-1) !} \frac{d^{k-1}}{d z^{k-1}}\left[\left(z-z_{0}\right)^{k} f(z)\right]_{z=z_{0}}\nonumber \\ & \equiv 2 \pi i \operatorname{Res}\left[f(z) ; z_{0}\right]\label{eq:24} \end{align} \]

If \(z_{0}\) is a simple pole, the residue is easily computed as

\[\operatorname{Res}\left[f(z) ; z_{0}\right]=\lim _{z \rightarrow z_{0}}\left(z-z_{0}\right) f(z)\nonumber \]

In fact, one can show (Problem 18) that for \(g\) and \(h\) analytic functions at \(z_{0}\), with \(g\left(z_{0}\right) \neq 0, h\left(z_{0}\right)=0\), and \(h^{\prime}\left(z_{0}\right) \neq 0\),

\[\operatorname{Res}\left[\frac{g(z)}{h(z)} ; z_{0}\right]=\frac{g\left(z_{0}\right)}{h^{\prime}\left(z_{0}\right)}\nonumber \]

Find the residues of \(f(z)=\frac{z-1}{(z+1)^{2}\left(z^{2}+4\right)}\).

###### Solution

\(f(z)\) has poles at \(z=-1, z=2\) i, and \(z=-2\) i. The pole at \(z=-1\) is a double pole (pole of order 2). The other poles are simple poles. We compute those residues first:

\[\begin{align} \operatorname{Res}[f(z) ; 2 i] &=\lim _{z \rightarrow 2 i}(z-2 i) \frac{z-1}{(z+1)^{2}(z+2 i)(z-2 i)}\nonumber \\ &=\lim _{z \rightarrow 2 i} \frac{z-1}{(z+1)^{2}(z+2 i)}\nonumber \\ &=\frac{2 i-1}{(2 i+1)^{2}(4 i)}=-\frac{1}{50}-\frac{11}{100} i .\label{eq:25} \\ \operatorname{Res}[f(z) ;-2 i] &=\lim _{z \rightarrow-2 i}(z+2 i) \frac{z-1}{(z+1)^{2}(z+2 i)(z-2 i)} \\ &=\lim _{z \rightarrow-2 i} \frac{z-1}{(z+1)^{2}(z-2 i)}\nonumber \\ &=\frac{-2 i-1}{(-2 i+1)^{2}(-4 i)}=-\frac{1}{50}+\frac{11}{100} i .\label{eq:26} \end{align} \]

For the double pole, we have to do a little more work.

\[\begin{align} \operatorname{Res}[f(z) ;-1] &=\lim _{z \rightarrow-1} \frac{d}{d z}\left[(z+1)^{2} \frac{z-1}{(z+1)^{2}\left(z^{2}+4\right)}\right]\nonumber \\ &=\lim _{z \rightarrow-1} \frac{d}{d z}\left[\frac{z-1}{z^{2}+4}\right]\nonumber \\ &=\lim _{z \rightarrow-1} \frac{d}{d z}\left[\frac{z^{2}+4-2 z(z-1)}{\left(z^{2}+4\right)^{2}}\right]\nonumber \\ &=\lim _{z \rightarrow-1} \frac{d}{d z}\left[\frac{-z^{2}+2 z+4}{\left(z^{2}+4\right)^{2}}\right]\nonumber \\ &=\frac{1}{25}\label{eq:27} \end{align} \]

Find the residue of \(f(z)=\cot z\) at \(z=0\).

###### Solution

We write \(f(z)=\cot z=\frac{\cos z}{\sin z}\) and note that \(z=0\) is a simple pole. Thus,

\[\operatorname{Res}[\cot z ; z=0]=\lim _{z \rightarrow 0} \frac{z \cos z}{\sin z}=\cos (0)=1 \text {. }\nonumber \]

Another way to find the residue of a function \(f(z)\) at a singularity \(z_{0}\) is to look at the Laurent series expansion about the singularity. This is because the residue of \(f(z)\) at \(z_{0}\) is the coefficient of the \(\left(z-z_{0}\right)^{-1}\) term, or \(c_{-1}=b_{1}\).

The residue of \(f(z)\) at \(z_0\) is the coefficient of the \((z-z_0)^{-1}\) term, \(c_{-1}=b_1\), of the Laurent series expansion about \(z_0\).

Find the residue of \(f(z)=\frac{1}{z(3-z)}\) at \(z=0\) using a Laurent series expansion.

###### Solution

First, we need the Laurent series expansion about \(z=0\) of the form \(\sum_{-\infty}^{\infty} c_{n} z^{n}\). A partial fraction expansion gives

\[f(z)=\frac{1}{z(3-z)}=\frac{1}{3}\left(\frac{1}{z}+\frac{1}{3-z}\right) .\nonumber \]

The first term is a power of \(z\). The second term needs to be written as a convergent series for small \(z\). This is given by

\[\begin{align} \frac{1}{3-z} &=\frac{1}{3(1-z / 3)}\nonumber \\ &=\frac{1}{3} \sum_{n=0}^{\infty}\left(\frac{z}{3}\right)^{n}\label{eq:28} \end{align} \]

Thus, we have found

\[f(z)=\frac{1}{3}\left(\frac{1}{z}+\frac{1}{3} \sum_{n=0}^{\infty}\left(\frac{z}{3}\right)^{n}\right) .\nonumber \]

The coefficient of \(z^{-1}\) can be read off to give \(\operatorname{Res}[f(z) ; z=0]=\frac{1}{3}\).

Find the residue of \(f(z)=z \cos \frac{1}{z}\) at \(z=0\) using a Laurent series expansion.

###### Solution

In this case \(z=0\) is an essential singularity. The only way to find residues at essential singularities is to use Laurent series. Since

\[\cos z=1-\frac{1}{2 !} z^{2}+\frac{1}{4 !} z^{4}-\frac{1}{6 !} z^{6}+\ldots,\nonumber \]

then we have

\[\begin{align} f(z) &=z\left(1-\frac{1}{2 ! z^{2}}+\frac{1}{4 ! z^{4}}-\frac{1}{6 ! z^{6}}+\ldots\right)\nonumber \\ &=z-\frac{1}{2 ! z}+\frac{1}{4 ! z^{3}}-\frac{1}{6 ! z^{5}}+\ldots\label{eq:29} \end{align} \]

From the second term we have that \(\operatorname{Res}[f(z) ; z=0]=-\frac{1}{2}\).

We are now ready to use residues in order to evaluate integrals.

Evaluate \(\oint_{|z|=1} \frac{d z}{\sin z}\).

