Complex Integration
Fundamental Theorem of Calculus
Contour Integrals of Monomials With Singularities
Let us consider a complex function f(z), which can be decomposed into its real and imaginary parts:
\begin{aligned} & f(z) = u(x,y) + i v(x,y) \\ & u(x,y) = \Re(f(z)) \\ & v(x,y) = \Im(f(z)) \end{aligned}
The integral of f(z) is evaluated along a specified path, or contour, C, in the complex plane. This path connects an initial point z_0 to a terminal point z_1.
The differential element \mathrm{d}z is expressed as \mathrm{d}z = \mathrm{d}x + i\,\mathrm{d}y. Consequently, the complex contour integral can be expanded into two real line integrals.
\int _{C }f(z)\,\mathrm{d}z=\int _{C }(u+iv)(\mathrm{d}x+i\,\mathrm{d}y)=\int _{C}\left[(u\,\mathrm{d}x-v\,\mathrm{d}y)+i(v\,\mathrm{d}x+u\,\mathrm{d}y)\right]
We will use the Cauchy-Goursat theorem(here). This theorem states that if a function f(z) is analytic at all points within and on a simple closed contour C, then the integral of f(z) around that contour is zero:
\oint_{C}f(z)\,\mathrm{d}z = 0
This result has a direct implication: for an analytic function, the value of a contour integral between two points is independent of the path taken.
This property is analogous to the behavior of conservative vector fields in mechanics, where the work done moving between two points is independent of the trajectory.
To demonstrate this path independence, consider two distinct contours, C_1 and C_2, both connecting an initial point z_0 to a final point z_1 within a domain where f(z) is analytic.
Let us define the integrals along these paths as I_1 and I_2. By reversing the direction of contour C_2, denoted as -C_2, we form a single closed contour C = C_1 - C_2. Applying the Cauchy-Goursat theorem to this composite closed path yields the following.
\begin{aligned} & I_1 = \int_{C_1}f(z)\,\mathrm{d}z \\ & I_2 = \int_{C_2}f(z)\,\mathrm{d}z \end{aligned}
The integral over the combined closed loop C_1 followed by -C_2 (the path C_2 from z_1 to z_0) is zero.
\begin{aligned} & \oint_{C_1 - C_2}f(z)\,\mathrm{d}z = \int_{C_1}f(z)\,\mathrm{d}z + \int_{-C_2}f(z)\,\mathrm{d}z = 0 \\ & \int_{C_1}f(z)\,\mathrm{d}z - \int_{C_2}f(z)\,\mathrm{d}z = 0 \\ & \int_{C_1}f(z)\,\mathrm{d}z = \int_{C_2}f(z)\,\mathrm{d}z \end{aligned}
Therefore, I_1 = I_2, confirming that the integral’s value depends only on the endpoints z_0 and z_1, not on the specific contour connecting them, provided f(z) is analytic in the region containing the paths.
Fundamental theorem of calculus
The path independence of integrals for analytic functions leads to a tool for their evaluation: a version of the Fundamental Theorem of Calculus for complex analysis.
First, let us formalize the evaluation of a contour integral using a parametrization. If a contour C is parametrized by a function z(t) = x(t) + i y(t) for a \leq t \leq b, the integral is defined as:
\int_C f(z)\,\mathrm{d}z = \int_a^b f\left(z(t)\right)z^\prime(t)\,\mathrm{d}t
This transforms the complex contour integral into an integral of a complex-valued function of a real variable t. This can be separated into its real and imaginary components.
\int_a^b w(t)\,\mathrm{d}t = \int_a^b \Re(w(t))\,\mathrm{d}t + i \int_a^b \Im(w(t))\,\mathrm{d}t
The theorem establishes a direct method for evaluating complex integrals when an antiderivative of the integrand is known.
Theorem: Let f(z) be an analytic function in a domain \mathcal{D}. If there exists a single-valued function F(z) such that F'(z) = f(z) for all z in \mathcal{D}, then for any contour C lying within \mathcal{D} from an initial point z_0 to a final point z_1, the following holds:
\int_C f(z) \mathrm{d}z = F(z_1) - F(z_0)
Proof: the proof of this theorem is a direct consequence of the definition of the parametrized integral and the chain rule for complex derivatives.
Let the contour C be parametrized by z(t) for t \in [a, b], with z(a) = z_0 and z(b) = z_1.
\begin{aligned} \int_C f(z) \mathrm{d}z &= \int_a^b f(z(t)) z^\prime(t) \mathrm{d}t \\ \end{aligned}
Since f(z) = F^\prime(z), we can substitute this into the integrand.
\begin{aligned} \int_C f(z) \mathrm{d}z &= \int_a^b F^\prime(z(t)) z^\prime(t) \mathrm{d}t \\ \end{aligned}
By the chain rule, the integrand is the total derivative of the composite function F(z(t)) with respect to t.
