Contour integration is a powerful technique, based on complex analysis, that allows us to solve certain integrals that are otherwise hard or impossible to solve. Contour integrals also have important applications in physics, particularly in the study of waves and oscillations.
Recall that for a real function $f(x)$, the definite integral from $x=a$ to $x=b$ is the area under the curve between those two points. As previously discussed, the integral can be expressed as a limit expression: we divide the interval into $N$ segments of width $\Delta x$, take the sum of $\Delta x\, f(x_n)$, and go to the $N \rightarrow \infty$ limit:
$$\int_a^b dx\; f(x) \;=\; \lim_{N \rightarrow 0} \, \sum_{n=0}^{N} \Delta x\; f(x_n), \;\;\;\mathrm{where}\;\;\;x_n = a + n\Delta x, \;\;\Delta x = \frac{b-a}{N}.$$Now suppose $f$ is a complex function of a complex variable. A straightfoward way to define the integral of $f(z)$ is to adopt an analogous expression:
$$\lim_{N \rightarrow 0} \, \sum_{n=0}^{N} \Delta z\; f(z_n)$$But there's a conceptual snag: since $f$ takes complex inputs, the values of $z_n$ need not lie along the real line. In general, the complex numbers $z_n$ form a set of points in the complex plane. To accomodate this, we can imagine chaining together a sequence of points $z_1, z_2, \dots, z_N$, separated by displacements $\Delta z_1, \Delta z_2, \Delta z_3, \dots, \Delta z_{N-1}$:
Then the sum we are interested in is
$$\sum_{n=1}^{N-1} \Delta z_n\; f(z_n) \;=\; \Delta z_1\, f(z_1) + \Delta z_2\, f(z_2) + \cdots + \Delta z_{N-1}\, f(z_{N-1}).$$In the limit $N \rightarrow \infty$, each displacement $\Delta z_{n}$ becomes infinitesimal, and the sequence of points $\{z_1, z_2, \dots, z_N\}$ becomes a continuous trajectory in the complex plane. Such a trajectory is called a contour. Let us denote a given contour by an abstract symbol, such as $\Gamma$. Then the contour integral over $\Gamma$ is defined as
$$\int_\Gamma \, f(z)\, dz = \lim_{N \rightarrow \infty} \sum_{n=1}^{N-1} \Delta z_n\; f(z_n).$$The symbol $\Gamma$ in the subscript of the integral sign indicates that the integral takes place over the contour $\Gamma$. When defining a contour integral, it is always necessary to specify which contour we are integrating over. This is analogous to specifying the end-points of the interval over which to perform a definite real integral. In the complex case, the integration variable $z$ lies in a two-dimensional plane (the complex plane), not a line; therefore we cannot just specify two end-points, and must specify an entire contour.
Also, note that in defining a contour $\Gamma$ we must specify not just a curve in the complex plane, but also the direction along which to traverse the curve. If we integrate along the same curve in the opposite direction, the value of the contour integral switches sign (this is similar to swapping the end-points of a definite real integral).
| Additional Notes |
| A contour integral generally cannot be interpreted as the area under a curve, the way a definite real integral can. In particular, the contour should not be mistakenly interpreted the graph of the integrand! Always remember that in a contour integral, the integrand $f(z)$ and the integration variable $z$ are both complex numbers. Moreover, the concept of an indefinite integral cannot be usefully generalized to the complex case. |
Simple contour integrals can be calculated by parameterizing the contour. Consider a contour integral
$$\int_\Gamma \, dz \; f(z),$$where $f$ is a complex function of a complex variable and $\Gamma$ is a given contour. As previously discussed, we can describe a trajectory in the complex plane by a complex function of a real variable, $z(t)$:
$$\Gamma \equiv \Big\{z(t) \;\Big|\; t_1 < t < t_2\Big\}, \quad \mathrm{where}\;\; t \in \mathbb{R}, \,z(t) \in \mathbb{C}.$$The real numbers $t_1$ and $t_2$ specify two complex numbers, $z(t_1)$ and $z(t_2)$, which are the end-points of the contour. The rest of the contour consists of the values of $z(t)$ between those end-points. Provided we can parameterize $\Gamma$ in such a manner, the complex displacement $dz$ in the contour integral can be written as
$$dz \rightarrow dt\, \frac{dz}{dt}.$$Then we can express the contour integral over $\Gamma$ as a definite integral over $t$:
$$\int_\Gamma dz\; f(z) = \int_{t_1}^{t_2} \; dt\; \frac{dz}{dt}\, f\big(z(t)\big).$$This can then be calculated using standard integration techniques. A simple example is given in the next section.
