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Evaluating Improper Integrals Using the Residue Theorem

Table of Contents

1. Foundations of Complex Integration
   • Complex Functions and Contours: A Brief Review
   • Cauchy’s Theorem and Integral Formula
   • Classification of Isolated Singularities
   • The Residue Theorem: Statement and Intuition
   • How the Residue Theorem Evaluates Real Integrals
2. Chapter 1: Rational Functions over the Real Line
   • The Standard Semi-Circular Contour
   • Vanishing of the Arc Integral: Degree Condition
   • Identifying Appropriate Contours for Poles off the Real Axis
   • Simple Poles, Higher-Order Poles, and Multiple Poles
   • Worked Examples
3. Chapter 2: Oscillatory Integrals and Jordan’s Lemma
   • Motivation: Why Standard Estimations Fail
   • Jordan’s Lemma: Statement, Proof, and Applications
   • Estimating Arc Contributions: Strategy and ML Inequality
   • Worked Examples
4. Chapter 3: Trigonometric Integrals over $[0, 2\pi]$ via Substitution
   • The Unit Circle Parametrization: $z = e^{i\theta}$
   • When and How to Transform: Checklist and Heuristics
   • Handling $\cos\theta$, $\sin\theta$, and Rational Trig Expressions
   • Worked Examples
5. Chapter 4: Integrals with Singularities on the Real Line
   • Principal Value Integrals and the Cauchy Principal Value
   • The Indented Contour Technique: Justification and Lemma
   • The Sokhotski–Plemelj Theorem (with Proof Sketch)
   • Worked Examples
6. Chapter 5: Branch Points and Keyhole Contours
   • Multivalued Functions: Logarithms and Power Laws
   • How to Choose and Place Branch Cuts
   • The Keyhole Contour: Anatomy and Application
   • Worked Examples
7. Chapter 6: Advanced Contour Strategies and Non-Standard Paths
   • Rectangular Contours for Periodic Integrands
   • Wedge and Sector Contours for Root and Angular Symmetries
   • Spiral and Logarithmic Contours (Optional, Advanced)
   • Reflection Principle and Even-Odd Extensions
   • Problem-Solving Flowchart and Contour Selection Guide
8. Appendix
   • Appendix A: Essential Lemmas and Estimation Tools
   • Appendix B: Quick Reference Table of Common Integrals

Foundations of Complex Integration

Welcome to this comprehensive guide on leveraging the power of the residue theorem for evaluating improper integrals. As an advanced undergraduate or early graduate student, you likely possess a foundational understanding of complex analysis. This introductory chapter serves as a crucial bridge, a robust review of the core concepts that underpin the residue theorem, and a preliminary look at how this elegant theorem provides a systematic approach to evaluating real-valued integrals that often prove intractable by traditional real-variable calculus methods. We will build a solid understanding, moving from the definitions of complex functions and contours to the profound implications of Cauchy’s theorems, the classification of singularities, and finally, the statement and intuitive interpretation of the residue theorem itself.

Complex Functions and Contours: A Brief Review

At the heart of complex analysis lies the concept of a complex function, $f: \mathbb{C} \to \mathbb{C}$, which maps a complex number $z = x + iy$ to another complex number $w = u + iv$, where $u$ and $v$ are real-valued functions of $x$ and $y$. That is, $f(z) = u(x, y) + iv(x, y)$.

The integration of such functions is not over an interval on the real line, but rather along a path or contour in the complex plane. A contour $\gamma$ is a piecewise smooth, oriented curve in the complex plane. If a contour is closed (its start and end points coincide) and does not intersect itself, it is called a simple closed contour. We parameterize a contour by $z(t) = x(t) + iy(t)$ for $a \leq t \leq b$. The complex line integral of a function $f(z)$ along a contour $\gamma$ is defined as:

This definition extends the familiar concept of line integrals from multivariable calculus, incorporating the rich structure of complex numbers. The direction of integration is paramount; reversing the direction of a contour changes the sign of the integral. We often denote the contour integral over a simple closed contour $C$ as $\oint_C f(z) \, dz$, with the understanding that the integration is performed in the positive (counter-clockwise) direction.

Cauchy’s Theorem and Integral Formula

The elegance and power of complex analysis often stem from the remarkable properties of analytic functions. A complex function $f(z)$ is said to be analytic (or holomorphic) at a point $z_0$ if its derivative $f’(z_0)$ exists in a neighborhood of $z_0$. If $f(z)$ is analytic throughout an open set, it possesses infinitely many derivatives and can be represented by a convergent Taylor series.

Cauchy’s Theorem (Cauchy-Goursat Theorem): If a function $f(z)$ is analytic everywhere inside and on a simple closed contour $C$, then the integral of $f(z)$ around $C$ is zero:

This theorem is profoundly significant. It implies that for an analytic function, the integral between two points is path-independent, a characteristic reminiscent of conservative vector fields in real analysis. The vanishing of the integral signifies the absence of any “sources” or “sinks” within the enclosed region, a concept crucial for understanding what happens when a function is not analytic.

When a function $f(z)$ is not analytic at certain points, these points are called singularities. The most important direct consequence of non-analytic behavior within a contour is captured by Cauchy’s Integral Formula.

Cauchy’s Integral Formula: If $f(z)$ is analytic everywhere inside and on a simple closed contour $C$, and $z_0$ is any point inside $C$, then:

This formula is astounding because it shows that the value of an analytic function at any point inside a contour is completely determined by its values on the boundary of that contour. Moreover, by repeatedly differentiating with respect to $z_0$, we can obtain a generalized version for the $n$-th derivative:

Cauchy’s Integral Formula (and its generalized form) is not just a theoretical result; it forms the bedrock for understanding the concept of a residue, which is precisely what we need to evaluate integrals when singularities are present. Notice the structure: an integral whose integrand has a singularity of the form $1/(z-z_0)^{n+1}$ yields $2\pi i / n!$ times a derivative of $f(z)$ evaluated at $z_0$. This foreshadows the definition of a residue.