\frac{\mathrm d}{\mathrm d t} F(z(t)) = F^\prime(z(t)) z^\prime(t)
Therefore, the integral becomes:
\begin{aligned} \int_C f(z) \mathrm{d}z &= \int_{a}^{b} \frac{\mathrm{d}}{\mathrm{d}t} F(z(t)) \mathrm{d}t \\ &= \bigg[ F(z(t)) \bigg]_a^b \\ &= F(z(b)) - F(z(a)) \\ &= F(z_1) - F(z_0) \end{aligned}
This completes the demonstration.
The existence of a single-valued antiderivative F(z) is guaranteed if f(z) is analytic throughout a simply connected domain containing the contour C.
This theorem provides a method for evaluating complex integrals, bypassing the need for explicit parametrization and integration, provided an antiderivative can be found.
Contour integrals of monomials with singularities
We begin by examining the contour integral of the function f(z) = (z-a)^n for integer values of n. This function is analytic everywhere for n \ge 0. For negative integers, n < 0, the function possesses a singularity at the point z=a and is therefore not analytic at that point. We will investigate the value of the closed contour integral \oint_C (z-a)^n \, \mathrm d z, where the contour C is a simple closed curve that encloses the point a in a counterclockwise direction.
A key property derived from the Cauchy-Goursat theorem is that this contour C may be continuously deformed into any other simple closed contour enclosing a, such as a circle, without altering the value of the integral. This is because the region between the two contours is free of singularities. For convenience, we will consider the contour to be a circle C^\prime of constant radius r centered at a.
Let us analyze two distinct cases based on the value of the integer exponent n.
Case 1: n \ne -1
For any integer n except for -1, the function (z-a)^n has a single-valued antiderivative given by \frac{(z-a)^{n+1}}{n+1}. The fundamental theorem of calculus for contour integrals states that if a function has a single-valued antiderivative in a domain, its integral around any closed loop in that domain is zero. If we parametrize our path starting and ending at a point z_0, the integral evaluates to:
\oint_C (z-a)^n \, \mathrm d z = \left[ \frac{(z-a)^{n+1}}{n+1} \right]_{z_0}^{z_0} = 0
Alternatively, parametrizing along the circular contour C' where z-a = r e^{i\theta} and thus \mathrm d z = i r e^{i\theta} \mathrm d\theta:
\begin{aligned} \oint_C (z-a)^n \, \mathrm d z &= \int_0^{2\pi} (r e^{i\theta})^n (i r e^{i\theta} \mathrm d\theta) \\ &= i r^{n+1} \int_0^{2\pi} e^{i(n+1)\theta} \mathrm d\theta \\ &= i r^{n+1} \left[ \frac{e^{i(n+1)\theta}}{i(n+1)} \right]_0^{2\pi} \\ &= \frac{r^{n+1}}{n+1} (e^{i2\pi(n+1)} - e^0) = \frac{r^{n+1}}{n+1} (1 - 1) = 0 \end{aligned}
Case 2: n = -1
This specific case is unique. The integral is \oint_C \frac{1}{z-a} \mathrm d z. The antiderivative of \frac{1}{z-a} is the complex logarithm, \operatorname{Log}(z-a), which is a multi-valued function.
\operatorname{Log}(w) = \ln|w| + i \arg(w)
As we traverse the closed contour C once in the counterclockwise direction, the argument of z-a increases by 2\pi. If we begin and end at a point z_0, the value of the logarithm changes:
\oint_C \frac{1}{z-a} \mathrm d z = \operatorname{Log}(z-a) \bigg|_{z_0}^{z_0, \text{one loop}} = (\ln|z_0-a| + i(\arg(z_0-a)+2\pi)) - (\ln|z_0-a| + i\arg(z_0-a)) = 2\pi i
Using the polar parametrization confirms this result directly. With z-a = r e^{i\theta} and \mathrm d z = i r e^{i\theta} \mathrm d\theta:
\oint_C \frac{1}{z-a} \mathrm d z = \int_0^{2\pi} \frac{1}{r e^{i\theta}} (i r e^{i\theta} \mathrm d\theta) = \int_0^{2\pi} i \, \mathrm d\theta = i[\theta]_0^{2\pi} = 2\pi i
In summary, the integral of (z-a)^n around a contour enclosing a is non-zero only for the case n=-1.