Let us use the method of parameterizing the contour to calculate the contour integral
$$ \int_{\Gamma[R, \theta_1,\theta_2]} dz\; z^n,\; n\in\mathbb{Z},$$where the trajectory $\Gamma[R, \theta_1,\theta_2]$ consists of a counter-clockwise arc of radius $R > 0$, from the point $z_1 = R e^{i\theta_1}$ to the point $z_2 = R e^{i\theta_2}$:
We can parameterize the contour as follows:
$$\Gamma[R, \theta_1,\theta_2] = \big\{z(\theta) \;\big|\; \theta_1 \le \theta \le \theta_2\big\}, \quad \mathrm{where}\;\, z(\theta) = R e^{i\theta}.$$Then the contour integral can be converted into an integral over the real parameter $\theta$:
$$\begin{align}\int_{\Gamma[R, \theta_1,\theta_2]} dz\; z^n &= \int_{\theta_1}^{\theta_2} d\theta \; z^n \; \frac{dz}{d\theta} \\ &= \int_{\theta_1}^{\theta_2} d\theta \; \left( Re^{i\theta}\right)^n\, \left(iR\, e^{i\theta}\right) \\ &= i R^{n+1} \, \int_{\theta_1}^{\theta_2} d\theta \; e^{i(n+1)\theta}.\end{align}$$To proceed, there are two cases that we must treat separately. First, for $n \ne -1$,
$$\int_{\theta_1}^{\theta_2} d\theta \; e^{i(n+1)\theta} = \left[\frac{e^{i(n+1)\theta}}{i(n+1)}\right]_{\theta_1}^{\theta_2} = \frac{e^{i(n+1)\theta_2} - e^{i(n+1)\theta_1}}{i(n+1)}.$$Second, we have the case $n = -1$. This cannot be handled by the above equations, since the factor of $n + 1$ in the denominator would vanish. Instead,
$$\int_{\theta_1}^{\theta_2} d\theta \; \left[e^{i(n+1)\theta} \right]_{n=-1} \; = \int_{\theta_1}^{\theta_2} d\theta \;=\; \theta_2 - \theta_1.$$Putting the two cases together, we arrive at the result
$$\int_{\Gamma[\theta_1,\theta_2]} dz\; z^n = \left\{\begin{array}{ll}i(\theta_2 - \theta_1), & \mathrm{if}\;n = -1 \\ \displaystyle R^{n+1}\frac{e^{i(n+1)\theta_2} - e^{i(n+1)\theta_1}}{n+1},&\mathrm{if}\;n\ne -1.\end{array}\right.$$The case where $\theta_2 = \theta_1 + 2\pi$ is of particular interest. Here, $\Gamma$ forms a complete loop, and the result simplifies to
$$\oint_{\Gamma} dz\; z^n = \left\{\begin{array}{ll}2\pi i, & \mathrm{if}\;n = -1 \\ \displaystyle 0,&\mathrm{if}\;n\ne -1,\end{array}\right.$$which is independent of $R$ as well as the choice of $\theta_1$ and $\theta_2$. (Here, the special integration symbol $\oint$ is used to indicate that the contour integral is taken over a loop.) This is a very important result that we will make ample use of later.