Classification of Isolated Singularities

A point $z_0$ is an isolated singularity of $f(z)$ if $f(z)$ is not analytic at $z_0$ but is analytic at every point in some punctured disk $0 < \lvert z - z_0 \rvert < \rho$ for some $\rho > 0$. Isolated singularities are classified based on the behavior of the function’s Laurent series expansion around $z_0$. A function $f(z)$ with an isolated singularity at $z_0$ can be represented by a Laurent series of the form:

where the coefficient $a_n$ is given by $a_n = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz$ for any simple closed contour $C$ enclosing $z_0$ and lying within the annulus of convergence.

There are three types of isolated singularities:

1. Removable Singularities: If $a_n = 0$ for all $n < 0$, the principal part of the Laurent series (the terms with negative powers) is zero. In this case, $\lim_{z \to z_0} f(z)$ exists, and we can redefine $f(z_0)$ to make $f(z)$ analytic at $z_0$. For example, $f(z) = \frac{\sin z}{z}$ has a removable singularity at $z=0$.

2. Poles: If there is a finite number of negative power terms in the Laurent series, meaning $a_{-m} \neq 0$ for some positive integer $m$ but $a_n = 0$ for all $n < -m$, then $z_0$ is called a pole of order $m$.

   • If $m=1$, it is a simple pole.
   • For example, $f(z) = \frac{1}{z^2}$ has a pole of order 2 at $z=0$. $f(z) = \frac{1}{z-i}$ has a simple pole at $z=i$. A pole of order $m$ can be characterized by $\lim_{z \to z_0} (z - z_0)^m f(z) = L \neq 0$ and finite.

3. Essential Singularities: If there are infinitely many negative power terms in the Laurent series, then $z_0$ is an essential singularity. For example, $f(z) = e^{1/z}$ has an essential singularity at $z=0$, as its Laurent series is $e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \dots$. The behavior of a function near an essential singularity is highly erratic and complex (e.g., Picard’s Great Theorem).

The classification of singularities is crucial because the value of the integral around a singularity depends entirely on the coefficient $a_{-1}$ in the Laurent series. This coefficient is known as the residue of $f(z)$ at $z_0$.

Definition: Residue The residue of a function $f(z)$ at an isolated singularity $z_0$, denoted $\text{Res}(f, z_0)$ is the coefficient $a_{-1}$ of the term $(z - z_0)^{-1}$ in the Laurent series expansion of $f(z)$ about $z_0$.

From the formula for $a_n$ and setting $n=-1$, we see that:

This fundamental relationship directly links the residue to the contour integral. Rearranging this gives us the core of the residue theorem.

Calculating Residues:

• For a simple pole ($m=1$): If $f(z) = \frac{P(z)}{Q(z)}$ where $P(z_0) \neq 0$ and $Q(z)$ has a simple zero at $z_0$ (i.e., $Q(z_0)=0$ but $Q’(z_0) \neq 0$), then:

  \[\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z)\]

  Alternatively, for $f(z) = \frac{P(z)}{Q(z)}$ with $P(z_0) \neq 0$ and $Q(z_0) = 0, Q’(z_0) \neq 0$:

  \[\text{Res}(f, z_0) = \frac{P(z_0)}{Q'(z_0)}\]

• For a pole of order $m > 1$: If $f(z)$ has a pole of order $m$ at $z_0$, then:

  \[\text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} [(z - z_0)^m f(z)]\]

  This formula can be derived directly from the Laurent series coefficients. For $m=2$, it’s $\lim_{z \to z_0} \frac{d}{dz} [(z - z_0)^2 f(z)]$.

The Residue Theorem: Statement and Intuition

The Residue Theorem is the pinnacle of this foundational review, providing an extraordinarily powerful tool for evaluating contour integrals.

The Residue Theorem: Let $f(z)$ be analytic inside and on a simple closed contour $C$, except for a finite number of isolated singularities $z_1, z_2, \ldots, z_n$ inside $C$. Then the integral of $f(z)$ around $C$ in the positive sense is given by:

Intuition: The Residue Theorem states that the total “flux” or “circulation” of a function around a closed path is determined entirely by the “strengths” of the singularities enclosed by that path. Each singularity acts like a “source” or “vortex” in the complex plane, and its residue quantifies its contribution to the overall integral. If a singularity is outside the contour, it contributes nothing to the integral, consistent with Cauchy’s Theorem.

Imagine the complex plane as a fluid flow. An analytic function represents a smooth flow without any sources or sinks, so the net flow around any closed loop is zero (Cauchy’s Theorem). Singularities, on the other hand, represent points where “fluid” is either emerging or disappearing. The residue at such a point quantifies the strength and direction of this “source” or “sink.” The Residue Theorem then says that the total circulation around a closed path is simply the sum of the “strengths” of all the sources/sinks enclosed by that path, scaled by $2\pi i$.

This scaling factor $2\pi i$ often surprises newcomers. It arises naturally from the definition of the residue and the geometry of winding around a point in the complex plane. Think of it as the fundamental integral $\oint_C \frac{1}{z - z_0} dz = 2\pi i$ for $C$ encircling $z_0$ once counter-clockwise. The residue $a_{-1}$ for a general function $f(z)$ is precisely the coefficient that determines this $2\pi i$ contribution from each singularity.

How the Residue Theorem Evaluates Real Integrals

The primary goal of this guide is to demonstrate how the Residue Theorem can be systematically applied to evaluate various types of improper real integrals, many of which are challenging or impossible using only real calculus techniques. The general strategy involves transforming a real integral into a contour integral in the complex plane.