###### Solution

We begin by looking for the singularities of the integrand. These are located at values of \(z\) for which \(\sin z=0\). Thus, \(z=0, \pm \pi, \pm 2 \pi, \ldots\), are the singularities. However, only \(z=0\) lies inside the contour, as shown in Figure \(\PageIndex{18}\). We note further that \(z=0\) is a simple pole, since

\[\lim _{z \rightarrow 0}(z-0) \frac{1}{\sin z}=1 .\nonumber \]

Therefore, the residue is one and we have

\[\oint_{|z|=1} \frac{d z}{\sin z}=2 \pi i .\nonumber \]

In general, we could have several poles of different orders. For example, we will be computing

\[\oint_{|z|=2} \frac{d z}{z^{2}-1} .\nonumber \]

The integrand has singularities at \(z^{2}-1=0\), or \(z=\pm 1\). Both poles are inside the contour, as seen in Figure \(\PageIndex{20}\). One could do a partial fraction decomposition and have two integrals with one pole each integral. Then, the result could be found by adding the residues from each pole.

In general, when there are several poles, we can use the Residue Theorem.

Let \(f(z)\) be a function which has poles \(z_{j}, j=1, \ldots, N\) inside a simple closed contour \(\mathrm{C}\) and no other singularities in this region. Then,

\[\oint_{C} f(z) d z=2 \pi i \sum_{j=1}^{N} \operatorname{Res}\left[f(z) ; z_{j}\right],\label{eq:30} \]

where the residues are computed using Equation \(\eqref{eq:23}\),

\[\operatorname{Res}\left[f(z) ; z_{0}\right]=\lim _{z \rightarrow z_{0}} \frac{1}{(k-1) !} \frac{d^{k-1}}{d z^{k-1}}\left[\left(z-z_{0}\right)^{k} f(z)\right] .\nonumber \]

The proof of this theorem is based upon the contours shown in Figure \(\PageIndex{19}\). One constructs a new contour \(C^{\prime}\) by encircling each pole, as show in the figure. Then one connects a path from \(C\) to each circle. In the figure two separated paths along the cut are shown only to indicate the direction followed on the cut. The new contour is then obtained by following \(C\) and crossing each cut as it is encountered. Then one goes around a circle in the negative sense and returns along the cut to proceed around \(C\). The sum of the contributions to the contour integration involve two integrals for each cut, which will cancel due to the opposing directions. Thus, we are left with

\[\oint_{C^{\prime}} f(z) d z=\oint_{C} f(z) d z-\oint_{C_{1}} f(z) d z-\oint_{C_{2}} f(z) d z-\oint_{C_{3}} f(z) d z=0 .\nonumber \]

Of course, the sum is zero because \(f(z)\) is analytic in the enclosed region, since all singularities have been cut out. Solving for \(\oint_{C} f(z) d z\), one has that this integral is the sum of the integrals around the separate poles, which can be evaluated with single residue computations. Thus, the result is that \(\oint_{C} f(z) d z\) is \(2 \pi i\) times the sum of the residues.

Evaluate \(\oint_{|z|=2} \frac{d z}{z^{2}-1}\).

###### Solution

We first note that there are two poles in this integral since

\[\frac{1}{z^{2}-1}=\frac{1}{(z-1)(z+1)} \text {. }\nonumber \]

In Figure \(\PageIndex{20}\) we plot the contour and the two poles, denoted by an "x." Since both poles are inside the contour, we need to compute the residues for each one. They are each simple poles, so we have

\[\begin{align} \operatorname{Res}\left[\frac{1}{z^2-1};z=1\right]&=\lim_{z\to 1}(z-1)\frac{1}{z^2-1}\nonumber \\ &=\lim_{z\to 1}\frac{1}{z+1}=\frac{1}{2},\label{eq:31}\end{align} \]

and

\[\begin{align} \operatorname{Res}\left[\frac{1}{z^{2}-1} ; z=-1\right] &=\lim _{z \rightarrow-1}(z+1) \frac{1}{z^{2}-1}\nonumber \\ &=\lim _{z \rightarrow-1} \frac{1}{z-1}=-\frac{1}{2}\label{eq:32} \end{align} \]

Then,

\[\oint_{|z|=2} \frac{d z}{z^{2}-1}=2 \pi i\left(\frac{1}{2}-\frac{1}{2}\right)=0 .\nonumber \]

Evaluate \(\oint_{|z|=3} \frac{z^{2}+1}{(z-1)^{2}(z+2)} d z\).

###### Solution

In this example there are two poles \(z=1,-2\) inside the contour. [See Figure \(\PageIndex{21}\).] \(z=1\) is a second order pole and \(z=-2\) is a simple pole. Therefore, we need to compute the residues at each pole of \(f(z)=\frac{z^{2}+1}{(z-1)^{2}(z+2)}\) :

\[\begin{align} \operatorname{Res}[f(z) ; z=1] &=\lim _{z \rightarrow 1} \frac{1}{1 !} \frac{d}{d z}\left[(z-1)^{2} \frac{z^{2}+1}{(z-1)^{2}(z+2)}\right]\nonumber \\ &=\lim _{z \rightarrow 1}\left(\frac{z^{2}+4 z-1}{(z+2)^{2}}\right)\nonumber \\ &=\frac{4}{9} .\label{eq:33} \\ \operatorname{Res}[f(z) ; z=-2] &=\lim _{z \rightarrow-2}(z+2) \frac{z^{2}+1}{(z-1)^{2}(z+2)} \\ &=\lim _{z \rightarrow-2} \frac{z^{2}+1}{(z-1)^{2}}\nonumber \\ &=\frac{5}{9} .\label{eq:34} \end{align} \]

The evaluation of the integral is found by computing \(2 \pi i\) times the sum of the residues:

\[\oint_{|z|=3} \frac{z^{2}+1}{(z-1)^{2}(z+2)} d z=2 \pi i\left(\frac{4}{9}+\frac{5}{9}\right)=2 \pi i \text {. }\nonumber \]

Compute \(\oint_{|z|=2} z^{3} e^{2 / z} d z\).

###### Solution

In this case, \(z=0\) is an essential singularity and is inside the contour. A Laurent series expansion about \(z=0\) gives

\[\begin{align} z^{3} e^{2 / z} &=z^{3} \sum_{n=0}^{\infty} \frac{1}{n !}\left(\frac{2}{z}\right)^{n}\nonumber \\ &=\sum_{n=0}^{\infty} \frac{2^{n}}{n !} z^{3-n}\nonumber \\ &=z^{3}+\frac{2}{2 !} z^{2}+\frac{4}{3 !} z+\frac{8}{4 !}+\frac{16}{5 ! z}+\ldots\label{eq:35} \end{align} \]

The residue is the coefficient of \(z^{-1}\), or \(\operatorname{Res}\left[z^{3} e^{2 / z} ; z=0\right]=-\frac{2}{15}\). Therefore,

\[\oint_{|z|=2} z^{3} e^{2 / z} d z=\frac{4}{15} \pi i .\nonumber \]

Evaluate \(\int_{0}^{2 \pi} \frac{d \theta}{2+\cos \theta}\).