In [ ]:
## As an exercise, let's evaluate the same integral numerically.
from numpy import linspace, exp, pi, trapz
n = -1 # Try varying this...
theta1 = 0.0
theta2 = 2*pi
R = 1.0
theta = linspace(theta1, theta2, 1000) # Values of the parameter theta
z = R * exp(1j * theta) # The corresponding values of z...
dz_dtheta = 1j * z # ... and dz/dtheta
result = trapz(z**n * dz_dtheta, theta) # Trapezium rule numerical integral.
print("Numerical integral result = {}".format(result))
prediction = 2j*pi if n == -1 else 0
print("Predicted result = {}".format(prediction))
By the way, what if $n$ is not an integer? In that case, the integrand $z^n$ is a multi-valued operation, whereas the definition of a contour integral assumes the integrand is a well-defined function. To get around this problem, we can specify a branch cut and perform the contour integral with any of the branches of $z^n$ (this is fine since the branches are well-defined functions). So long as the branch cut avoids intersecting with the contour $\Gamma$, the result obtained above by parametrizing the contour integral remain valid. However, $\Gamma$ cannot properly be taken along a complete loop, as that would entail crossing the branch cut.
A loop integral is a contour integral taken over a loop in the complex plane; i.e., with the same starting and ending point. In the previous example, we encountered the case of a circular loop integral. More generally, however, loop contours do not be circular but can have other shapes.
Loop integrals play an important role in complex analysis. This importance stems from the following property, known as Cauchy's integral theorem:
If $f(z)$ is analytic everywhere inside a loop $\Gamma$, then $\displaystyle\oint_\Gamma dz\; f(z) = 0.$
Cauchy's integral theorem can be derived from Stokes' theorem, which states that for any differentiable vector field $\vec{A}(x,y,z)$ defined within a three-dimensional space, its line integral around a loop $\Gamma$ is equal to the flux of its curl through any surface enclosed by the loop. Mathematically, this is stated as
$$\oint_\Gamma \vec{d\ell} \cdot \vec{A} = \int_{S(\Gamma)} d^2r \; \hat{n} \cdot \left(\nabla \times \vec{A}\right),$$where $S(\Gamma)$ denotes a two-dimensional surface enclosed by the loop $\Gamma$, and $\hat{n}$ denotes a normal vector sticking out of the surface at each integration point.
We only need the 2D version of Stokes' theorem, in which both the loop $\Gamma$ and the enclosed surface $S(\Gamma)$ are restricted to the $x\mathrm{-}y$ plane, and $\vec{A}(x,y)$ likewise has no $z$ component. Then Stokes' theorem simplifies to
$$\oint_\Gamma \vec{d\ell} \cdot \vec{A} = \int\!\!\!\int_{S(\Gamma)}\! dx \,dy \; \left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right).$$Now consider a loop integral
$$\oint_\Gamma dz \; f(z),$$where $\Gamma$ is a loop trajectory and $f(z)$ is some analytic function that is analytic in the domain $S(\Gamma)$. Let us decompose $f$ into its real and imaginary parts,
$$f(x+iy) = u(x,y) + iv(x,y).$$The analyticity of $f$ implies that the real functions $u$ and $v$ are differentiable and obey the Cauchy-Riemann equations. Now we can manipulate the loop integral as follows:
$$\begin{align}\oint_\Gamma dz \, f(z) &= \oint_\Gamma \left(dx + i dy\right) \; \left(u + i v\right) \\ &= \oint_\Gamma \Bigg(dx\, (u+iv) + dy\, (iu - v) \Bigg) \\ &= \oint_\Gamma \begin{bmatrix}dx\\dy\end{bmatrix} \cdot \begin{bmatrix}u + i v \\ iu - v\end{bmatrix}\\ &= \oint_\Gamma \vec{d\ell} \cdot \vec{A}.\end{align}$$The last expression is a line integral involving the complex vector field
$$\vec{A}(x,y) = \begin{bmatrix}u(x,y) + i v(x,y) \\ iu(x,y) - v(x,y)\end{bmatrix}.$$Using Stokes' theorem in 2D, we convert this into the area integral
$$\begin{align}\int\!\!\!\int_{S(\Gamma)} dx \,dy \; \left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right) &= \int\!\!\!\int_{S(\Gamma)} dx \,dy \; \left[\left(i\frac{\partial u}{\partial x} - \frac{\partial v}{\partial x}\right) - \left(\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}\right)\right] \\ &= \int\!\!\!\int_{S(\Gamma)} dx \,dy \; \left[- \left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right) + i\left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right)\right].\end{align}$$On the last line, the two terms in parenthesis are both zero because, according to the Cauchy-Riemann equations,
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\;\;\mathrm{and}\;\; \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.$$Hence, the loop integral is zero. Q.E.D.