Consider a real integral of the form $\int_{-\infty}^{\infty} f(x) \, dx$. The core idea is to construct a closed contour $C$ in the complex plane such that a portion of $C$ lies along the real axis, corresponding to the real integral we wish to evaluate.

The typical process is as follows:

1. Formulate the Complex Function: Identify a complex function $f(z)$ such that its restriction to the real axis, $f(x)$, matches the integrand of the real integral.

2. Choose a Suitable Contour: Select a closed contour $C$ that encloses the relevant singularities of $f(z)$ and includes a segment along the real axis. The most common choice for integrals over $(-\infty, \infty)$ is a large semicircular contour in the upper half-plane (or lower half-plane, depending on the singularities).

   • A semicircular contour in the upper half-plane: This contour consists of two parts: a line segment along the real axis from $-R$ to $R$, denoted as $C_R$, and a large semi-circular arc $\Gamma_R$ of radius $R$ in the upper half-plane, centered at the origin, going from $R$ to $-R$. The full contour is $C = [-R, R] \cup \Gamma_R$.
   \[\begin{array}{l} \quad \quad \quad \quad \Gamma_R \\ \quad \quad \quad \quad \text{Arc in upper half-plane} \\ \quad \quad \text{----------- } \bullet \text{ ----------- } \\ \text{Real axis: -R to R} \end{array}\]

3. Apply the Residue Theorem: Evaluate the contour integral $\oint_C f(z) \, dz$ using the Residue Theorem. This involves finding all singularities of $f(z)$ inside the chosen contour $C$ and calculating their residues.

   \[\oint_C f(z) \, dz = 2\pi i \sum_{k=1}^n \text{Res}(f, z_k)\]

4. Decompose the Contour Integral: Break the contour integral into parts corresponding to the segments of the contour. For the semi-circular contour:

   \[\oint_C f(z) \, dz = \int_{-R}^R f(x) \, dx + \int_{\Gamma_R} f(z) \, dz\]

5. Take the Limit as $R \to \infty$: The integral along the real axis becomes the desired improper integral:

   \[\lim_{R \to \infty} \int_{-R}^R f(x) \, dx = \int_{-\infty}^{\infty} f(x) \, dx\]

   The crucial step is to show that the integral over the arc $\Gamma_R$ vanishes as $R \to \infty$. This is where various estimation lemmas (like the ML-inequality and Jordan’s Lemma, which we will explore in detail) come into play. If this arc integral goes to zero, then:

   \[\int_{-\infty}^{\infty} f(x) \, dx = 2\pi i \sum_{k=1}^n \text{Res}(f, z_k)\]

   This establishes the direct link between a real improper integral and the sum of residues of an associated complex function.

This framework forms the core strategy for evaluating improper integrals using the residue theorem. The subsequent chapters will delve into the specific conditions, techniques, and common pitfalls associated with each step, equipping you with the expertise to tackle a wide variety of integral forms. You will learn to choose appropriate contours, master estimation techniques, and skillfully calculate residues to solve problems that might otherwise seem insurmountable.

Chapter 1: Rational Functions over the Real Line

Having established the foundational concepts, we now embark on our first journey into applying the Residue Theorem to evaluate improper integrals. This chapter focuses specifically on integrals of rational functions over the entire real line, i.e., integrals of the form $\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} \, dx$, where $P(x)$ and $Q(x)$ are polynomials with real coefficients, and $Q(x)$ has no zeros on the real axis. This class of integrals provides an ideal starting point due to the predictable nature of their singularities (poles) and the relatively straightforward contour choice.

The Standard Semi-Circular Contour

To evaluate $\int_{-\infty}^{\infty} f(x) \, dx$ where $f(x) = P(x)/Q(x)$, our strategy is to convert this real integral into a contour integral in the complex plane. The standard and most common choice for such integrals is a large semi-circular contour.

Let $f(z) = P(z)/Q(z)$ be the complex extension of $f(x)$. We consider a closed contour $C_R$ consisting of two parts:

1. A line segment along the real axis from $-R$ to $R$, denoted $L_R$.
2. A semi-circular arc in the upper half-plane, denoted $\Gamma_R$, with radius $R$ and centered at the origin, traversed from $R$ to $-R$ in the counter-clockwise direction.

This contour is visualized as:

        Im(z)
          ^
          |
       .:::::.  <-- Gamma_R (semi-circle)
    .::       ::.
  .:             :.
  |               |
--+---------------+----> Re(z)
 -R             R
        L_R (real axis segment)

The contour $C_R = L_R \cup \Gamma_R$ encloses any poles of $f(z)$ that lie in the upper half-plane. By the Residue Theorem, the integral over $C_R$ is:

where the sum is over all poles $z_k$ of $f(z)$ located inside $C_R$ (i.e., in the upper half-plane, with $ \lvert z_k \rvert < R$).

We can decompose this integral:

As $R \to \infty$, the integral over $L_R$ approaches the desired real integral:

Therefore, if we can show that the integral over the semi-circular arc $\Gamma_R$ vanishes as $R \to \infty$, then we have:

This makes the vanishing of the arc integral absolutely critical for this method to work.

Why Choose the Upper Half-Plane?

The choice of the upper half-plane is often arbitrary. You could equally well choose a semi-circular contour in the lower half-plane, enclosing poles with negative imaginary parts. The key is to consistently select one half-plane and ensure you identify all poles within that chosen region. Sometimes, the specific form of the integrand (especially with exponential terms, as we will see in Chapter 2) might suggest a preferred half-plane. For pure rational functions, both typically work, but the upper half-plane is the default convention. If you choose the lower half-plane, remember that the contour must be traversed clockwise to enclose the poles, which introduces a negative sign: $\oint_{C_R} f(z) \, dz = -2\pi i \sum \text{Res}(f, z_k)$ (or traverse counter-clockwise, for which the real line segment would go from $R$ to $-R$, which is less common for real integrals from $-\infty$ to $\infty$). Sticking to the upper half-plane for now simplifies things.

