Chapter 7 • Theory & Derivations

Advanced Contour Integration Techniques: Real Integrals, Jordan's Lemma & Branch Cuts

Evaluation of real trigonometric and improper rational integrals, Jordan's lemma, indented contours, and multivalued functions with branch cuts and keyhole contours.

§7.1 Rational Trigonometric Integrals ∫_0^(2π) R(cos θ, sin θ) dθ via Unit Circle Substitution

1. Unit Circle Substitution

Integrals of the form $I = \int_0^{2\pi} R(\cos\theta, \sin\theta)\,d\theta$ where $R$ is a rational function without singularities on $[0, 2\pi]$ are evaluated by parameterizing the unit circle $C = \{|z| = 1\}$ via $z = e^{i\theta}$:

$$dz = i e^{i\theta} d\theta = i z\,d\theta \implies d\theta = \frac{dz}{iz}$$ $$\cos\theta = \frac{z + z^{-1}}{2} = \frac{z^2 + 1}{2z}, \qquad \sin\theta = \frac{z - z^{-1}}{2i} = \frac{z^2 - 1}{2iz}$$

The integral transforms directly into a rational contour integral over the unit circle:

$$I = \oint_{|z|=1} R\left( \frac{z^2 + 1}{2z}, \frac{z^2 - 1}{2iz} \right) \frac{dz}{iz} = 2\pi i \sum_{|z_k| < 1} \text{Res}(F, z_k)$$

§7.2 Improper Real Integrals ∫_(-∞)^∞ P(x)/Q(x) dx & Semi-Circular Contours

1. Rational Integrals over the Real Line

To evaluate $\int_{-\infty}^\infty \frac{P(x)}{Q(x)}\,dx$ where $P, Q$ are polynomials, $Q(x) \ne 0$ for $x \in \mathbb{R}$, and $\deg Q \ge \deg P + 2$:

Integrate $f(z) = \frac{P(z)}{Q(z)}$ over the closed contour $\Gamma_R = [-R, R] \cup C_R$, where $C_R$ is the upper semicircle $z = R e^{i\theta}, \theta \in [0, \pi]$. By Cauchy's Residue Theorem:

$$\int_{-R}^R f(x)\,dx + \int_{C_R} f(z)\,dz = 2\pi i \sum_{\text{Im}(z_k) > 0} \text{Res}(f, z_k)$$

By the $ML$-inequality, as $R \to \infty$:

$$\left| \int_{C_R} f(z)\,dz \right| \le \frac{M}{R^{\deg Q - \deg P}} \cdot (\pi R) \to 0$$ $$\int_{-\infty}^\infty \frac{P(x)}{Q(x)}\,dx = 2\pi i \sum_{\text{Im}(z_k) > 0} \text{Res}\left( \frac{P}{Q}, z_k \right)$$

§7.3 Fourier-Type Integrals, Indented Contours & Jordan's Lemma

1. Jordan's Lemma

Theorem 7.1 (Jordan's Lemma): Let $C_R$ be the upper semicircle $z = R e^{i\theta}, 0 \le \theta \le \pi$. If $g(z)$ is continuous on $C_R$ and $M(R) = \max_{z \in C_R} |g(z)| \to 0$ as $R \to \infty$, then for any $m > 0$: $$\lim_{R \to \infty} \int_{C_R} g(z) e^{imz}\,dz = 0$$

This allows evaluation of Fourier transforms $\int_{-\infty}^\infty f(x) e^{imx}\,dx$ under the milder degree condition $\deg Q \ge \deg P + 1$!

2. Indented Semicircular Contours

When the integrand has simple poles on the real axis (such as $\frac{e^{iz}}{z}$ at $z = 0$), we indent the contour around the singularity using a small semicircle $\gamma_\epsilon$ of radius $\epsilon$:

Fractional Residue Theorem: If $\gamma_\epsilon$ is a clockwise circular arc of angle $\alpha$ around a simple pole $z_0$: $$\lim_{\epsilon \to 0} \int_{\gamma_\epsilon} f(z)\,dz = -\alpha i\, \text{Res}(f, z_0)$$ For a clockwise semicircle on the real axis ($\alpha = \pi$), the contribution is $-\pi i \text{Res}(f, z_0)$.