If the integrand $f(z)$ is non-analytic somewhere inside the loop, Cauchy's integral theorem does not apply, and the loop integral need not vanish. In particular, suppose $f(z)$ vanishes at one or more discrete points inside the loop, $\{z_1, z_2, \dots\}$. Then we can show that
$$\oint_\Gamma dz\; f(z) \,=\, \sum_{n} \oint_{z_n} dz\; f(z),$$where $\oint_{z_n}$ denotes an integral over a loop of infinitesimal radius around the $n$-th point of non-analyticity, in the same direction (i.e., clockwise or counter-clockwise) as $\Gamma$.
The proof of this result is based on the figure below. The red loop, $\Gamma$, is the contour we want to integrate over. The integrand is analytic throughout the enclosed area except at several discrete points, say $\{z_1, z_2, z_3\}$. Let us define a new loop contour, $\Gamma'$, shown by the blue loop. It follows the same curve as $\Gamma$ but with the following differences: (i) it circles in the opposite direction from $\Gamma$, (ii) it contains tendrils that extend from the outer curve to each point of non-analyticity, and (iii) each tendril is attached to an infinitesimal loop encircling a point of non-analyticity.
The loop $\Gamma'$ encloses no points of non-analyticity, so Cauchy's integral theorem says that the integral over it is zero. But the contour integral over $\Gamma'$ can be broken up into three pieces: (i) the part that follows $\Gamma$ but in the opposite direction, (ii) the tendrils, and (iii) the infinitesimal inner loops:
$$\begin{align}\oint_{\Gamma'}& dz\; f(z) \;\;= \;\; 0 \quad (\mathrm{by}\;\mathrm{Cauchy's}\;\mathrm{Integral}\;\mathrm{Theorem}) \\ &= \int_{\mathrm{big}\,\mathrm{loop}} dz\; f(z) + \int_{\mathrm{tendrils}} dz\; f(z) +\; \;\sum_{\mathrm{small~loop}~n}\;\; \oint_{z_n} dz\; f(z) \end{align}$$The first term is equal to the negative of $\oint_\Gamma dz \, f(z)$, since it follows a contour that is just like $\Gamma$ except going the other way. The second term is zero, because each tendril consists of two contours taken in opposite directions, which cancel. Thus, the above equation reduces to
$$\oint_\Gamma dz\; f(z) \,=\, \sum_n \oint_{z_n} dz\; f(z).$$The loop contour integral over $\Gamma$ is equal to the sum of infinitesimal loop contour integrals encircling each point of non-analyticity. Notably, each of the infinitesimal loops circles in the same direction as $\Gamma$ (e.g., counter-clockwise in the above figure).
Another way of thinking about this is that Cauchy's integral theorem says regions of analyticity don't count towards the value of a loop integral. Hence, we can contract a loop across any domain in which $f(z)$ is analytic, until the contour becomes as small as possible. This contraction replaces $\Gamma$ with a discrete set of infinitesimal loops enclosing the points of non-analyticity.
In the previous section, we referred to situations where $f(z)$ is non-analytic at discrete points. "Discrete", in this context, means that each point of non-analyticity is surrounded by a finite region over which $f(z)$ is analytic, isolating it from other points of non-analyticity. Such situations commonly arise from functions of the form
$$f(z) \approx \frac{A}{(z-z_0)^n}, \quad \mathrm{where}\;\; n\in\{1,2,3,\dots\}.$$For $z = z_0$, the function is non-analytic because its value is singular. Such a function is said to have a pole at $z_0$. The integer $n$ is called the order of the pole.