Vanishing of the Arc Integral: Degree Condition

The condition for the integral over $\Gamma_R$ to vanish as $R \to \infty$ is perhaps the most important detail to master in this chapter. It relies on bounding the integrand’s magnitude on the arc.

Derivation using ML Inequality

The primary tool for estimating the magnitude of a contour integral is the ML Inequality:

If $f(z)$ is continuous on a contour $C$ of length $L$, and $\lvert f(z) \rvert \leq M$ for all $z \in C$, then:

For our semi-circular arc $\Gamma_R$:

• The length $L$ is the length of a semi-circle of radius $R$, so $L = \pi R$.
• We need to find an upper bound $M$ for $\lvert f(z)\rvert$ on $\Gamma_R$.

Let $f(z) = \frac{P(z)}{Q(z)}$, where $P(z) = a_n z^n + \dots + a_0$ and $Q(z) = b_m z^m + \dots + b_0$ are polynomials. For large $\lvert z\rvert$ (i.e., on $\Gamma_R$ where $\lvert z\rvert =R$), the highest power terms dominate the behavior of the polynomials. So, as $\lvert z\rvert \to \infty$: $\lvert P(z)\rvert \approx \lvert a_n z^n\rvert$ $\lvert Q(z)\rvert \approx \lvert b_m z^m\rvert$

Therefore, for sufficiently large $R$:

where $C$ is some constant. So, we can set $M \approx C/R^{m-n}$.

Now, applying the ML Inequality to $\int_{\Gamma_R} f(z) \, dz$:

For this expression to go to zero as $R \to \infty$, the exponent in the denominator must be positive:

This is the famous degree condition.

What if the Condition Fails? Techniques and Alternate Strategies

If $\deg Q < \deg P + 2$, the standard semi-circular contour method with the simple ML-inequality bound will not work directly. This means the integral over $\Gamma_R$ might not vanish, and the method fails.

Common Scenarios where it fails:

• $\deg Q = \deg P + 1$: The integrand decays like $1/z$ for large $\lvert z\rvert $. The product $M L \approx (1/R) \cdot (\pi R) = \pi$, which does not go to zero.
• $\deg Q \leq \deg P$: The integrand does not decay, or even grows, for large $\lvert z\rvert $. The integral $\int_{-\infty}^{\infty} f(x) \, dx$ might not converge in the first place.

What to do if the condition fails?

1. Check for convergence: If $\deg Q \leq \deg P$, the real integral $\int_{-\infty}^{\infty} f(x) \, dx$ itself often diverges (e.g., $\int_{-\infty}^{\infty} \frac{x}{x^2+1} dx$ diverges). The residue theorem cannot help evaluate a divergent integral.
2. Jordan’s Lemma (Chapter 2): If the integrand contains oscillatory terms like $e^{ikx}$, even if $\deg Q = \deg P + 1$, Jordan’s Lemma might allow the arc integral to vanish. This is a common situation for Fourier transforms.
3. Indented Contours (Chapter 4): If poles lie directly on the real axis, the semi-circular contour cannot be used as is. We must indent the contour around these poles, leading to principal value integrals.
4. Other Contour Shapes (Chapter 6): For more complex integrands or specific symmetries, other contours (rectangular, wedge, keyhole) might be necessary.

Identifying Appropriate Contours for Poles off the Real Axis

For rational functions $P(x)/Q(x)$ where $Q(x)$ has no real roots, all poles of $f(z) = P(z)/Q(z)$ are off the real axis. The choice of the semi-circular contour (upper or lower half-plane) determines which poles are enclosed. You must identify all poles of $f(z)$ by finding the roots of $Q(z)$. Then, select the half-plane that contains a non-empty set of poles.

For example, if $Q(z)$ has roots $z_1, z_2, \ldots, z_m$, calculate their imaginary parts. If $\text{Im}(z_k) > 0$, the pole is in the upper half-plane. If $\text{Im}(z_k) < 0$, it’s in the lower half-plane. For our standard upper semi-circular contour, we only include poles with positive imaginary parts.

Simple Poles, Higher-Order Poles, and Multiple Poles

Once the contour is chosen and the vanishing arc integral confirmed, the next step is to calculate the residues of $f(z)$ at all poles enclosed by the contour.

Simple Poles (Order $m=1$)

If $z_0$ is a simple pole, the residue is:

If $f(z) = P(z)/Q(z)$ where $Q(z_0)=0$ and $Q’(z_0) \neq 0$ (i.e., $z_0$ is a simple root of $Q(z)$), this simplifies to:

This formula is very convenient and often faster than the limit definition.

Higher-Order Poles (Order $m > 1$)

If $z_0$ is a pole of order $m$, the residue is given by:

This formula requires computing $(m-1)$ derivatives. For $m=2$ (a pole of order 2), you need to compute the first derivative:

Multiple Poles vs. Higher-Order Poles

It’s important to distinguish between “multiple poles” and “higher-order poles.”

• A “higher-order pole” (e.g., a pole of order 2 or 3) refers to the multiplicity of a single singularity.
• “Multiple poles” (e.g., $z_1, z_2, z_3$) simply refers to having several distinct singularities within the contour, each of which might be simple or higher-order. You calculate the residue for each enclosed pole and sum them up.

Worked Examples

Let’s now apply these concepts to evaluate some common improper integrals.

Example 1.1: Integral with Simple Poles

Evaluate $\int_{-\infty}^{\infty} \frac{1}{x^2 + a^2} dx$ where $a > 0$.

