The Complex Integral

Let’s discuss integration in the complex world. We will use Lebesgue-integration in formulating our definitions, since we have painstakingly developed that technology elsewhere in this blog already, and can therefore leverage its many useful convergence properties, especially the Fubini–Tonelli theorem.

Definition 1. Let $f = u + iv: [a, b] \to \mathbb C$ , where $u, v :[a,b] \to \mathbb R$ . Equip $[a, b] \subseteq \mathbb R$ with the usual Lebesgue measure. If $u, v$ are Lebesgue-integrable, then we define

$\displaystyle \int_{[a, b]} f \, \mathrm d \lambda := \int_{[a, b]} u \, \mathrm d \lambda + i \int_{[a, b]} v \, \mathrm d \lambda.$

To be explicit with the variables, we make the notation $\mathrm d\lambda(t) = \mathrm dt$ to obtain

$\displaystyle \int_{[a, b]} f(t) \, \mathrm dt := \int_{[a, b]} u(t) \, \mathrm d t + i \int_{[a, b]} v(t) \, \mathrm d t.$

In the case $u, v$ are Riemann-integrable,

$\displaystyle \int_a^b f(t) \, \mathrm dt := \int_a^b u(t) \, \mathrm d t + i \int_a^b v(t) \, \mathrm d t.$

Occasionally, when there is no ambiguity, we even suppress the dummy variables for brevity:

$\displaystyle \int_a^b f = \int_a^b u + i \int_a^b v.$

Note that $u,v$ are Lebesgue-integrable if they are Riemann-integrable, and that their integrals coincide.

Lemma 1. For $f : [a, b] \to \mathbb C$ , whenever defined, $\displaystyle \left| \int_a^b f \right| \leq \int_a^b |f|$ .

Proof. First, define

$\displaystyle \theta := \mathrm{Arg} \int_a^b f\quad \Rightarrow \quad e^{-i\theta} \int_a^b f \in \mathbb R.$

Since $|e^{-i \theta}| = 1$ ,

$\begin{aligned}\left| \int_a^b f \right| &= |e^{-i\theta}| \cdot \left| \int_a^b f \right| \\ &= \left| e^{-i\theta} \int_a^b f \right| = \left| \mathrm{Re} \left( e^{-i\theta} \int_a^b f \right) \right| \\ &= \left| \int_a^b \mathrm{Re} ( e^{-i\theta} f ) \right| \leq \int_a^b \left| \mathrm{Re} ( e^{-i\theta} f ) \right| \leq \int_a^b| e^{-i\theta} f | = \int_a^b |f|.\end{aligned}$

That sounds easy enough! What about complex integration $f : \mathbb C \to \mathbb C$ ? Our idea will require the use of curves.

Definition 2. A subset $\gamma \subseteq \mathbb C$ is a curve if there exists a continuous function $r : [a, b] \to \mathbb C$ such that $\gamma = r([a, b])$ . Furthermore, we say that $\gamma$ is smooth if $r$ is continuously differentiable.

Our first line of business would be the compute the length $\| \gamma \|$ of $\gamma$ . To that end, let $P = \{ t_0, t_1,\dots, t_n\}$ be a partition of $[a, b]$ with $( t_0, t_n ) = (a, b)$ . Approximate $\|\gamma \|$ using straight-lines:

$\displaystyle L_n(\gamma) := \sum_{k=1}^n | r(t_{k+1}) - r(t_k) | = \sum_{k=1}^n \frac{ | r(t_{k+1}) - r(t_k) | }{\Delta t_k} \cdot \Delta t_k.$

Taking $n \to \infty$ so that $\Delta t_k \to 0$ (we can formalise this idea using Darboux integrals), we obtain

$\displaystyle \lim_{n \to \infty} L_n(\gamma) = \int_a^b \left| \lim_{\Delta t \to 0} \frac{ r(t + \Delta t) - r(t) }{\Delta t} \right| \mathrm dt = \int_a^b | r'(t) |\, \mathrm dt.$

Definition 3. Define the length of a smooth curve $\gamma = r([a,b])$ by

$\displaystyle \| \gamma \| := \int_a^b | r' |.$

Example 1. Let $C := r([0, 2\pi])$ where $r(t) = e^{it}$ . Then

$| r'(t) | = | i e^{ i t} | = | i | \cdot | e^{2\pi it} | = 1$

implies that

$\begin{aligned} \| C \| = \int_0^{2\pi} | r'(t) |\, \mathrm dt &= \int_0^{2\pi} 1 = 2\pi - 0 = 2\pi, \end{aligned}$

which intuitively agrees with the circumference of a unit circle equaling $2\pi$ .