Poles of order 1 are called simple poles, and they are of special interest. Near a simple pole, the function has the form
$$f(z) \approx \frac{A}{z-z_0}.$$In this case, the complex numerator $A$ is called the residue of the pole (so-called because it's what's left-over if we take away the singular factor corresponding to the pole.) The residue of a function at a point $z_0$ is commonly denoted $\mathrm{Res}[f(z_0)]$. Note that if a function is analytic at $z_0$, then $\mathrm{Res}[f(z_0)] = 0$.
| Example |
| Consider the function $$f(z) = \frac{5}{i-3z}.$$To find the pole and residue, divide the numerator and denominator by $-3$: $$f(z) = \frac{-5/3}{z-i/3}.$$ Thus, there is a simple pole at $z = i/3$ with residue $-5/3$. |
| Example |
| Consider the function $$f(z) = \frac{z}{z^2 + 1}.$$ To find the poles and residues, we factorize the denominator: $$f(z) = \frac{z}{(z+i)(z-i)}.$$ Hence, there are two simple poles, at $z = \pm i$. To find the residue at $z = i$, separate the divergent part from the rest of the expression: $$\begin{align}f(z) &= \frac{\left(\frac{z}{z+i}\right)}{z-i} \\ \Rightarrow\quad \mathrm{Res}\big[\,f(z)\,\big]_{z=i} &= \left[\frac{z}{z+i}\right]_{z=i} = 1/2.\end{align}$$ Similarly, for the other pole, $$\mathrm{Res}\big[\,f(z)\,\big]_{z=-i} = \left[\frac{z}{z-i}\right]_{z=-i} = 1/2.$$ |
Previously, we used contour parameterization to calculate
$$\oint_{\Gamma} \frac{dz}{z} = 2\pi i,$$where $\Gamma$ is a counter-clockwise circular loop centered on the origin. This holds for any (non-zero) loop radius. By combining this result with the previously-discussed implications of Cauchy's integral theorem, we can obtain the residue theorem:
For any analytic function $f(z)$ with a simple pole at $z_0$, $$\oint_{\Gamma[z_0]} dz \; f(z) = \pm 2\pi i \, \mathrm{Res}\big[\,f(z)\,\big]_{z = z_0},$$ where $\Gamma[z_0]$ denotes an infinitesimal loop around $z_0$. The $+$ sign holds for a counter-clockwise loop, and the $-$ sign for a clockwise loop.</td></tr></table>
By combining the residue theorem with the results of the last few sections, we arrive at a technique for integrating a function $f(z)$ over a loop $\Gamma$, called the calculus of residues:
Consider
$$f(z) = \frac{1}{z^2 + 1}.$$This can be re-written as
$$f(z) = \frac{1}{(z + i)\,(z-i)}.$$By inspection, we can identify two poles: one at $+i$, with residue $1/2i$, and the other at $-i$, with residue $-1/2i$. The function is analytic everywhere else.
Suppose we integrate $f(z)$ around a counter-clockwise contour $\Gamma_1$ that encloses only the pole at $+i$, as indicated by the blue curve in the figure below:
According to the residue theorem, the result is
$$\begin{align} \oint_{\Gamma_1}dz \; f(z) &= 2\pi i\, \mathrm{Res}\big[\,f(z)\,\big]_{z = i} \\ &= 2\pi i \cdot \frac{1}{2i} \\& = \pi. \end{align}$$On the other hand, suppose we integrate around a contour $\Gamma_2$ that encloses \emph{both} poles, as shown by the purple curve. Then the result is
$$\oint_{\Gamma_2}dz \; f(z) = 2\pi i \cdot \left[\frac{1}{2i} - \frac{1}{2i}\right] = 0.$$The calculus of residues allows us to employ contour integration for solving definite integrals over the real domain. The trick is to convert the definite integral into a contour integral, and then solve the contour integral using the residue theorem.
As an example, consider the definite integral
$$\int_{-\infty}^\infty \frac{dx}{x^2 + 1}.$$This integral is taken over real values of $x$, and we have previously solved it using a change of variables. Now let's see how to solve it using contour integration.