1. Complex Function and Contour: Let $f(z) = \frac{1}{z^2 + a^2}$. We use the standard semi-circular contour $C_R$ in the upper half-plane.

2. Check Degree Condition: $P(z) = 1$ ($\deg P = 0$), $Q(z) = z^2 + a^2$ ($\deg Q = 2$). $\deg Q = 2 \geq \deg P + 2 = 0 + 2 = 2$. The condition is met, so $\lim_{R \to \infty} \int_{\Gamma_R} f(z) \, dz = 0$.

3. Identify Poles: Set $z^2 + a^2 = 0 \implies z^2 = -a^2 \implies z = \pm ia$. The pole in the upper half-plane is $z_0 = ia$. This is a simple pole.

4. Calculate Residue: Using the formula $\text{Res}(f, z_0) = \frac{P(z_0)}{Q’(z_0)}$: $P(z) = 1$, $Q(z) = z^2 + a^2 \implies Q’(z) = 2z$. $\text{Res}(f, ia) = \frac{1}{2(ia)} = \frac{1}{2ia}$.

5. Apply Residue Theorem:

   \[\oint_{C_R} f(z) \, dz = 2\pi i \cdot \text{Res}(f, ia) = 2\pi i \cdot \frac{1}{2ia} = \frac{\pi}{a}\]

6. Decompose and Take Limit:

   \[\lim_{R \to \infty} \left( \int_{-R}^R \frac{dx}{x^2 + a^2} + \int_{\Gamma_R} \frac{dz}{z^2 + a^2} \right) = \frac{\pi}{a}\]

   Since the arc integral vanishes:

   \[\int_{-\infty}^{\infty} \frac{1}{x^2 + a^2} dx = \frac{\pi}{a}\]

   This matches the result from real calculus, confirming our method.

Example 1.2: Integral with a Higher-Order Pole

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

1. Complex Function and Contour: Let $f(z) = \frac{1}{(z^2 + 1)^2}$. Use the standard semi-circular contour $C_R$ in the upper half-plane.

2. Check Degree Condition: $P(z) = 1$ ($\deg P = 0$), $Q(z) = (z^2 + 1)^2 = z^4 + 2z^2 + 1$ ($\deg Q = 4$). $\deg Q = 4 \geq \deg P + 2 = 0 + 2 = 2$. The condition is met, so the arc integral vanishes.

3. Identify Poles: Set $(z^2 + 1)^2 = 0 \implies z^2 + 1 = 0 \implies z = \pm i$. The pole in the upper half-plane is $z_0 = i$. This is a pole of order 2 (since $(z^2+1)^2 = ((z-i)(z+i))^2 = (z-i)^2(z+i)^2$).

4. Calculate Residue: For a pole of order $m=2$ at $z_0 = i$:

   \[\text{Res}(f, i) = \lim_{z \to i} \frac{d}{dz} [(z - i)^2 f(z)]\] \[= \lim_{z \to i} \frac{d}{dz} \left[ (z - i)^2 \frac{1}{(z - i)^2 (z + i)^2} \right]\] \[= \lim_{z \to i} \frac{d}{dz} \left[ \frac{1}{(z + i)^2} \right]\] \[= \lim_{z \to i} \left[ -2 (z + i)^{-3} \right]\] \[= -2 (i + i)^{-3} = -2 (2i)^{-3} = -2 \frac{1}{8i^3} = -2 \frac{1}{8(-i)} = -2 \frac{i}{8} = -\frac{i}{4}\]

5. Apply Residue Theorem:

   \[\oint_{C_R} f(z) \, dz = 2\pi i \cdot \text{Res}(f, i) = 2\pi i \left( -\frac{i}{4} \right) = -2\pi \frac{i^2}{4} = -2\pi \frac{-1}{4} = \frac{\pi}{2}\]

6. Decompose and Take Limit:

   \[\int_{-\infty}^{\infty} \frac{1}{(x^2 + 1)^2} dx = \frac{\pi}{2}\]

Example 1.3: Integral with Multiple Simple Poles

Evaluate $\int_{-\infty}^{\infty} \frac{1}{x^4 + 1} dx$.

1. Complex Function and Contour: Let $f(z) = \frac{1}{z^4 + 1}$. Use the standard semi-circular contour $C_R$ in the upper half-plane.

2. Check Degree Condition: $P(z) = 1$ ($\deg P = 0$), $Q(z) = z^4 + 1$ ($\deg Q = 4$). $\deg Q = 4 \geq \deg P + 2 = 0 + 2 = 2$. The condition is met, so the arc integral vanishes.

3. Identify Poles: Set $z^4 + 1 = 0 \implies z^4 = -1 = e^{i(\pi + 2k\pi)}$. The roots are $z_k = e^{i(\pi + 2k\pi)/4}$ for $k = 0, 1, 2, 3$.

   • $k=0: z_0 = e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$ (Upper Half-Plane)
   • $k=1: z_1 = e^{i3\pi/4} = \cos(3\pi/4) + i\sin(3\pi/4) = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$ (Upper Half-Plane)
   • $k=2: z_2 = e^{i5\pi/4} = \cos(5\pi/4) + i\sin(5\pi/4) = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$ (Lower Half-Plane)
   • $k=3: z_3 = e^{i7\pi/4} = \cos(7\pi/4) + i\sin(7\pi/4) = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$ (Lower Half-Plane)

   The poles enclosed by $C_R$ are $z_0 = e^{i\pi/4}$ and $z_1 = e^{i3\pi/4}$. All are simple poles.