###### Solution

Here we have a real integral in which there are no signs of complex functions. In fact, we could apply simpler methods from a calculus course to do this integral, attempting to write \(1+\cos \theta=2 \cos ^{2} \frac{\theta}{2}\). However, we do not get very far.

One trick, useful in computing integrals whose integrand is in the form \(f(\cos \theta, \sin \theta)\), is to transform the integration to the complex plane through the transformation \(z=e^{i \theta}\). Then,

\[\begin{aligned} &\cos \theta=\frac{e^{i \theta}+e^{-i \theta}}{2}=\frac{1}{2}\left(z+\frac{1}{z}\right), \\ &\sin \theta=\frac{e^{i \theta}-e^{-i \theta}}{2 i}=-\frac{i}{2}\left(z-\frac{1}{z}\right) . \end{aligned} \nonumber \]

Computation of integrals of functions of sines and cosines, \(f(\cos \theta, \sin \theta)\).

Under this transformation, \(z=e^{i \theta}\), the integration now takes place around the unit circle in the complex plane. Noting that \(d z=i e^{i \theta} d \theta=i z d \theta\), we have

\[\begin{align} \int_{0}^{2 \pi} \frac{d \theta}{2+\cos \theta} &=\oint_{|z|=1} \frac{\frac{d z}{i z}}{2+\frac{1}{2}\left(z+\frac{1}{z}\right)}\nonumber \\ &=-i \oint_{|z|=1} \frac{d z}{2 z+\frac{1}{2}\left(z^{2}+1\right)}\nonumber \\ &=-2 i \oint_{|z|=1} \frac{d z}{z^{2}+4 z+1} .\label{eq:36} \end{align} \]

We can apply the Residue Theorem to the resulting integral. The singularities occur at the roots of \(z^{2}+4 z+1=0\). Using the quadratic formula, we have the roots \(z=-2 \pm \sqrt{3}\).

The location of these poles are shown in Figure \(\PageIndex{22}\). Only \(z=-2+\sqrt{3}\) lies inside the integration contour. We will therefore need the residue of \(f(z)=\frac{-2 i}{z^{2}+4 z+1}\) at this simple pole:

\[\begin{align} \operatorname{Res}[f(z) ; z=-2+\sqrt{3}] &=\lim _{z \rightarrow-2+\sqrt{3}}(z-(-2+\sqrt{3})) \frac{-2 i}{z^{2}+4 z+1}\nonumber \\ &=-2 i \lim _{z \rightarrow-2+\sqrt{3}} \frac{z-(-2+\sqrt{3})}{(z-(-2+\sqrt{3}))(z-(-2-\sqrt{3}))}\nonumber \\ &=-2 i \lim _{z \rightarrow-2+\sqrt{3}} \frac{1}{z-(-2-\sqrt{3})}\nonumber \\ &=\frac{-2 i}{-2+\sqrt{3}-(-2-\sqrt{3})}\nonumber \\ &=\frac{-i}{\sqrt{3}}\nonumber \\ &=\frac{-i \sqrt{3}}{3} .\label{eq:37} \end{align} \]

Therefore, we have

\[\int_{0}^{2 \pi} \frac{d \theta}{2+\cos \theta}=-2 i \oint_{|z|=1} \frac{d z}{z^{2}+4 z+1}=2 \pi i\left(\frac{-i \sqrt{3}}{3}\right)=\frac{2 \pi \sqrt{3}}{3} .\label{eq:38} \]

Before moving on to further applications, we note that there is another way to compute the integral in the last example. Karl Theodor Wilhelm Weierstraß (1815-1897) introduced a substitution method for computing integrals involving rational functions of sine and cosine. One makes the substitution \(t=\tan \frac{\theta}{2}\) and converts the integrand into a rational function of \(t\). One can show that this substitution implies that

\[\sin \theta=\frac{2 t}{1+t^{2}}, \quad \cos \theta=\frac{1-t^{2}}{1+t^{2}},\nonumber \]

and

\[d \theta=\frac{2 d t}{1+t^{2}} .\nonumber \]

The details are left for Problem 8 and apply the method. In order to see how it works, we will redo the last problem.

Apply the Weierstraj substitution method to compute \(\int_{0}^{2 \pi} \frac{d \theta}{2+\cos \theta}\).

###### Solution

\[\begin{align} \int_{0}^{2 \pi} \frac{d \theta}{2+\cos \theta} &=\int_{-\infty}^{\infty} \frac{1}{2+\frac{1-t^{2}}{1+t^{2}}} \frac{2 d t}{1+t^{2}}\nonumber \\ &=2 \int_{-\infty}^{\infty} \frac{d t}{t^{2}+3}\nonumber \\ &=\frac{2}{3} \sqrt{3}\left[\tan ^{-1}\left(\frac{\sqrt{3}}{3} t\right)\right]_{-\infty}^{\infty}=\frac{2 \pi \sqrt{3}}{3} .\label{eq:39} \end{align} \]

## Infinite Integrals

Infinite integrals of the Form \(\int_{-\infty}^{\infty} f(x) d x\) occur often in physics. They can represent wave packets, wave diffraction, Fourier transforms, and arise in other applications. In this section we will see that such integrals may be computed by extending the integration to a contour in the complex plane.

Recall from your calculus experience that these integrals are improper integrals and the way that one determines if improper integrals exist, or converge, is to carefully compute these integrals using limits such as

\[\int_{-\infty}^{\infty} f(x) d x=\lim _{R \rightarrow \infty} \int_{-R}^{R} f(x) d x .\nonumber \]

For example, we evaluate the integral of \(f(x)=x\) as

\[\int_{-\infty}^{\infty} x d x=\lim _{R \rightarrow \infty} \int_{-R}^{R} x d x=\lim _{R \rightarrow \infty}\left(\frac{R^{2}}{2}-\frac{(-R)^{2}}{2}\right)=0 .\nonumber \]

One might also be tempted to carry out this integration by splitting the integration interval, \((-\infty, 0] \cup[0, \infty)\). However, the integrals \(\int_{0}^{\infty} x d x\) and \(\int_{-\infty}^{0} x d x\) do not exist. A simple computation confirms this.

\[\int_{0}^{\infty} x d x=\lim _{R \rightarrow \infty} \int_{0}^{R} x d x=\lim _{R \rightarrow \infty}\left(\frac{R^{2}}{2}\right)=\infty .\nonumber \]