First, generalize the integrand from a function of $x$ to an analytic function of $z$. (This procedure is called analytic continuation.) Usually, we choose the new (complex) integrand so that it reduces to the old integrand for $z \in \mathbb{R}$, and is analytic over a broad domain. In this case, let
$$\frac{1}{x^2 + 1} \rightarrow \frac{1}{z^2 + 1}.$$This is just the integrand we dealt with in the previous section.
We now have to choose the contour. The usual procedure is to define a closed (loop) contour, such that one segment of the loop is the real line (from $-\infty$ to $+\infty$), and the other segment of the loop "doubles back" in the complex plane to close the loop. This is called closing the contour.
Here, we choose to close the contour along an anticlockwise semicircular arc in the upper half of the complex plane, as shown below:
The resulting loop contour encloses the pole at $z = +i$, so
$$\oint \frac{dz}{z^2+1} = 2\pi i \; \mathrm{Res}\left[\frac{1}{z^2 + 1}\right]_{z = +i} = \pi.$$Note that the loop is counterclockwise, so we take the positive sign for the residue theorem. The loop integral can also be written as a sum of two integrals:
$$\oint \frac{dz}{z^2 + 1} = \int_{-\infty}^\infty \frac{dx}{x^2 + 1} \;+\; \int_{\mathrm{arc}} \frac{dz}{z^2 + 1}.$$The first term is the integral we're interested in. The second term, the contour integral along the arc, goes to zero. To see why, observe that along an arc of radius $R$, the magnitude of the integrand goes as $1/R^{2}$, while the $dz$ gives another factor of $R$ (see the earlier example of parameterizing a contour integral over an arc), so the overall integral goes as $1/R$, which vanishes as $R \rightarrow \infty$.
We thus obtain the result
$$\int_{-\infty}^\infty \frac{dx}{x^2 + 1} = \pi.$$As an exercise, you can verify that closing the contour in the lower half-plane leads to exactly the same result.
Before proceeding to more complicated uses of contour integration, we must discuss an important result called Jordan's lemma:
Let $$I = \int_C dz \; e^{iqz} \,g(z),$$ where $q$ is any positive real constant, and the contour $C$ which is a semi-circular arc of radius $R$ in the upper half-plane, centered at the origin. Then $$\text{If}\;\; \big|\,g(z)\,\big| < g_{\mathrm{max}} \;\;\;\text{for all}\;\;z \in C \;\;\;\Rightarrow \;\;\; I \rightarrow 0 \;\;\mathrm{as}\;\; g_{\mathrm{max}} \rightarrow 0.$$
In other words, if the factor of $g(z)$ in the integrand does not blow up along the arc contour (i.e., its value is bounded), then in the limit where the bounding value goes to zero, the value of the entire integral vanishes.
Usually, the limiting case of interest is when the radius of the arc goes to infinity. Even if the integrand vanishes in that limit, it may not be obvious that the integral $I$ vanishes, because the integration is taken along an arc of infinite length (so we have a $0\times\infty$ sort of situation). Jordan's lemma then proves useful, as it provides a set of criteria that can let us instantly conclude that $I$ should vanish.
The proof for Jordan's lemma is tedious, and we will not go into its details (see the Wikipedia page if you're interested).
For integrands containing a prefactor of $e^{-iqz}$ rather than $e^{iqz}$ (again, where $q \in \mathbb{R}^+$), a different version of Jordan's lemma holds, referring to an arc contour $C'$ in the lower half-plane:
Let $$I = \int_C dz \; e^{-iqz} \,g(z),$$ where $q$ is any positive real constant, and the contour $C$ which is a semi-circular arc of radius $R$ in the lower half-plane, centered at the origin. Then $$\text{If}\;\; \big|\,g(z)\,\big| < g_{\mathrm{max}} \;\;\;\text{for all}\;\;z \in C \;\;\;\Rightarrow \;\;\; I \rightarrow 0 \;\;\mathrm{as}\;\; g_{\mathrm{max}} \rightarrow 0.$$
This is easily seen by doing the change of variable $z \rightarrow -z$ on the original form of Jordan's lemma.