4. Calculate Residues: Using $\text{Res}(f, z_k) = \frac{P(z_k)}{Q’(z_k)}$ where $P(z)=1$ and $Q’(z) = 4z^3$.

   • For $z_0 = e^{i\pi/4}$:

     \[\text{Res}(f, e^{i\pi/4}) = \frac{1}{4(e^{i\pi/4})^3} = \frac{1}{4e^{i3\pi/4}} = \frac{1}{4 (\cos(3\pi/4) + i\sin(3\pi/4))}\] \[= \frac{1}{4(-\sqrt{2}/2 + i\sqrt{2}/2)} = \frac{1}{2\sqrt{2}(-1 + i)} = \frac{1}{2\sqrt{2}(-1 + i)} \cdot \frac{-1 - i}{-1 - i}\] \[= \frac{-1 - i}{2\sqrt{2}(1 - i^2)} = \frac{-1 - i}{2\sqrt{2}(2)} = \frac{-1 - i}{4\sqrt{2}}\]

   • For $z_1 = e^{i3\pi/4}$:

     \[\text{Res}(f, e^{i3\pi/4}) = \frac{1}{4(e^{i3\pi/4})^3} = \frac{1}{4e^{i9\pi/4}} = \frac{1}{4e^{i(\pi/4 + 2\pi)}} = \frac{1}{4e^{i\pi/4}}\] \[= \frac{1}{4 (\cos(\pi/4) + i\sin(\pi/4))} = \frac{1}{4 (\sqrt{2}/2 + i\sqrt{2}/2)} = \frac{1}{2\sqrt{2}(1 + i)}\] \[= \frac{1}{2\sqrt{2}(1 + i)} \cdot \frac{1 - i}{1 - i} = \frac{1 - i}{2\sqrt{2}(1 - i^2)} = \frac{1 - i}{2\sqrt{2}(2)} = \frac{1 - i}{4\sqrt{2}}\]

5. Sum of Residues:

   \[\sum \text{Res} = \frac{-1 - i}{4\sqrt{2}} + \frac{1 - i}{4\sqrt{2}} = \frac{-2i}{4\sqrt{2}} = -\frac{i}{2\sqrt{2}}\]

6. Apply Residue Theorem:

   \[\oint_{C_R} f(z) \, dz = 2\pi i \sum \text{Res} = 2\pi i \left( -\frac{i}{2\sqrt{2}} \right) = -2\pi \frac{i^2}{2\sqrt{2}} = -2\pi \frac{-1}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}} = \frac{\pi\sqrt{2}}{2}\]

7. Decompose and Take Limit:

   \[\int_{-\infty}^{\infty} \frac{1}{x^4 + 1} dx = \frac{\pi\sqrt{2}}{2}\]

Pitfalls and Error Patterns

Even with a clear understanding, common mistakes can derail your calculations. Be vigilant about the following:

1. Incorrect Pole Identification:

   • Missing poles: Ensure you find all roots of the denominator $Q(z)$. Complex roots always come in conjugate pairs, so if $z_0$ is a root, $\overline{z_0}$ is also a root.
   • Including poles outside the contour: Only sum residues for poles inside your chosen contour. For the upper semi-circle, this means poles with positive imaginary parts.
   • Misclassifying pole order: Double-check if a pole is simple, double, or higher order, as this affects the residue calculation formula.

2. Failure to Check Degree Condition:

   • Never assume the arc integral vanishes. Always verify $\deg Q \geq \deg P + 2$. If it doesn’t hold, the standard method for rational functions is insufficient, and you’ll need more advanced techniques (like Jordan’s Lemma in the next chapter) or the integral might diverge.

3. Errors in Residue Calculation:

   • Algebraic mistakes: Complex arithmetic can be tricky. Be careful with powers of $i$, adding/subtracting complex numbers, and rationalizing denominators.
   • Derivative errors: If using the higher-order pole formula, ensure your derivatives are correct. Remember to differentiate with respect to $z$.
   • Sign errors from contour direction: While less common for the standard semi-circular contour (which is almost always taken counter-clockwise), if you ever choose a clockwise contour (e.g., lower half-plane with a negative $2\pi i$), be mindful of the resulting sign change.

4. Misinterpreting $i$ vs. $1/i$:

   • Remember $1/i = -i$. This comes up frequently in residue calculations.

By systematically following the steps outlined in this chapter – defining the complex function, choosing the right contour, rigorously checking the vanishing arc integral, identifying and calculating all relevant residues, and summing them – you can reliably evaluate a broad class of improper real integrals. The next chapter will build on this foundation by addressing integrals that include oscillatory terms, requiring a more specialized lemma for arc integral estimation.

Chapter 2: Oscillatory Integrals and Jordan’s Lemma

In Chapter 1, we mastered the evaluation of improper integrals of rational functions, where the vanishing of the semi-circular arc integral was guaranteed by the condition $\deg Q \geq \deg P + 2$. However, many integrals of significant practical importance involve oscillatory terms, such as $\cos(kx)$ or $\sin(kx)$. These integrals, often encountered in Fourier analysis and signal processing, require a more sophisticated approach for the arc integral estimation. This chapter introduces Jordan’s Lemma, a specialized and powerful tool that allows the arc integral to vanish under less stringent conditions on the rational part of the integrand, specifically designed for these “oscillatory integrals.”

Motivation: Why Standard Estimations Fail

Consider an integral like $\int_{-\infty}^{\infty} \frac{\cos(kx)}{x^2 + b^2} dx$, where $k$ is a positive real constant. Our initial thought might be to extend the integrand directly to the complex plane as $f(z) = \frac{\cos(kz)}{z^2 + b^2}$ and apply the semi-circular contour method. Let’s analyze the behavior of $\cos(kz)$ on the semi-circular arc $\Gamma_R$ in the upper half-plane, where $z = x + iy$ and $y \geq 0$.