Therefore,

\[\int_{-\infty}^{\infty} f(x) d x=\int_{-\infty}^{0} f(x) d x+\int_{0}^{\infty} f(x) d x\nonumber \]

does not exist while \(\lim _{R \rightarrow \infty} \int_{-R}^{R} f(x) d x\) does exist. We will be interested in computing the latter type of integral. Such an integral is called the Cauchy Principal Value Integral and is denoted with either a \(P, P V\), or a bar through the integral:

\[P \int_{-\infty}^{\infty} f(x) d x=P V \int_{-\infty}^{\infty} f(x) d x=f_{-\infty}^{\infty} f(x) d x=\lim _{R \rightarrow \infty} \int_{-R}^{R} f(x) d x .\label{eq:40} \]

If there is a discontinuity in the integral, one can further modify this definition of principal value integral to bypass the singularity. For example, if \(f(x)\) is continuous on \(a \leq x \leq b\) and not defined at \(x=x_{0} \in[a, b]\), then

\[\int_{a}^{b} f(x) d x=\lim _{\epsilon \rightarrow 0}\left(\int_{a}^{x_{0}-\epsilon} f(x) d x+\int_{x_{0}+\epsilon}^{b} f(x) d x\right) .\nonumber \]

In our discussions we will be computing integrals over the real line in the Cauchy principal value sense.

Compute \(\int_{-1}^{1} \frac{d x}{x^{3}}\) in the Cauchy Principal Value sense.

###### Solution

In this case, \(f(x)=\frac{1}{x^{3}}\) is not defined at \(x=0\). So, we have

\[\begin{align} \int_{-1}^{1} \frac{d x}{x^{3}} &=\lim _{\epsilon \rightarrow 0}\left(\int_{-1}^{-\epsilon} \frac{d x}{x^{3}}+\int_{\epsilon}^{1} \frac{d x}{x^{3}}\right)\nonumber \\ &=\lim _{\epsilon \rightarrow 0}\left(-\left.\frac{1}{2 x^{2}}\right|_{-1} ^{-\epsilon}-\left.\frac{1}{2 x^{2}}\right|_{\epsilon} ^{1}\right)=0 .\label{eq:41} \end{align} \]

Computation of real integrals by embedding the problem in the complex plane.

We now proceed to the evaluation of principal value integrals using complex integration methods. We want to evaluate the integral \(\int_{-\infty}^{\infty} f(x) d x\). We will extend this into an integration in the complex plane. We extend \(f(x)\) to \(f(z)\) and assume that \(f(z)\) is analytic in the upper half plane \((\operatorname{Im}(z)>0)\) except at isolated poles. We then consider the integral \(\int_{-R}^{R} f(x) d x\) as an integral over the interval \((-R, R)\). We view this interval as a piece of a larger contour \(C_{R}\) obtained by completing the contour with a semicircle \(\Gamma_{R}\) of radius \(R\) extending into the upper half plane as shown in Figure \(\PageIndex{23}\). Note, a similar construction is sometimes needed extending the integration into the lower half plane \((\operatorname{Im}(z)<0)\) as we will later see.

The integral around the entire contour \(C_{R}\) can be computed using the Residue Theorem and is related to integrations over the pieces of the contour by

\[\oint_{C_{R}} f(z) d z=\int_{\Gamma_{R}} f(z) d z+\int_{-R}^{R} f(z) d z .\label{eq:42} \]

Taking the limit \(R \rightarrow \infty\) and noting that the integral over \((-R, R)\) is the desired integral, we have

\[P \int_{-\infty}^{\infty} f(x) d x=\oint_{C} f(z) d z-\lim _{R \rightarrow \infty} \int_{\Gamma_{R}} f(z) d z,\label{eq:43} \]

where we have identified \(C\) as the limiting contour as \(R\) gets large.

Now the key to carrying out the integration is that the second integral vanishes in the limit. This is true if \(R|f(z)| \rightarrow 0\) along \(\Gamma_{R}\) as \(R \rightarrow \infty\). This can be seen by the following argument. We parametrize the contour \(\Gamma_{R}\) using \(z=R e^{i \theta}\). Then, when \(|f(z)|<M(R)\),

\[\begin{align} \left|\int_{\Gamma_{R}} f(z) d z\right| &=\left|\int_{0}^{2 \pi} f\left(R e^{i \theta}\right) R e^{i \theta} d \theta\right|\nonumber \\ & \leq R \int_{0}^{2 \pi}\left|f\left(R e^{i \theta}\right)\right| d \theta\nonumber \\ &<R M(R) \int_{0}^{2 \pi} d \theta\nonumber \\ &=2 \pi R M(R) .\label{eq:44} \end{align} \]

So, if \(\lim _{R \rightarrow \infty} R M(R)=0\), then \(\lim _{R \rightarrow \infty} \int_{\Gamma_{R}} f(z) d z=0\).

We now demonstrate how to use complex integration methods in evaluating integrals over real valued functions.

Evaluate \(\int_{-\infty}^{\infty} \frac{d x}{1+x^{2}}\).

###### Solution

We already know how to do this integral using calculus without complex analysis. We have that

\[\int_{-\infty}^{\infty} \frac{d x}{1+x^{2}}=\lim _{R \rightarrow \infty}\left(2 \tan ^{-1} R\right)=2\left(\frac{\pi}{2}\right)=\pi .\nonumber \]

We will apply the methods of this section and confirm this result. The needed contours are shown in Figure \(\PageIndex{24}\) and the poles of the integrand are at \(z=\pm i\). We first write the integral over the bounded contour \(C_{R}\) as the sum of an integral from \(-R\) to \(R\) along the real axis plus the integral over the semicircular arc in the upper half complex plane,

\[\int_{C_{R}} \frac{d z}{1+z^{2}}=\int_{-R}^{R} \frac{d x}{1+x^{2}}+\int_{\Gamma_{R}} \frac{d z}{1+z^{2}} .\nonumber \]

Next, we let \(R\) get large.

We first note that \(f(z)=\frac{1}{1+z^{2}}\) goes to zero fast enough on \(\Gamma_{R}\) as \(R\) gets large.

\[R|f(z)|=\frac{R}{\mid 1+R^{2} e^{2 i \theta \mid}}=\frac{R}{\sqrt{1+2 R^{2} \cos \theta+R^{4}}} .\nonumber \]

Thus, as \(R \rightarrow \infty, R|f(z)| \rightarrow 0\) and \(C_{R} \rightarrow\) C. So,

\[\int_{-\infty}^{\infty} \frac{d x}{1+x^{2}}=\oint_{C} \frac{d z}{1+z^{2}} .\nonumber \]

We need only compute the residue at the enclosed pole, \(z=i\).

\[\operatorname{Res}[f(z) ; z=i]=\lim _{z \rightarrow i}(z-i) \frac{1}{1+z^{2}}=\lim _{z \rightarrow i} \frac{1}{z+i}=\frac{1}{2 i} \text {. }\nonumber \]

Then, using the Residue Theorem, we have

\[\int_{-\infty}^{\infty} \frac{d x}{1+x^{2}}=2 \pi i\left(\frac{1}{2 i}\right)=\pi .\nonumber \]

Evaluate \(P \int_{-\infty}^{\infty} \frac{\sin x}{x} d x\).