\oint_{C} (z - a)^n\,\mathrm{d}z = \begin{cases} 0 & n \ne -1 \\ 2 \pi i & n = -1 \end{cases}
Integration of Laurent series
This result has implications for integrating more general complex functions. A function f(z) with an isolated singularity at z=a can be represented in an circular region around a by its Laurent series:
f(z) = \sum _{n=-\infty }^{\infty } c_{n}(z-a)^{n}
If we integrate this series term-by-term around a closed contour C within the circular region of convergence and enclosing a, every term of the form (z-a)^n for n \ne -1 integrates to zero.
The only term that contributes to the integral is the n=-1 term.
\oint_{C}f(z)\,\mathrm{d}z = \oint_{C}\sum _{n=-\infty }^{\infty }c_{n}(z-a)^{n}\,\mathrm{d}z = \sum _{n=-\infty }^{\infty } c_{n} \oint_{C} (z-a)^{n}\,\mathrm{d}z = c_{-1} (2\pi i)
The coefficient c_{-1} is known as the residue of the function f(z) at the point a. This forms the basis of the Residue Theorem.
The Cauchy integral formula
The Cauchy Integral Formula provides a method to determine the value of an analytic function at any point inside a closed contour by evaluating an integral along that contour.
Theorem: If f(z) is an analytic function within and on a simple closed contour C, taken in the positive (counterclockwise) direction, and if a is any point in the interior of C, then:
\oint_C \frac{f(z)}{z-a}\, \mathrm{d}z = 2\pi i\,f(a)
Proof: we rewrite the integral by adding and subtracting f(a) in the numerator of the integrand.
\oint_C \frac{f(z)}{z-a}\, \mathrm{d}z = \oint_C \frac{f(z) - f(a) + f(a)}{z-a}\, \mathrm{d}z = \oint_C \frac{f(a)}{z-a}\, \mathrm{d}z + \oint_C \frac{f(z)-f(a)}{z-a}\, \mathrm{d}z
Let us analyze the two resulting integrals separately.
The first integral is straightforward. Since f(a) is a constant with respect to the integration variable z, it can be factored out of the integral:
\oint_C \frac{f(a)}{z-a}\, \mathrm{d}z = f(a) \oint_C \frac{1}{z-a}\, \mathrm{d}z = f(a) (2\pi i)
This gives us the expected result. The proof is complete if we can demonstrate that the second integral is equal to zero.
I_2 = \oint_C \frac{f(z)-f(a)}{z-a}\, \mathrm{d}z
The integrand \frac{f(z)-f(a)}{z-a} is analytic everywhere inside C except for the point z=a. However, since f(z) is analytic at a, the limit \lim_{z \to a} \frac{f(z)-f(a)}{z-a} = f^\prime(a) exists.
This implies the singularity at z=a is removable. By the Cauchy-Goursat theorem, we can deform the contour C to an arbitrarily small circle C_\delta with radius \delta centered at a, without changing the value of the integral I_2.
Since f(z) is analytic, it is also continuous at z=a. This means that for any chosen \varepsilon > 0, we can find a \delta > 0 such that if |z-a| = \delta, then |f(z) - f(a)| < \varepsilon.
We now apply the ML-inequality to bound the magnitude of I_2 on the circular contour C_\delta. The length of the contour is L = 2\pi\delta. The maximum value M of the magnitude of the integrand on C_\delta is:
M = \max_{z \in C_\delta} \left| \frac{f(z)-f(a)}{z-a} \right| = \max_{z \in C_\delta} \frac{|f(z)-f(a)|}{|z-a|} < \frac{\varepsilon}{\delta}
Therefore, the magnitude of the integral is bounded:
|I_2| = \left| \oint_{C_\delta} \frac{f(z)-f(a)}{z-a}\, \mathrm{d}z \right| \le M L < \frac{\varepsilon}{\delta} (2\pi\delta) = 2\pi\varepsilon
Since |I_2| can be made smaller than any arbitrarily small positive number 2\pi\varepsilon, its value must be exactly zero.
With the second integral shown to be zero, we are left with the result from the first integral, which completes the proof.
\oint_C \frac{f(z)}{z-a}\, \mathrm{d}z = 2\pi i f(a) + 0 = 2\pi i f(a)
Examples
Example 1
Considering f(z) = \frac{1}{z^2 -3 z +2}. This function has two poles at z=1,2. As a note, \frac{1}{z-1} is analytic at z=2 and vice versa.
Four contours will be considered. It is already clear that:
\oint_{C_4}f(z)\,\mathrm{d}z = 0
using the Cauchy-Goursat theorem (as there aren’t singularities).