Using Euler’s formula, $\cos(kz) = \frac{e^{ikz} + e^{-ikz}}{2}$. Let $z = R e^{i\theta} = R(\cos\theta + i\sin\theta)$ for $0 \leq \theta \leq \pi$ (on $\Gamma_R$). Then

So, $\lvert e^{ikz}\rvert = \lvert e^{ikR\cos\theta} e^{-kR\sin\theta}\rvert = e^{-kR\sin\theta}$.

Since $k>0$ and $\sin\theta \geq 0$ for $\theta \in [0, \pi]$ (upper half-plane), the term $e^{-kR\sin\theta}$ is bounded by 1 and actually decays (or stays constant at 1 for $\theta=0, \pi$) as $R \to \infty$. This part is well-behaved.

Now consider the second term, $e^{-ikz}$:

Since $k>0$ and $\sin\theta \geq 0$ in the upper half-plane, this term grows exponentially as $R \to \infty$ (unless $\sin\theta = 0$, i.e., on the real axis). For example, at $\theta = \pi/2$ (top of the semi-circle), $\sin\theta = 1$, and $\lvert e^{-ikz}\rvert = e^{kR}$, which grows without bound.

This exponential growth means that the maximum value $M$ of $\lvert f(z)\rvert $ on $\Gamma_R$ will also grow exponentially, making the ML Inequality bound ($M \cdot L$) diverge. Thus, the standard ML Inequality approach from Chapter 1 fails for integrands containing $\cos(kz)$ or $\sin(kz)$ when directly extended to the complex plane.

The Solution: The key insight is to work with the complex exponential $e^{ikz}$ (or $e^{-ikz}$) directly, and then take the real or imaginary part of the final result.

• For integrals involving $\cos(kx)$, we evaluate $\text{Re} \left( \int_{-\infty}^{\infty} \frac{P(x)e^{ikx}}{Q(x)} dx \right)$.
• For integrals involving $\sin(kx)$, we evaluate $\text{Im} \left( \int_{-\infty}^{\infty} \frac{P(x)e^{ikx}}{Q(x)} dx \right)$.

The complex function we consider is now $F(z) = \frac{P(z)e^{ikz}}{Q(z)}$. The term $e^{ikz}$ (for $k>0$) decays in the upper half-plane, and this decay is precisely what Jordan’s Lemma exploits.

Jordan’s Lemma: Statement, Proof, and Applications

Jordan’s Lemma provides the rigorous justification for the vanishing of arc integrals when an exponential factor is present. It is crucial for integrals of the form $\int_{-\infty}^{\infty} f(x) e^{iax} dx$.

Jordan’s Lemma (Formal Statement): Let $\Gamma_R$ be a semi-circular contour of radius $R$ in the upper half-plane, defined by $z = R e^{i\theta}$ for $0 \leq \theta \leq \pi$. If $f(z)$ is a function such that $\lim_{R \to \infty} f(z) = 0$ uniformly for $z \in \Gamma_R$, then for any constant $a > 0$:

Analogous Statement for Lower Half-Plane: Let $\Gamma_R’$ be a semi-circular contour of radius $R$ in the lower half-plane, defined by $z = R e^{i\theta}$ for $\pi \leq \theta \leq 2\pi$ (or $-\pi \leq \theta \leq 0$). If $f(z)$ is a function such that $\lim_{R \to \infty} f(z) = 0$ uniformly for $z \in \Gamma_R’$, then for any constant $a < 0$:

Key Conditions for Applying Jordan’s Lemma:

1. Form of Integrand: The integrand must be of the form $e^{iaz} f(z)$.
2. Decay of $f(z)$: The function $f(z)$ (the non-exponential part) must tend to zero uniformly as $\lvert z\rvert \to \infty$ on the chosen arc. For rational functions $f(z) = P(z)/Q(z)$, this condition is satisfied if $\deg Q \geq \deg P + 1$. This is a weaker condition than the $\deg Q \geq \deg P + 2$ required in Chapter 1 for purely rational functions.
3. Sign of $a$ and Half-Plane Choice: This is paramount.
   • If $a > 0$, you must use the upper half-plane contour ($\text{Im}(z) > 0$). This ensures $e^{-a \text{Im}(z)}$ decays.
   • If $a < 0$, you must use the lower half-plane contour ($\text{Im}(z) < 0$). This ensures $e^{-a \text{Im}(z)}$ decays (since $a$ is negative, $-a$ is positive, and $\text{Im}(z)$ is negative, their product is negative).

Proof Sketch of Jordan’s Lemma (for $a>0$, upper half-plane):

The proof relies on carefully bounding the integral using the ML Inequality and a clever bound for $\sin\theta$. Let $M_R = \max_{z \in \Gamma_R} \lvert f(z)\rvert $. Since $\lim_{R \to \infty} f(z) = 0$ uniformly, we know $M_R \to 0$ as $R \to \infty$. On $\Gamma_R$, $z = R e^{i\theta}$, so $dz = iR e^{i\theta} d\theta$. Also, $\text{Im}(z) = R\sin\theta$. The integral is:

Due to the symmetry of $\sin\theta$ about $\theta = \pi/2$, we can write $\int_0^\pi e^{-aR\sin\theta} d\theta = 2 \int_0^{\pi/2} e^{-aR\sin\theta} d\theta$.

For $0 \leq \theta \leq \pi/2$, the inequality $\sin\theta \geq \frac{2\theta}{\pi}$ holds. This is a crucial geometric inequality; it states that the sine curve lies above the straight line connecting $(0,0)$ and $(\pi/2,1)$. Using this, we get $-aR\sin\theta \leq -aR \frac{2\theta}{\pi}$. Therefore, $e^{-aR\sin\theta} \leq e^{-aR \frac{2\theta}{\pi}}$.