###### Solution

For this example the integral is unbounded at \(z=0\). Constructing the contours as before we are faced for the first time with a pole lying on the contour. We cannot ignore this fact. We can proceed with the computation by carefully going around the pole with a small semicircle of radius \(\epsilon\), as shown in Figure \(\PageIndex{25}\). Then the principal value integral computation becomes

\[P \int_{-\infty}^{\infty} \frac{\sin x}{x} d x=\lim _{\epsilon \rightarrow 0, R \rightarrow \infty}\left(\int_{-R}^{-\epsilon} \frac{\sin x}{x} d x+\int_{\epsilon}^{R} \frac{\sin x}{x} d x\right) .\label{eq:45} \]

We will also need to rewrite the sine function in term of exponentials in this integral. There are two approaches that we could take. First, we could employ the definition of the sine function in terms of complex exponentials. This gives two integrals to compute:

\[P \int_{-\infty}^{\infty} \frac{\sin x}{x} d x=\frac{1}{2 i}\left(P \int_{-\infty}^{\infty} \frac{e^{i x}}{x} d x-P \int_{-\infty}^{\infty} \frac{e^{-i x}}{x} d x\right) .\label{eq:46} \]

The other approach would be to realize that the sine function is the imaginary part of an exponential, \(\operatorname{Im} e^{i x}=\sin x\). Then, we would have

\[P \int_{-\infty}^{\infty} \frac{\sin x}{x} d x=\operatorname{Im}\left(P \int_{-\infty}^{\infty} \frac{e^{i x}}{x} d x\right) .\label{eq:47} \]

We first consider \(P \int_{-\infty}^{\infty} \frac{e^{i x}}{x} d x\), which is common to both approaches. We use the contour in Figure \(\PageIndex{25}\). Then we have

\[\oint_{\mathcal{C}_{R}} \frac{e^{i z}}{z} d z=\int_{\Gamma_{R}} \frac{e^{i z}}{z} d z+\int_{-R}^{-\epsilon} \frac{e^{i z}}{z} d z+\int_{\mathcal{C}_{\epsilon}} \frac{e^{i z}}{z} d z+\int_{\epsilon}^{R} \frac{e^{i z}}{z} d z .\nonumber \]

The integral \(\oint_{C_{R}} \frac{e^{i z}}{z} d z\) vanishes since there are no poles enclosed in the contour! The sum of the second and fourth integrals gives the integral we seek as \(\epsilon \rightarrow 0\) and \(R \rightarrow \infty\). The integral over \(\Gamma_{R}\) will vanish as \(R\) gets large according to Jordan’s Lemma.

Jordan’s Lemma give conditions as when integrals over \(\Gamma_{R}\) will vanish as \(R\) gets large. We state a version of Jordan’s Lemma here for reference and give a proof is at the end of this chapter.

If \(f(z)\) converges uniformly to zero as \(z \rightarrow \infty\), then

\[\lim_{R\to\infty}\int_{C_R} f(z)e^{ikz}dz=0\nonumber \]

where \(k>0\) and \(C_{R}\) is the upper half of the circle \(|z|=R .\)

A similar result applies for \(k<0\), but one closes the contour in the lower half plane. [See Section 8.5.8 for the proof of Jordan’s Lemma.]

The remaining integral around the small semicircular arc has to be done separately. We have

\[\int_{C_{\epsilon}} \frac{e^{i z}}{z} d z=\int_{\pi}^{0} \frac{\exp \left(i \epsilon e^{i \theta}\right)}{\epsilon e^{i \theta}} i \epsilon e^{i \theta} d \theta=-\int_{0}^{\pi} i \exp \left(i \epsilon e^{i \theta}\right) d \theta\nonumber \]

Taking the limit as \(\epsilon\) goes to zero, the integrand goes to \(i\) and we have

\[\int_{C_{\epsilon}} \frac{e^{i z}}{z} d z=-\pi i .\nonumber \]

So far, we have that

\[P \int_{-\infty}^{\infty} \frac{e^{i x}}{x} d x=-\lim _{\epsilon \rightarrow 0} \int_{C_{\epsilon}} \frac{e^{i z}}{z} d z=\pi i .\nonumber \]

At this point we can get the answer using the second approach in Equation \(\eqref{eq:47}\). Namely,

\[P \int_{-\infty}^{\infty} \frac{\sin x}{x} d x=\operatorname{Im}\left(P \int_{-\infty}^{\infty} \frac{e^{i x}}{x} d x\right)=\operatorname{Im}(\pi i)=\pi\label{eq:48} \]

It is instructive to carry out the first approach in Equation \(\eqref{eq:46}\). We will need to compute \(P \int_{-\infty}^{\infty} \frac{e^{-i x}}{x} d x\). This is done in a similar to the above computation, being careful with the sign changes due to the orientations of the contours as shown in Figure \(\PageIndex{26}\).

We note that the contour is closed in the lower half plane. This is because \(k<0\) in the application of Jordan’s Lemma. One can understand why this is the case from the following observation. Consider the exponential in Jordan’s Lemma. Let \(z=z_{R}+i z_{I}\). Then,

\[e^{i k z}=e^{i k\left(z_{R}+i z_{l}\right)}=e^{-k z_{I}} e^{i k z_{R}} .\nonumber \]

As \(|z|\) gets large, the second factor just oscillates. The first factor would go to zero if \(k z_{I}>0\). So, if \(k>0\), we would close the contour in the upper half plane. If \(k<0\), then we would close the contour in the lower half plane. In the current computation, \(k=-1\), so we use the lower half plane.

Working out the details, we find the same value for

\[P \int_{-\infty}^{\infty} \frac{e^{-i x}}{x} d x=\pi i .\nonumber \]

Finally, we can compute the original integral as

\[\begin{align} P \int_{-\infty}^{\infty} \frac{\sin x}{x} d x &=\frac{1}{2 i}\left(P \int_{-\infty}^{\infty} \frac{e^{i x}}{x} d x-P \int_{-\infty}^{\infty} \frac{e^{-i x}}{x} d x\right)\nonumber \\ &=\frac{1}{2 i}(\pi i+\pi i)\nonumber \\ &=\pi\label{eq:49} \end{align} \]

This is the same result as we obtained using Equation \(\eqref{eq:47}\).