§7.4 Multivalued Functions, Branch Points, Branch Cuts, Keyhole & Dogbone Contours

1. Multivalued Functions and Branch Points

Functions like $\log z = \ln|z| + i(\arg z + 2\pi k)$ and $z^\alpha = e^{\alpha \log z}$ are multivalued. A point $z_0$ is a branch point if traversing a small closed loop around $z_0$ changes the branch value of the function.

A branch cut is a curve introduced in $\mathbb{C}$ to prevent paths from encircling the branch point, rendering the function single-valued and holomorphic on the cut domain.

2. The Keyhole Contour

To evaluate integrals of the form $\int_0^\infty x^{\alpha - 1} f(x)\,dx$ ($0 < \alpha < 1$), we cut the plane along the positive real axis $[0, \infty)$ and integrate over a keyhole contour consisting of:

The upper edge of the cut: $z = x + i0^+ \implies z^\alpha = x^\alpha$.
Large outer circle $C_R$ ($R \to \infty$).
The lower edge of the cut: $z = x - i0^+ \implies z^\alpha = x^\alpha e^{2\pi i \alpha}$.
Small inner circle $C_\epsilon$ around $0$ ($\epsilon \to 0$).

The difference between the upper and lower edges yields $(1 - e^{2\pi i \alpha}) \int_0^\infty x^{\alpha - 1} f(x)\,dx = 2\pi i \sum \text{Res}$.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 • Foundational Example 7.1: Real Trigonometric Definite Integral via Residue Calculus

Evaluate using contour integration:

\int_0^{2\pi} \frac{d\theta}{5 + 4\cos\theta}

Step 1: Unit circle transformation
Let $z = e^{i\theta} \implies d\theta = \frac{dz}{iz}$, and $\cos\theta = \frac{z + z^{-1}}{2} = \frac{z^2 + 1}{2z}$.

\int_0^{2\pi} \frac{d\theta}{5 + 4\cos\theta} = \oint_{|z|=1} \frac{1}{5 + 4\left(\frac{z^2 + 1}{2z}\right)} \frac{dz}{iz} = \oint_{|z|=1} \frac{dz}{i z \left(5 + \frac{2z^2 + 2}{z}\right)}

= \oint_{|z|=1} \frac{dz}{i(2z^2 + 5z + 2)}

Step 2: Factor denominator
$2z^2 + 5z + 2 = (2z + 1)(z + 2) = 2(z + 1/2)(z + 2)$. Poles: $z_1 = -1/2$ and $z_2 = -2$.

Step 3: Poles inside unit circle $|z| = 1$

$z_1 = -1/2$: $| -1/2 | = 1/2 < 1$ (inside).
$z_2 = -2$: $| -2 | = 2 > 1$ (outside).

Step 4: Residue at $z_1 = -1/2$

\text{Res}\left( \frac{1}{i(2z^2 + 5z + 2)}, -\frac{1}{2} \right) = \lim_{z \to -1/2} \frac{z + 1/2}{2i(z + 1/2)(z + 2)} = \frac{1}{2i(-1/2 + 2)} = \frac{1}{2i(3/2)} = \frac{1}{3i}

Step 5: Apply Residue Theorem

\int_0^{2\pi} \frac{d\theta}{5 + 4\cos\theta} = 2\pi i \left( \frac{1}{3i} \right) = \frac{2\pi}{3}

Final Answer & Physical Insight

$\frac{2\pi}{3}$

Tier 2 • Intermediate Exam Example 7.2: Dirichlet Integral ∫_0^∞ (sin x)/x dx via Indented Contour

Evaluate the Dirichlet integral $\int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2}$ using an indented contour around the pole at $z = 0$.

Step 1: Complex integrand and contour
Consider $f(z) = \frac{e^{iz}}{z}$. Contour $\Gamma$:

$[-R, -\epsilon]$ along real axis.
Clockwise semicircle $\gamma_\epsilon: z = \epsilon e^{i\theta}, \theta: \pi \to 0$.
$[\epsilon, R]$ along real axis.
Counter-clockwise semicircle $C_R: z = R e^{i\theta}, \theta: 0 \to \pi$.

Since $f(z)$ has no poles inside $\Gamma$, Cauchy's Theorem gives $\oint_\Gamma f(z)\,dz = 0$.