To give the expression $\displaystyle \int f(z)\, \mathrm dz$ a useful meaning depends on what values of $z$ we are effectively doing an infinite summation over. If $z \in \gamma = r([a, b])$ , then $z = r(t)$ , which implies at least informally that $\mathrm dz = r'(t)\, \mathrm dt$ . This is how we define the complex integral over $\gamma$ , and establish what goes wrong if we don’t do it this way.

For a sanity check, suppose $\gamma = r_1([a, b])$ . Let $\varphi : [a, b] \to [c, d]$ denote the canonical homemorphism. Suppose $\gamma = r_2([c, d])$ where $r_2 = \varphi \circ r_1$ . Using the substitution $s = \varphi(t)$ so that $s(a) = c$ and $s(b) = d$ ,

$\begin{aligned} \int_a^b (f \circ r_1) \cdot r_1' &= \int_a^b f(r_1(t)) \cdot r_1'(t)\, \mathrm dt \\ &= \int_a^b f(r_2(\varphi (t) ))\cdot r_2'(\varphi(t)) \cdot \varphi'(t)\, \mathrm dt \\ &= \int_c^d f(r_2( s ))\cdot r_2'( s )\, \mathrm ds \\ &= \int_c^d (f \circ r_2) \cdot r_2' \end{aligned}$

Definition 4. The contour integral of $f : \mathbb C \to \mathbb C$ over $\gamma = r([a, b])$ is defined by

$\displaystyle \oint_{\gamma} f(z)\, \mathrm dz := \int_a^b f(r(t)) \cdot r'(t)\, \mathrm dt \equiv \int_a^b (f \circ r) \cdot r',$

whenever the right-hand side exists. For instance, this definition makes sense when $f$ is continuous.

Example 2. Using the circle $C$ in Example 1,

$\displaystyle \oint_C \frac 1{z}\, \mathrm dz = \int_0^{2\pi} \frac 1{e^{it}} \cdot i \cdot e^{it}\, \mathrm dt = \int_0^{2\pi} i\, \mathrm dt = 2 \pi i.$

Example 2 will play a crucial role later on when discussing Cauchy’s integral formula.

Why not just do it the usual calculus way? That is, given $f$ , compute an antiderivative $F$ such that $F' = f$ , then use the fundamental theorem of calculus to compute

$\displaystyle \int_a^b f = F(b) - F(a).$

Define the upper-half semicircle by $C^+ := r([0, \pi])$ and the lower-half semicircle by $C^- := \tilde r([0, \pi])$ , where $\tilde r := r(-\cdot)$ . Then

$\displaystyle \oint_{C^+} \frac 1z\, \mathrm dz = \pi i,\quad \oint_{C^-} \frac 1z\, \mathrm dz = -\pi i.$

However, using the endpoints of the curves $r(0) = 1$ and $r(\pi) = -1$ and the calculation $\mathrm{Log}'(z) = 1/z$ ,

$\displaystyle \pi i = \mathrm{Log}(-1) - \mathrm{Log}(1) = \int_1^{-1} \frac 1z\, \mathrm dz = \oint_{C^-} \frac 1z\, \mathrm dz = - \pi i,$

a blatant contradiction. Therefore, these definitions change depending on the paths taken between the endpoints.

Remark 1. If, instead, we took $\mathrm{Log}(-1) := \lim_{t \to \pi} \mathrm{Log}(e^{-it})$ , then we would get

$\mathrm{Log}(-1) - \mathrm{Log}(1) = -i \pi.$

Nevertheless, these various paths are not necessarily unrelated to one another.