Note that we have not previously done integrals in which a singularity lies on the contour. One can show, as in this example, that points on the contour can be accounted for by using half of a residue (times \(2 \pi i\) ). For the semicircle \(C_{e}\) you can verify this. The negative sign comes from going clockwise around the semicircle.

Evaluate \(\oint_{|z|=1} \frac{d z}{z^{2}+1}\).

###### Solution

In this example there are two simple poles, \(z=\pm i\) lying on the contour, as seen in Figure \(\PageIndex{27}\). This problem is similar to Problem 1c, except we will do it using contour integration instead of a parametrization. We bypass the two poles by drawing small semicircles around them. Since the poles are not included in the closed contour, then the Residue Theorem tells us that the integral over the path vanishes. We can write the full integration as a sum over three paths, \(C_{\pm}\)for the semicircles and \(\mathrm{C}\) for the original contour with the poles cut out. Then we take the limit as the semicircle radii go to zero. So,

\[0=\int_{C} \frac{d z}{z^{2}+1}+\int_{C_{+}} \frac{d z}{z^{2}+1}+\int_{C_{-}} \frac{d z}{z^{2}+1} .\nonumber \]

The integral over the semicircle around i can be done using the parametrization \(z=i+\epsilon e^{i \theta}\). Then \(z^{2}+1=2 i \epsilon e^{i \theta}+\epsilon^{2} e^{2 i \theta}\). This gives

\[\int_{C_{+}} \frac{d z}{z^{2}+1}=\lim _{\epsilon \rightarrow 0} \int_{0}^{-\pi} \frac{i \epsilon e^{i \theta}}{2 i \epsilon e^{i \theta}+\epsilon^{2} e^{2 i \theta}} d \theta=\frac{1}{2} \int_{0}^{-\pi} d \theta=-\frac{\pi}{2} .\nonumber \]

As in the last example, we note that this is just \(\pi i\) times the residue, \(\operatorname{Res}\left[\frac{1}{z^{2}+1} ; z=i\right]=\) \(\frac{1}{21}\). Since the path is traced clockwise, we find the contribution is \(-\pi i \operatorname{Res}=-\frac{\pi}{2}\), which is what we obtained above. A Similar computation will give the contribution from \(z=-i\) as \(\frac{\pi}{2}\). Adding these values gives the total contribution from \(C_{\pm}\)as zero. So, the final result is that

\[\oint_{|z|=1} \frac{d z}{z^{2}+1}=0 .\nonumber \]

Evaluate \(\int_{-\infty}^{\infty} \frac{e^{a x}}{1+e^{x}} d x\), for \(0<a<1\).

###### Solution

In dealing with integrals involving exponentials or hyperbolic functions it is sometimes useful to use different types of contours. This example is one such case. We will replace \(x\) with \(z\) and integrate over the contour in Figure \(\PageIndex{28}\). Letting \(R \rightarrow \infty\), the integral along the real axis is the integral that we desire. The integral along the path for \(y=2 \pi\) leads to a multiple of this integral since \(z=x+2 \pi i\) along this path. Integration along the vertical paths vanish as \(R \rightarrow \infty\). This is captured in the following integrals:

\[\begin{align} \oint_{C_{R}} \frac{e^{a z}}{1+e^{z}} d z=& \int_{-R}^{R} \frac{e^{a x}}{1+e^{x}} d x+\int_{0}^{2 \pi} \frac{e^{a(R+i y)}}{1+e^{R+i y}} d y\nonumber \\ &+\int_{R}^{-R} \frac{e^{a(x+2 \pi i)}}{1+e^{x+2 \pi i}} d x+\int_{2 \pi}^{0} \frac{e^{a(-R+i y)}}{1+e^{-R+i y}} d y\label{eq:50} \end{align} \]

We can now let \(R \rightarrow \infty\). For large \(R\) the second integral decays as \(e^{(a-1) R}\) and the fourth integral decays as \(e^{-a R}\). Thus, we are left with

\[\begin{align} \oint_{C} \frac{e^{a z}}{1+e^{z}} d z &=\lim _{R \rightarrow \infty}\left(\int_{-R}^{R} \frac{e^{a x}}{1+e^{x}} d x-e^{2 \pi i a} \int_{-R}^{R} \frac{e^{a x}}{1+e^{x}} d x\right)\nonumber \\ &=\left(1-e^{2 \pi i a}\right) \int_{-\infty}^{\infty} \frac{e^{a x}}{1+e^{x}} d x .\label{eq:51} \end{align} \]

We need only evaluate the left contour integral using the Residue Theorem. The poles are found from

\[1+e^{z}=0 .\nonumber \]

Within the contour, this is satisfied by \(z=i \pi\). So,

\[\operatorname{Res}\left[\frac{e^{a z}}{1+e^{z}} ; z=i \pi\right]=\lim _{z \rightarrow i \pi}(z-i \pi) \frac{e^{a z}}{1+e^{z}}=-e^{i \pi a} .\nonumber \]

Applying the Residue Theorem, we have

\[\left(1-e^{2 \pi i a}\right) \int_{-\infty}^{\infty} \frac{e^{a x}}{1+e^{x}} d x=-2 \pi i e^{i \pi a} .\nonumber \]

Therefore, we have found that

\[\int_{-\infty}^{\infty} \frac{e^{a x}}{1+e^{x}} d x=\frac{-2 \pi i e^{i \pi a}}{1-e^{2 \pi i a}}=\frac{\pi}{\sin \pi a}, \quad 0<a<1 .\nonumber \]

## Integration Over Multivalued Functions

We have seen that some complex functions inherently possess multivaluedness; i.e., such "functions" do not evaluate to a single value, but have many values. The key examples were \(f(z)=z^{1 / n}\) and \(f(z)=\ln z\). The \(n\)th roots have \(n\) distinct values and logarithms have an infinite number of values as determined by the range of the resulting arguments. We mentioned that the way to handle multivaluedness is to assign different branches to these functions, introduce a branch cut and glue them together at the branch cuts to form Riemann surfaces. In this way we can draw continuous paths along the Riemann surfaces as we move from one Riemann sheet to another.