The integral on C_1 can be written as:
\begin{aligned} \oint_{C_1}\frac{1}{z^2 -3 z +2} \,\mathrm{d}z & = \oint_{C_1}\frac{\frac{1}{z-2}}{z-1} \,\mathrm{d}z = \oint_{C_1}\frac{f(z)}{z-1} \,\mathrm{d}z \\[12pt] & = 2\pi i f(1) = -2 \pi i \end{aligned}
Similar for C_2:
\begin{aligned} \oint_{C_2}\frac{1}{z^2 -3 z +2} \,\mathrm{d}z & = \oint_{C_2}\frac{\frac{1}{z-1}}{z-2} \,\mathrm{d}z = \oint_{C_2}\frac{f(z)}{z-2} \,\mathrm{d}z \\[12pt] & = 2\pi i f(2) = 2 \pi i \end{aligned}
Finally, for C_3 the region can be deformed as a region which is just containing the two poles:
C_3 = C_1 + C_2 = 0
Example 2
It is possible to solve a real integral of the type:
\int_{-\infty}^{\infty} f(x) \mathrm{d}x
For example, considering the integral:
I_1 = \int_{-\infty}^{\infty} \frac{1}{x^4 + a^4} \mathrm{d}x = 2 \int_{0}^{\infty} \frac{1}{x^4 + a^4} \mathrm{d}x
as f(x) is an even function.
It is possible to consider the complex function
f(z) = \frac{1}{z^4 + a^4}
and the integral along a semicircle of radius R in the complex plane and take the limit for R \to \infty:
\lim_{R \to \infty} \oint_{C} f(z) \mathrm{d}z = \lim_{R \to \infty} \int_{C_1} f(z) \mathrm{d}z + \lim_{R \to \infty} \int_{C_2} f(z) \mathrm{d}z = I_1 + I_2
This integral is equal to I_1 if \lim_{R \to \infty} I_2 = 0.
The function f(z) has poles where:
\begin{aligned} & z^4 + a^4 = 0 \\ & z^4 = a^4 \\ & z = a \sqrt[4]{-1} \\ & z = \frac{\pm 1 \pm i}{\sqrt 2}a = a_{1,2,3,4} \end{aligned}
f(z) can be split as a product of 4 binomials:
\oint_{C} f(z) \mathrm{d}z = \oint_{C} \frac{1}{z^4 + a^4}\mathrm{d}z = \oint_{C} \frac{1}{(z-a_1)(z-a_2)(z-a_3)(z-a_4)}\mathrm{d}z
As in example 1, at a_1 = \frac{+1 + i}{\sqrt 2}a, the function \frac{1}{(z-a_2)(z-a_3)(z-a_4)} is analytic, and similarly in a_2 \frac{1}{(z-a_1)(z-a_3)(z-a_4)}. Therefore:
\begin{aligned} &\oint_{C} \frac{1}{z^4 + a^4}\mathrm{d}z = \oint_{C_3} \frac{\frac{1}{(z-a_2)(z-a_3)(z-a_4)}}{(z-a_1)}\mathrm{d}z + \oint_{C_4} \frac{\frac{1}{(z-a_1)(z-a_3)(z-a_4)}}{(z-a_2)}\mathrm{d}z \\ & = 2\pi i \left(\frac{1}{(z-a_2)(z-a_3)(z-a_4)} \right) \bigg|_{a_1} + 2\pi i \left(\frac{1}{(z-a_1)(z-a_3)(z-a_4)} \right) \bigg|_{a_2} \\ & = 2\pi i \left(\frac{1}{(a_1-a_2)(a_1-a_3)(a_1-a_4)} \right) + 2\pi i \left(\frac{1}{(a_2-a_1)(a_2-a_3)(a_2-a_4)} \right) \\ & = \frac{2 \pi i }{2 \sqrt 2 a^3 i} = \frac{\pi}{\sqrt 2 a^3} \end{aligned}
This is true \forall R > a. The last item is to show \lim_{R \to \infty} I_2 = 0.
\lim_{R \to \infty} \int_{C_2} f(z) \mathrm{d}z = \lim_{R \to \infty} \int_{C_2} \frac{1}{z^4 + a^4}\mathrm{d}z
It is possible to show it using the ML bound property, considering L=\pi\,R the length of the curve.
\begin{aligned} &\left| \int_{C_2} \frac{1}{z^4 + a^4}\mathrm{d}z \right| \le M\,L = \frac{\pi \, R}{R^4} = \frac{\pi}{R^3}\\ & L = \pi \, R \\ & M = \max\limits_{C_2} \frac{1}{z^4+a^4} \le \frac{1}{R^4} \end{aligned}
Taking the \lim_{R \to \infty}, then the bound goes to 0 and therefore I_2=0. Furthermore, I_1 equals the real integral that needed to be computed.