As a convenient way to remember which variant of Jordan's lemma to use, think about which end of imaginary axis causes the exponential factor to vanish:
$$\begin{aligned} e^{iqz}\big|_{z = i\infty}\;\; = e^{-\infty} = 0\quad &\Rightarrow \;\; e^{iqz} \;\;\;\,\text{vanishes far above the origin}. \\ e^{-iqz}\big|_{z = -i\infty} = e^{-\infty} = 0\quad &\Rightarrow \;\; e^{-iqz} \;\;\textrm{vanishes far below the origin}.\end{aligned}$$Hence, for $e^{iqz}$ (where $q$ is any positive real number), the suppression occurs in the upper-half-plane. For $e^{-iqz}$, the suppression occurs in the lower-half-plane.
Consider the integral
$$I = \int_{-\infty}^\infty dx\; \frac{\cos(x)}{4x^2 + 1}.$$One possible approach is to break the cosine up into $(e^{ix} + e^{-ix})/2$, and do the contour integral on each piece separately. Another approach, which saves a bit of effort, is to write
$$I = \mathrm{Re} \; \int_{-\infty}^\infty dx\; \frac{e^{ix}}{4x^2 + 1}.$$To do the integral, close the contour in the upper half-plane:
Then
$$\oint dz \; \frac{e^{iz}}{4z^2 + 1} = \int_{-\infty}^\infty dx\; \frac{e^{ix}}{4x^2 + 1} + \int_{\mathrm{arc}} dz \; \frac{e^{iz}}{4z^2 + 1}.$$On the right-hand side, the first term is what we want. The second term is a counter-clockwise arc in the upper half-plane. According to Jordan's lemma, this term goes to zero as the arc radius goes to infinity, since the rest of the integrand goes to zero for large $|z|$:
$$\left|\frac{1}{4z^2 + 1}\right| \sim \frac{1}{4|z|^2} \rightarrow 0 \quad \mathrm{as} \;|z| \rightarrow \infty.$$As for the loop contour, it can be evaluated using the residue theorem:
$$\begin{align}\oint dz \; \frac{e^{iz}}{4z^2 + 1} &= \mathrm{Res}\left[\frac{e^{iz}}{4z^2 + 1}\right]_{\mathrm{enclosed}\;\mathrm{poles}}\\ &= 2\pi i \; \mathrm{Res}\left[\frac{1}{4}\, \frac{e^{iz}}{(z+i/2)(z-i/2)}\right]_{z = i/2} \\ &= 2\pi i \; \frac{e^{-1/2}}{4i}.\end{align}$$Hence,
$$I = \mathrm{Re}\;\left[\frac{\pi}{2\sqrt{e}}\right]= \frac{\pi}{2\sqrt{e}}.$$In solving the integral this way, we must close the contour in the upper half-plane because our choice of complex integrand was bounded in the upper half-plane. Alternatively, we could have chosen to write
$$I = \mathrm{Re} \; \int_{-\infty}^\infty dx\; \frac{e^{-ix}}{4x^2 + 1},$$i.e., with $e^{-ix}$ rather than $e^{ix}$ in the numerator. In that case, Jordan's lemma tells us to close the contour in the lower half-plane. The arc in the lower half-plane vanishes, as before, while the loop contour is clockwise (contributing an extra minus sign) and encloses the lower pole:
$$\begin{align}\oint dz \frac{e^{-iz}}{4z^2 + 1} &= -2\pi i \, \mathrm{Res}\left[ \frac{e^{-iz}}{4z^2 + 1} \right]_{z = -i/2} \\ &= - 2\pi i \frac{e^{-1/2}}{-4i} \\ &= \frac{\pi}{2\sqrt{e}}. \end{align}$$Taking the real part, we obtain the same result as before.
In [1]:
## As an exercise, let's verify the integral numerically.
from numpy import linspace, cos, trapz, sqrt, pi, e
x = linspace(-500, 500, 10000)
y = cos(x) / (4*x**2 + 1)
I = trapz(y, x)
print("Numerical result = {}".format(I))
print("Contour integral = {}".format(pi/2/sqrt(e)))
Sometimes, we come across integrals that have poles lying on the desired integration contour.