Before we do examples of contour integration involving multivalued functions, lets first try to get a handle on multivaluedness in a simple case. We will consider the square root function,

\[w=z^{1 / 2}=r^{1 / 2} e^{i\left(\frac{\theta}{2}+k \pi\right)}, \quad k=0,1 .\nonumber \]

There are two branches, corresponding to each \(k\) value. If we follow a path not containing the origin, then we stay in the same branch, so the final argument \((\theta)\) will be equal to the initial argument. However, if we follow a path that encloses the origin, this will not be true. In particular, for an initial point on the unit circle, \(z_{0}=e^{i \theta_{0}}\), we have its image as \(w_{0}=e^{i \theta_{0} / 2}\). However, if we go around a full revolution, \(\theta=\theta_{0}+2 \pi\), then

\[z_{1}=e^{i \theta_{0}+2 \pi i}=e^{i \theta_{0}},\nonumber \]

but

\[w_{1}=e^{\left(i \theta_{0}+2 \pi i\right) / 2}=e^{i \theta_{0} / 2} e^{\pi i} \neq w_{0} .\nonumber \]

Here we obtain a final argument \((\theta)\) that is not equal to the initial argument! Somewhere, we have crossed from one branch to another. Points, such as the origin in this example, are called branch points. Actually, there are two branch points, because we can view the closed path around the origin as a closed path around complex infinity in the compactified complex plane. However, we will not go into that at this time.

We can demonstrate this in the following figures. In Figure \(\PageIndex{29}\) we show how the points A-E are mapped from the \(z\)-plane into the \(w\)-plane under the square root function for the principal branch, \(k=0\). As we trace out the unit circle in the \(z\)-plane, we only trace out a semicircle in the \(w\)-plane. If we consider the branch \(k=1\), we then trace out a semicircle in the lower half plane, as shown in Figure \(\PageIndex{30}\) following the points from \(\mathrm{F}\) to \(\mathrm{J}\).

We can combine these into one mapping depicting how the two complex planes corresponding to each branch provide a mapping to the w-plane. This is shown in Figure \(\PageIndex{31}\).

A common way to draw this domain, which looks like two separate complex planes, would be to glue them together. Imagine cutting each plane along the positive \(x\)-axis, extending between the two branch points, \(z=0\) and \(z=\infty\). As one approaches the cut on the principal branch, then one can move onto the glued second branch. Then one continues around the origin on this branch until one once again reaches the cut. This cut is glued to the principal branch in such a way that the path returns to its starting point. The resulting surface we obtain is the Riemann surface shown in Figure \(\PageIndex{32}\). Note that there is nothing that forces us to place the branch cut at a particular place. For example, the branch cut could be along the positive real axis, the negative real axis, or any path connecting the origin and complex infinity.

We now look at examples involving integrals of multivalued functions.

Evaluate \(\int_{0}^{\infty} \frac{\sqrt{x}}{1+x^{2}} d x\).

###### Solution

We consider the contour integral \(\oint_{C} \frac{\sqrt{z}}{1+z^{2}} d z\).

The first thing we can see in this problem is the square root function in the integrand. Being there is a multivalued function, we locate the branch point and determine where to draw the branch cut. In Figure \(\PageIndex{33}\) we show the contour that we will use in this problem. Note that we picked the branch cut along the positive \(x\)-axis.

We take the contour \(C\) to be positively oriented, being careful to enclose the two poles and to hug the branch cut. It consists of two circles. The outer circle \(C_{R}\) is a circle of radius \(R\) and the inner circle \(C_{e}\) will have a radius of \(\epsilon\). The sought answer will be obtained by letting \(R \rightarrow \infty\) and \(\epsilon \rightarrow 0\). On the large circle we have that the integrand goes to zero fast enough as \(R \rightarrow \infty\). The integral around the small circle vanishes as \(\epsilon \rightarrow 0\). We can see this by parametrizing the circle as \(z=\epsilon e^{i \theta}\) for \(\theta \in[0,2 \pi]\) :

\[\begin{align} \oint_{C_{\epsilon}} \frac{\sqrt{z}}{1+z^{2}} d z &=\int_{0}^{2 \pi} \frac{\sqrt{\epsilon e^{i \theta}}}{1+\left(\epsilon e^{i \theta}\right)^{2}} i \epsilon e^{i \theta} d \theta\nonumber \\ &=i \epsilon^{3 / 2} \int_{0}^{2 \pi} \frac{e^{3 i \theta / 2}}{1+\left(\epsilon^{2} e^{2 i \theta}\right)} d \theta\label{eq:52} \end{align} \]

It should now be easy to see that as \(\epsilon \rightarrow 0\) this integral vanishes.

The integral above the branch cut is the one we are seeking, \(\int_{0}^{\infty} \frac{\sqrt{x}}{1+x^{2}} d x\). The integral under the branch cut, where \(z=r e^{2 \pi i}\), is

\[\begin{align} \int \frac{\sqrt{z}}{1+z^{2}} d z &=\int_{\infty}^{0} \frac{\sqrt{r e^{2 \pi i}}}{1+r^{2} e^{4 \pi i}} d r\nonumber \\ &=\int_{0}^{\infty} \frac{\sqrt{r}}{1+r^{2}} d r .\label{eq:53} \end{align} \]

We note that this is the same as that above the cut.

Up to this point, we have that the contour integral, as \(R \rightarrow \infty\) and \(\epsilon \rightarrow 0\) is

\[\oint_{C} \frac{\sqrt{z}}{1+z^{2}} d z=2 \int_{0}^{\infty} \frac{\sqrt{x}}{1+x^{2}} d x .\nonumber \]

In order to finish this problem, we need the residues at the two simple poles.

\[\begin{aligned} \operatorname{Res}\left[\frac{\sqrt{z}}{1+z^{2}} ; z=i\right] &=\frac{\sqrt{i}}{2 i}=\frac{\sqrt{2}}{4}(1+i), \\ \operatorname{Res}\left[\frac{\sqrt{z}}{1+z^{2}} ; z=-i\right] &=\frac{\sqrt{-i}}{-2 i}=\frac{\sqrt{2}}{4}(1-i) . \end{aligned} \nonumber \]

So,

\[2 \int_{0}^{\infty} \frac{\sqrt{x}}{1+x^{2}} d x=2 \pi i\left(\frac{\sqrt{2}}{4}(1+i)+\frac{\sqrt{2}}{4}(1-i)\right)=\pi \sqrt{2} \text {. }\nonumber \]

Finally, we have the value of the integral that we were seeking,

\[\int_{0}^{\infty} \frac{\sqrt{x}}{1+x^{2}} d x=\frac{\pi \sqrt{2}}{2} \text {. }\nonumber \]

Compute \(\int_{a}^{\infty} f(x) d x\) using contour integration involving logarithms.\(^{2}\)

In this example we will apply contour integration to the integral

\[\oint_{C} f(z) \ln (a-z) d z\nonumber \]

for the contour shown in Figure \(\PageIndex{34}\).