Step 2: Real axis contribution

\int_{-R}^{-\epsilon} \frac{e^{ix}}{x}\,dx + \int_\epsilon^R \frac{e^{ix}}{x}\,dx = \int_\epsilon^R \frac{e^{ix} - e^{-ix}}{x}\,dx = 2i \int_\epsilon^R \frac{\sin x}{x}\,dx

Step 3: Small indentation $\gamma_\epsilon$ ($\epsilon \to 0$)
Near $z = 0$, $\frac{e^{iz}}{z} = \frac{1 + iz + \dots}{z} = \frac{1}{z} + i + \dots$.

\lim_{\epsilon \to 0} \int_{\gamma_\epsilon} \frac{e^{iz}}{z}\,dz = -\pi i\, \text{Res}\left(\frac{e^{iz}}{z}, 0\right) = -\pi i(1) = -\pi i

Step 4: Large semicircle $C_R$ ($R \to \infty$)
By Jordan's lemma, $\lim_{R \to \infty} \int_{C_R} \frac{e^{iz}}{z}\,dz = 0$.

Step 5: Total sum

2i \int_0^\infty \frac{\sin x}{x}\,dx - \pi i + 0 = 0 \implies 2i \int_0^\infty \frac{\sin x}{x}\,dx = \pi i \implies \int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2}

Final Answer & Physical Insight

$\int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2}$

Tier 3 • Honors Challenge Example 7.3: Schwarz-Pick Lemma & Hyperbolic Automorphisms of Unit Disk

Prove the Schwarz-Pick Lemma: For any holomorphic self-map $f: \mathbb{D} \to \mathbb{D}$ of the unit disk, $\left|\frac{f'(z)}{1 - |f(z)|^2}\right| \le \frac{1}{1 - |z|^2}$.

Step 1: Automorphisms of the unit disk $\mathbb{D}$
For any point $\alpha \in \mathbb{D}$, the Blaschke factor:

\phi_\alpha(z) = \frac{z - \alpha}{1 - \bar{\alpha} z}

is a holomorphic bijection from $\mathbb{D}$ to $\mathbb{D}$ with $\phi_\alpha(\alpha) = 0$ and $\phi_\alpha^{-1}(w) = \phi_{-\alpha}(w)$.

Step 2: Composition with Blaschke factors
Fix $z_0 \in \mathbb{D}$ and let $w_0 = f(z_0) \in \mathbb{D}$. Define the composite map:

F(z) = (\phi_{w_0} \circ f \circ \phi_{-z_0})(z)

Notice that $F(0) = \phi_{w_0}(f(\phi_{-z_0}(0))) = \phi_{w_0}(f(z_0)) = \phi_{w_0}(w_0) = 0$. Since $\phi_{w_0}$ and $\phi_{-z_0}$ map $\mathbb{D}$ into $\mathbb{D}$, $F: \mathbb{D} \to \mathbb{D}$.

Step 3: Apply Schwarz's Lemma
By Schwarz's Lemma (Theorem 4.7), $|F'(0)| \le 1$.
Using the chain rule:

F'(0) = \phi_{w_0}'(w_0) \cdot f'(z_0) \cdot \phi_{-z_0}'(0)

Compute derivatives of Blaschke factors:

\phi_\alpha'(z) = \frac{(1 - \bar{\alpha} z)(1) - (z - \alpha)(-\bar{\alpha})}{(1 - \bar{\alpha} z)^2} = \frac{1 - |\alpha|^2}{(1 - \bar{\alpha} z)^2}

At $z = \alpha$: $\phi_{w_0}'(w_0) = \frac{1 - |w_0|^2}{(1 - |w_0|^2)^2} = \frac{1}{1 - |w_0|^2} = \frac{1}{1 - |f(z_0)|^2}$.
At $z = 0$: $\phi_{-z_0}'(0) = 1 - |-z_0|^2 = 1 - |z_0|^2$.

Step 4: Combine into $|F'(0)| \le 1$

|F'(0)| = \frac{1}{1 - |f(z_0)|^2} \cdot |f'(z_0)| \cdot (1 - |z_0|^2) \le 1 \implies \frac{|f'(z_0)|}{1 - |f(z_0)|^2} \le \frac{1}{1 - |z_0|^2}

This proves that every holomorphic self-map of the unit disk is a contraction with respect to the Poincaré hyperbolic metric $ds = \frac{|dz|}{1 - |z|^2}$! $\blacksquare$

Final Answer & Physical Insight

Proved: $\frac{|f'(z)|}{1 - |f(z)|^2} \le \frac{1}{1 - |z|^2}$ (Poincaré Hyperbolic Contraction).