Lemma 2. Given a curve $\gamma = r ( [a, b] )$ , define the reverse $-\gamma$ of $\gamma$ by $-\gamma = \tilde r([a, b])$ , where $\tilde r = r(a+b- \cdot)$ . For suitable $f : \mathbb C \to \mathbb C$ ,

$\displaystyle \oint_{-\gamma} f(z)\, \mathrm dz = -\oint_{\gamma} f(z)\, \mathrm dz.$

Proof. By Definition 4, making the change-of-variables $s = \tilde r$ so that $s(a) = b, s(b)= a$ ,

$\begin{aligned} \oint_{-\gamma} f(z)\, \mathrm dz &= \int_a^b (f \circ \tilde r) \cdot \tilde r' \\ &= \int_b^a (f \circ s) \cdot s' \cdot (-1) \\ &= - \int_a^b (f \circ s) \cdot s' = -\oint_{\gamma}f(z)\, \mathrm dz. \end{aligned}$

Also rather intuitively, it is not hard to integrate over curves consisting of “sub-curves”: given $\gamma_1 = r_1([a, b])$ and $\gamma_2 = r_2([b, c])$ with $r_1(b) = r_2(b)$ , define $r = r_1 \cdot \mathbb I_{[a, b]} + r_2 \cdot \mathbb I_{[b, c]}$ and $\gamma_1 + \gamma_2 := r_3([a, c])$ , so that

$\displaystyle \oint_{\gamma_1 + \gamma_2} f(z)\, \mathrm dz = \oint_{\gamma_1} f(z)\, \mathrm dz + \oint_{\gamma_2} f(z)\, \mathrm dz.$

Combining these notions leads to our first powerful lemma in complex analysis: the ML-inequality. We call $\gamma$ a contour if there exist smooth curves $\gamma_k$ such that $\Gamma = \sum_{k=1}^n \gamma_k$ , and suitably obtain

$\displaystyle \| \Gamma \| := \sum_{k=1}^n \| \gamma_k \|,\quad \oint_{\Gamma} f(z)\, \mathrm dz = \sum_{k=1}^n \oint_{\gamma_k} f(z)\, \mathrm dz.$

Henceforth, we will let $\Gamma$ denote a general contour that is the sum of possibly more than one smooth curve $\gamma$ .

Theorem 1 (ML-Inequality). Given continuous $f : \mathbb C \to \mathbb C$ and a contour $\Gamma$ with $M := \max_{z \in \Gamma} |f(z)|$ and $L := \| \Gamma \|$ ,

$\displaystyle \left| \oint_{\Gamma} f(z)\, \mathrm dz \right| \leq M \cdot L.$

Informally, the ML-inequality is the complex version of the triangle inequality that therefore becomes an exceedingly useful bounding technique in proving subsequent theorems in complex analysis.

Proof. First suppose $\gamma$ is a smooth curve. By Lemma 1,

$\begin{aligned} \left| \oint_{\gamma} f(z)\, \mathrm dz \right| &= \left| \int_a^b f(r(t)) \cdot r'(t)\, \mathrm dt \right| \\ &\leq \int_a^b |f(r(t))| \cdot |r'(t)|\, \mathrm dt \\ &\leq \int_a^b M \cdot |r'(t)|\, \mathrm dt \\ &= M \cdot \int_a^b |r'(t)|\, \mathrm dt = M \cdot L. \end{aligned}$

For the general case, denote $M := \max\{M_1,\dots, M_n\}$ and $L := \sum_{k=1}^n L_k$ . Then

$\begin{aligned} \left| \oint_{\Gamma} f(z)\, \mathrm dz \right| &= \left| \sum_{k=1}^n \oint_{\gamma_k} f(z)\, \mathrm dz \right| \\ &\leq \sum_{k=1}^n \left| \oint_{\gamma_k} f(z)\, \mathrm dz \right| \\ &\leq \sum_{k=1}^n M \cdot L_k = M \cdot \sum_{k=1}^n L_k = M \cdot L. \end{aligned}$

Let’s also wrap things up by investigating the situations when complex integration resembles real-variable integration, in terms of the fundamental theorem of calculus.

Definition 5. Let $D \subseteq \mathbb C$ be a domain and $f : D \to \mathbb C$ be continuous.

The function $F : D \to \mathbb C$ an antiderivative of $f$ on $D$ if $F' = f$ .
A contour $\Gamma = r([a,b])$ is closed if $r(a) = r(b)$ .

Furthermore, we call the contour integral from $z_1$ to $z_2$ is path-independent if for any $\gamma_1,\gamma_2$ with starting point $z_1$ and ending point $z_2$ ,

$\displaystyle \oint_{\gamma_1} f(z)\, \mathrm dz = \oint_{\gamma_2} f(z)\, \mathrm dz.$

Lemma 3. Let $f : [a, b] \to \mathbb C$ be continuous. If $F : [a, b] \to \mathbb C$ is an antiderivative of $f$ , then

$\displaystyle \int_a^b f' = F(b) - F(a).$

Theorem 2. Let $f : D \to \mathbb C$ be continuous. The following are equivalent:

$f$ has an antiderivative on $D$ .
$\oint_{\Gamma} f(z)\, \mathrm dz = 0$ for any closed contour $\Gamma \subseteq D$ .
For any $z_1,z_2 \in D$ , the contour integral from $z_1$ to $z_2$ is path-independent.