We will assume that \(f(z)\) is single valued and vanishes as \(|z| \rightarrow \infty\). We will choose the branch cut to span from the origin along the positive real axis. Employing the Residue Theorem and breaking up the integrals over the pieces of the contour in Figure \(\PageIndex{34}\), we have schematically that

\[2 \pi i \sum \operatorname{Res}[f(z) \ln (a-z)]=\left(\int_{C_{1}}+\int_{C_{2}}+\int_{C_{3}}+\int_{C_{4}}\right) f(z) \ln (a-z) d z\nonumber \]

First of all, we assume that \(f(z)\) is well behaved at \(z=a\) and vanishes fast enough as \(|z|=R \rightarrow \infty\). Then, the integrals over \(C_{2}\) and \(C_{4}\) will vanish. For example, for the path \(\mathrm{C}_{4}\), we let \(z=a+\epsilon e^{i \theta}, 0<\theta<2 \pi\). Then,

\[\int_{C_{4}} f(z) \ln (a-z) d z .=\lim _{\epsilon \rightarrow 0} \int_{2 \pi}^{0} f\left(a+\epsilon e^{i \theta}\right) \ln \left(\epsilon e^{i \theta}\right) i \epsilon e^{i \theta} d \theta .\nonumber \]

If \(f(a)\) is well behaved, then we only need to show that \(\lim _{\epsilon \rightarrow 0} \epsilon \ln \epsilon=0\). This is left to the reader.

Similarly, we consider the integral over \(C_{2}\) as \(R\) gets large,

\[\int_{C_{2}} f(z) \ln (a-z) d z=\lim _{R \rightarrow \infty} \int_{0}^{2 \pi} f\left(R e^{i \theta}\right) \ln \left(R e^{i \theta}\right) i \operatorname{Re}^{i \theta} d \theta \text {. }\nonumber \]

Thus, we need only require that

\[\lim _{R \rightarrow \infty} R \ln R\left|f\left(R e^{i \theta}\right)\right|=0 .\nonumber \]

Next, we consider the two straight line pieces. For \(C_{1}\), the integration along the real axis occurs for \(z=x\), so

\[\int_{C_{1}} f(z) \ln (a-z) d z=\int_{a}^{\infty} f(x) \ln (a-x) d z .\nonumber \]

However, integration over \(C_{3}\) requires noting that we need the branch for the logarithm such that \(\ln z=\ln (a-x)+2 \pi i\). Then,

\[\int_{\mathcal{C}_{3}} f(z) \ln (a-z) d z=\int_{\infty}^{a} f(x)[\ln (a-x)+2 \pi i] d z .\nonumber \]

Combining these results, we have

\[\begin{align} 2 \pi i \sum \operatorname{Res}[f(z) \ln (a-z)]=& \int_{a}^{\infty} f(x) \ln (a-x) d z\nonumber \\ &+\int_{\infty}^{a} f(x)[\ln (a-x)+2 \pi i] d z .\nonumber \\ =&-2 \pi i \int_{a}^{\infty} f(x) d z .\label{eq:54} \end{align} \]

Therefore,

\[\int_{a}^{\infty} f(x) d x=-\sum \operatorname{Res}[f(z) \ln (a-z)]\nonumber \]

Compute \(\int_{1}^{\infty} \frac{d x}{4 x^{2}-1}\).

###### Solution

We can apply the last example to this case. We see from Figure \(\PageIndex{35}\) that the two poles at \(z=\pm \frac{1}{2}\) are inside contour \(C\). So, we compute the residues of \(\frac{\ln (1-z)}{4 z^{2}-1}\) at these poles and find that

\[\begin{align} \int_{1}^{\infty} \frac{d x}{4 x^{2}-1} &=-\operatorname{Res}\left[\frac{\ln (1-z)}{4 z^{2}-1} ; \frac{1}{2}\right]-\operatorname{Res}\left[\frac{\ln (1-z)}{4 z^{2}-1} ;-\frac{1}{2}\right]\nonumber \\ &=-\frac{\ln \frac{1}{2}}{4}+\frac{\ln \frac{3}{2}}{4}=\frac{\ln 3}{4}\label{eq:55} \end{align} \]

## Appendix: Jordan’s Lemma

For completeness, we prove Jordan's Lemma

If \(f(z)\) converges uniformly to zero as \(z \rightarrow \infty\), then

\[\lim _{R \rightarrow \infty} \int_{\mathcal{C}_{R}} f(z) e^{i k z} d z=0\nonumber \]

where \(k>0\) and \(C_{R}\) is the upper half of the circle \(|z|=R\).

**Proof**

We consider the integral

\[I_{R}=\int_{C_{R}} f(z) e^{i k z} d z,\nonumber \]

where \(k>0\) and \(C_{R}\) is the upper half of the circle \(|z|=R\) in the complex plane. Let \(z=R e^{i \theta}\) be a parametrization of \(C_{R}\). Then,

\[I_{R}=\int_{0}^{\pi} f\left(R e^{i \theta}\right) e^{i k R \cos \theta-a R \sin \theta} i R e^{i \theta} d \theta .\nonumber \]

Since

\[\lim _{|z| \rightarrow \infty} f(z)=0, \quad 0 \leq \arg z \leq \pi\nonumber \]

then for large \(|R|,|f(z)|<\epsilon\) for some \(\epsilon>0\). Then,

\[\begin{align} \left|I_{R}\right| &=\left|\int_{0}^{\pi} f\left(R e^{i \theta}\right) e^{i k R \cos \theta-a R \sin \theta} i R e^{i \theta} d \theta\right|\nonumber \\ & \leq \int_{0}^{\pi}\left|f\left(R e^{i \theta}\right)\right|\left|e^{i k R \cos \theta}\right|\left|e^{-a R \sin \theta}\right|\left|i R e^{i \theta}\right| d \theta\nonumber \\ & \leq \epsilon R \int_{0}^{\pi} e^{-a R \sin \theta} d \theta\nonumber \\ &=2 \epsilon R \int_{0}^{\pi / 2} e^{-a R \sin \theta} d \theta .\label{eq:56} \end{align} \]

The last integral still cannot be computed, but we can get a bound on it over the range \(\theta \in[0, \pi / 2]\). Note from Figure \(\PageIndex{36}\) that

\[\sin \theta \geq \frac{2}{\pi} \theta, \quad \theta \in[0, \pi / 2] .\nonumber \]

Therefore, we have

\[\left|I_{R}\right| \leq 2 \epsilon R \int_{0}^{\pi / 2} e^{-2 a R \theta / \pi} d \theta=\frac{2 \epsilon R}{2 a R / \pi}\left(1-e^{-a R}\right) .\nonumber \]

For large \(R\) we have

\[\lim _{R \rightarrow \infty}\left|I_{R}\right| \leq \frac{\pi \epsilon}{a} .\nonumber \]

So, as \(\epsilon \rightarrow 0\), the integral vanishes.