Example 3
Taking another real integral:
\int_0^{2\pi} \sin(\theta)^2 \mathrm{d}\theta
This integral can be solved analytical using calculus integration formula and trigonometry:
\int_0^{2\pi} \sin(\theta)^2 \mathrm{d}\theta = \int_0^{2\pi} \frac{1-\cos(2\theta)}{2} \mathrm{d}\theta = \left[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}\right]\bigg|_0^{2\pi} = \pi
It is possible to solve it also as contour integral. Using the formula:
\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}
and transforming polar coordinates to Cartesian:
\begin{aligned} &z = e^{i\theta}\\ &\mathrm{d}z = ie^{i\theta}\mathrm{d}\theta\\ &\mathrm{d}\theta = \frac{\mathrm{d}z}{ie^{i\theta}} = \frac{\mathrm{d}z}{iz} \end{aligned}
Substituting:
\begin{aligned} \int_0^{2\pi} \sin(\theta)^2 \mathrm{d}\theta & = \oint_0^{2\pi} \left(\frac{e^{i\theta} - e^{-i\theta}}{2i}\right)^2 \mathrm{d}\theta \\ & = \oint_C \frac{\left(z - \frac{1}{z}\right)^2}{-4} \frac{\mathrm{d}z}{iz} = \frac{i}{4} \oint_C \frac{\left(z - \frac{1}{z}\right)^2}{z} \mathrm{d}z \\ & = \frac{i}{4} \oint_C \frac{z^2 - 2 - \frac{1}{z^2}}{z} \mathrm{d}z = \frac{i}{4} \oint_C z - \frac{2}{z} - \frac{1}{z^3} \mathrm{d}z \end{aligned}
The only term on the contour which is not null is -\frac{2}{z} and therefore:
\int_0^{2\pi} \sin(\theta)^2 \mathrm{d}\theta = \frac{i}{4} \oint_C - \frac{2}{z} \mathrm{d}z = \frac{i}{4} 2\pi i (-2) = \pi
Example 4
It is possible to solve complex integral as:
\oint_C \frac{\sin(z)}{z(z^2 + 2)} \mathrm{d}z
As a note, \frac{\sin(z)}{z} isn’t well-defined at z=0. It does, however, have an analytic continuation to the entire plane. \sin(z) has a zero of order 1 at z=0, and therefore the singularity of \frac{\sin(z)}{z} has order 0 at z=0, meaning it is removable and an analytic continuation at z=0 exists (and is given by the limit); when z \to 0, \frac{\sin(z)}{z} \to 1 and not to \pm \infty. Therefore the two singularities are at z = \pm \sqrt{2}i.
The integral on C_1 is zero as there are no singularities in this contour.
The integral on C_2 can be computed with the Cauchy formula with one pole in z=+\sqrt{2}i.
\begin{aligned} \oint_{C_2} \frac{\sin(z)}{z(z^2 + 2)} \mathrm{d}z & = \oint_{C_2} \frac{\sin(z)}{z(z+\sqrt{2}i)(z-\sqrt{2}i)} \mathrm{d}z \\ & 2\pi i \frac{\sin(z)}{z(z+\sqrt{2}i)}\bigg|_{\sqrt{2}i} = 2\pi i \frac{\sin(\sqrt{2}i)}{\sqrt{2}i(\sqrt{2}i+\sqrt{2}i)} \\ & = -\pi i \frac{\sin(\sqrt{2}i)}{2} \end{aligned}
The integral on C_3 can be computed as the sum of C_4 (equal to C_2) and C_5 which can be computed again with the Cauchy formula:
\begin{aligned} \oint_{C_5} \frac{\sin(z)}{z(z^2 + 2)} \mathrm{d}z & = \oint_{C_2} \frac{\sin(z)}{z(z+\sqrt{2}i)(z-\sqrt{2}i)} \mathrm{d}z \\ & 2\pi i \frac{\sin(z)}{z(z-\sqrt{2}i)}\bigg|_{-\sqrt{2}i} = 2\pi i \frac{\sin(-\sqrt{2}i)}{-\sqrt{2}i(-\sqrt{2}i-\sqrt{2}i)} \\ & -\pi i \frac{\sin(-\sqrt{2}i)}{2} = \pi i \frac{\sin(\sqrt{2}i)}{2} \end{aligned}
the latter applying \sin(-\theta)=\sin(\theta). Therefore C_5 = -C_4 and C_3 = C_4 + C_5 = 0.