As an example, consider
$$I = \int_{-\infty}^\infty dx\; \frac{\sin(x)}{x}.$$Because of the series expansion of the sine function, the integrand does not diverge at $x = 0$, and the integral is in fact convergent. The integral can be solved without using complex numbers by using the arcane trick of differentiating under the integral sign. But it can also be solved straightforwardly via contour integration, with just a few extra steps.
We start by writing
$$I = \mathrm{Im}(I'), \quad \mathrm{where}\;\;\; I' = \int_{-\infty}^\infty dx\; \frac{e^{ix}}{x}.$$We want to calculate $I'$ with the help of contour integration. But there's something strange about $I'$: the complex integrand has a pole at $z = 0$, right on the real line!
To handle this, we split $I'$ into two integrals, one going over $-\infty < x < -\epsilon$ (where $\epsilon$ is some positive infinitesimal), and the other over $\epsilon < x < \infty$:
$$\begin{align}I' &= \lim_{\epsilon \rightarrow 0} \left[ \int_{-\infty}^{-\epsilon} dx\; \frac{e^{ix}}{x} + \int_{\epsilon}^\infty dx\; \frac{e^{ix}}{x}\right] \\ &\equiv \mathcal{P} \int_{-\infty}^\infty dx\; \frac{e^{ix}}{x}. \end{align}$$In the last line, the notation $\mathcal{P}[\cdots]$ is short-hand for this procedure of "chopping away" an infinitesimal segment surrounding the pole. This is called taking the principal value of the integral. (Note: even though this bears the same name as the previously-discussed "principal values" for multi-valued complex operations, there is no connection between the two concepts.)
Now consider the loop contour shown in the figure below. The loop follows the principal-value contour along the real axis, skips over the pole at $z = 0$ and arcs back along the upper half-plane. Since it encloses no poles, the loop integral vanishes by Cauchy's integral theorem.
However, loop can also be decomposed into several sub-contours:
The integral over $\Gamma_1$ is the principal-value integral we are interested in. The integral over $\Gamma_2$ vanishes by Jordan's lemma. The integral over $\Gamma_3$ can be calculated by parameterization:
$$\begin{align} \int_{\Gamma_3} \frac{e^{iz}}{z} &= \lim_{\epsilon \rightarrow 0} \int_{\pi}^{0} \frac{e^{i\epsilon \exp(i\theta)}}{\epsilon e^{i\theta}} \left(i\epsilon e^{i\theta}\right) d\theta \\ &= \lim_{\epsilon \rightarrow 0} i \int_{\pi}^0 d\theta \\ &= - i\pi. \end{align}$$Intutively, since encircling a pole anticlockwise gives a factor of $2\pi i$ times the residue (which is 1 in this case), a clockwise semi-circle is associated with a factor of $- i \pi$. Finally, putting everything together,
$$\underbrace{\int_{\Gamma_1 + \Gamma_2 + \Gamma_3} f(z) dz}_{ =~0~(\text{Cauchy's integral theorem})} = \underbrace{\int_{\Gamma_1} f(z) dz}_{=~I'} + \underbrace{\int_{\Gamma_2} f(z) dz}_{=~0~(\text{Jordan's lemma})} + \underbrace{\int_{\Gamma_3} f(z) dz.}_{=~-i \pi}$$Hence,
$$I = \mathrm{Im}(I') = \mathrm{Im}(i\pi) = \pi.$$This agrees with the result obtained by the method of differentiating under the integral sign.
Alternatively, we could have chosen the loop contour so that it skips below the pole at $z = 0$. In that case, the loop integral would be non-zero, and can be evaluated using the residue theorem. The final result is the same.
In [6]:
## Let's verify the integral numerically.
from numpy import linspace, sin, trapz, pi
x1, x2 = linspace(-500, -1e-6, 5000), linspace(1e-6, 500, 5000) # left and right x segments
y1, y2 = sin(x1) / x1, sin(x2) / x2
I = trapz(y1, x1) + trapz(y2, x2)
print("Numerical result = {}".format(I))
print("Contour integral = {}".format(pi))