Proof. First, assume that $f$ has an antiderivative $F$ on $D$ . By the chain rule, $(F \circ r)' = (F' \circ r) \cdot r' = (f \circ r) \cdot r'$ . Fix any contour $\gamma = r([a, b])$ from $z_1$ to $z_2$ that is smooth without loss of generality. By Lemma 3,

$\displaystyle \begin{aligned} \oint_{\gamma} f(z)\, \mathrm dz &= \int_a^b (f \circ r) \cdot r' = \int_a^b (F \circ r)' \\ &= (F \circ r)(b) - (F \circ r)(a) \\ &= F(z_2) - F(z_1). \end{aligned}$

In particular, if $\gamma$ is closed, then $z_1 = r(a) = r(b) = z_2$ , so that

$\displaystyle \begin{aligned} \oint_{\gamma} f(z)\, \mathrm dz &= F(z_2) - F(z_1) = 0. \end{aligned}$

Now if this condition holds, then let $\gamma_1$ and $\gamma_2$ be two distinct contours from $z_1$ to $z_2$ . Then $\gamma_1 + (-\gamma_2)$ is a contour from $z_1$ to itself, i.e. it is a closed contour, so that

$\displaystyle \begin{aligned} \oint_{\gamma_1} f(z)\, \mathrm dz - \oint_{\gamma_2} f(z)\, \mathrm dz &= \oint_{\gamma_1} f(z)\, \mathrm dz + \oint_{-\gamma_2} f(z)\, \mathrm dz \\ &= \oint_{\gamma_1 + (-\gamma_2)} f(z)\, \mathrm dz = 0,\end{aligned}$

implying that

$\displaystyle \oint_{\gamma_1} f(z)\, \mathrm dz = \oint_{\gamma_2} f(z)\, \mathrm dz.$

Finally, suppose the path-independence of the contour integral. Fix $z_0 \in D$ . For any $\zeta \in D$ , let $\gamma_{\zeta}$ be a contour from $z_0$ to $\zeta$ . Define

$\displaystyle F(\zeta) := \oint_{\gamma_{\zeta}} f(z)\, \mathrm dz.$

We claim that $F'(z_0) = f(z_0)$ . Fix $\epsilon > 0$ . By construction $F(z_0) = 0$ . Consider the straight-line contour $[z_0, z_0 +w]$ from $z_0$ to $z_0 +w$ . Then

$\begin{aligned} F(z_0 + w) - F(z_0) - f(z_0) \cdot w &= \oint_{[z_0, z_0+w]} f(z)\, \mathrm dz - 0 - \oint_{[z_0, z_0 + w]} f(z_0)\, \mathrm dz \\ &= \oint_{[z_0, z_0+w]} (f(z) - f(z_0))\, \mathrm dz. \end{aligned}$

Since $f$ is continuous at $z_0$ , for any $k > 0$ , there exists $\delta > 0$ such that

$|z| < \delta \quad \Rightarrow \quad |f(z) - f(z_0)| < k \cdot \epsilon.$

Define

$\displaystyle M := \max_{z \in [z_0, z_0+w]} |f(z_0) - f(z_0)| \leq k \cdot \epsilon,\quad L := |w|.$

Then for $0 < |w| < \delta$ , by the ML-inequality,

$\begin{aligned} |F(z_0 + w) - F(z_0) - f(z_0) \cdot w| &= \left| \oint_{[z_0, z_0+w]} (f(z) - f(z_0))\, \mathrm dz \right| \leq (k \cdot \epsilon) \cdot |w|.\end{aligned}$

Setting $k = 1/2$ yields the desired result.

Next time, we revisit multivariable integration in order to establish Green’s theorem, which we will need to prove the Cauchy-Goursat theorem, which states that sufficiently nice functions over sufficiently nice contours automatically have zero integrals.

—Joel Kindiak, 14 Aug 25, 2252H

KindiakMath

The Complex Integral

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The Complex Integral

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