Now, we bound the integral:

This integral can be directly evaluated:

So, $\int_0^\pi e^{-aR\sin\theta} d\theta \leq 2 \cdot \frac{\pi}{2aR} (1 - e^{-aR}) = \frac{\pi}{aR} (1 - e^{-aR})$.

Substituting this back into our original bound for the arc integral:

As $R \to \infty$, $M_R \to 0$ (by assumption) and $e^{-aR} \to 0$ (since $a>0$). Therefore, $\lim_{R \to \infty} \left\lvert \int_{\Gamma_R} e^{iaz} f(z) \, dz \right\rvert = 0$.

This proof demonstrates how the exponential decay of $e^{iaz}$ in the appropriate half-plane, combined with the decay of $f(z)$, forces the entire arc integral to vanish, even when $f(z)$ only decays as $1/z$.

Estimating Arc Contributions: Strategy and ML Inequality

When dealing with oscillatory integrals, the strategy for estimating the arc contribution is modified as follows:

1. Convert to Complex Exponential: Always transform the trigonometric function into its complex exponential form:

   • For $\cos(kx)$, use $\text{Re}(e^{ikx})$.
   • For $\sin(kx)$, use $\text{Im}(e^{ikx})$. This means your complex integrand will be $F(z) = e^{ikz} \frac{P(z)}{Q(z)}$ (or $e^{ikz} f(z)$ where $f(z)$ is the rational part).

2. Verify Jordan’s Lemma Condition on $f(z)$: For the rational part $f(z) = P(z)/Q(z)$, ensure that $\lim_{z \to \infty} f(z) = 0$. This is satisfied if $\deg Q \geq \deg P + 1$. If this condition holds, Jordan’s Lemma applies. If $\deg Q < \deg P + 1$, Jordan’s Lemma does not apply, and the integral might diverge.

3. Choose Contour Half-Plane based on $k$:

   • If $k > 0$, use the standard semi-circular contour in the upper half-plane. This ensures $e^{ikz}$ decays (as $e^{-ky}$ where $y>0$).

   • If $k < 0$, use a semi-circular contour in the lower half-plane. This ensures $e^{ikz}$ decays (as $e^{-ky}$ where $y<0$ and $k$ is negative, so $-ky$ is negative). Remember that for a lower half-plane contour traversed clockwise, the integral is $-2\pi i \sum \text{Res}$. If traversed counter-clockwise, the real axis part would go from $R$ to $-R$. A common alternative for $k<0$ is to use $e^{-i\lvert k\rvert z}$ and use the upper half-plane, then take the complex conjugate of the result. However, the most direct way is to use the lower half-plane for $k<0$.

   • Important Note: If $k=0$, the exponential term is $e^0 = 1$. In this case, Jordan’s Lemma is not applicable, and you revert to the Chapter 1 degree condition ($\deg Q \geq \deg P + 2$) for the arc integral to vanish.

Worked Examples

Example 2.1: Cosine Integral with Simple Poles

Evaluate $\int_{-\infty}^{\infty} \frac{\cos(kx)}{x^2 + b^2} dx$ where $k > 0$ and $b > 0$.

1. Complex Function and Contour: We consider the complex integral $\oint_{C_R} \frac{e^{ikz}}{z^2 + b^2} dz$. The desired real integral will be the real part of this result. Let $F(z) = \frac{e^{ikz}}{z^2 + b^2}$. The rational part is $f(z) = \frac{1}{z^2 + b^2}$. Here, $\deg P = 0$ (for $P(z)=1$) and $\deg Q = 2$ (for $Q(z)=z^2+b^2$). Since $\deg Q = 2 \geq \deg P + 1 = 0 + 1 = 1$, the condition for Jordan’s Lemma is met ($\lim_{z \to \infty} f(z) = 0$). As $k > 0$, we use the standard semi-circular contour $C_R$ in the upper half-plane.

2. Vanishing of Arc Integral: Because $k > 0$ and $\lim_{z \to \infty} \frac{1}{z^2 + b^2} = 0$, by Jordan’s Lemma, we have:

   \[\lim_{R \to \infty} \int_{\Gamma_R} \frac{e^{ikz}}{z^2 + b^2} dz = 0\]

3. Identify Poles: The poles of $F(z)$ are the roots of $z^2 + b^2 = 0$, which are $z = \pm ib$. The only pole in the upper half-plane is $z_0 = ib$. This is a simple pole.

4. Calculate Residue: Using the formula $\text{Res}(F, z_0) = \frac{\text{Numerator}(z_0)}{\text{Denominator}’(z_0)}$ for simple poles: Numerator is $e^{ikz}$, Denominator is $z^2 + b^2$. Denominator’s derivative is $2z$.

   \[\text{Res}\left( \frac{e^{ikz}}{z^2 + b^2}, ib \right) = \frac{e^{ik(ib)}}{2(ib)} = \frac{e^{-kb}}{2ib}\]

5. Apply Residue Theorem:

   \[\oint_{C_R} \frac{e^{ikz}}{z^2 + b^2} dz = 2\pi i \cdot \text{Res}(F, ib) = 2\pi i \cdot \frac{e^{-kb}}{2ib} = \frac{\pi e^{-kb}}{b}\]

6. Decompose and Take Limit:

   \[\lim_{R \to \infty} \left( \int_{-R}^R \frac{e^{ikx}}{x^2 + b^2} dx + \int_{\Gamma_R} \frac{e^{ikz}}{z^2 + b^2} dz \right) = \frac{\pi e^{-kb}}{b}\]

   Since the arc integral vanishes:

   \[\int_{-\infty}^{\infty} \frac{e^{ikx}}{x^2 + b^2} dx = \frac{\pi e^{-kb}}{b}\]

7. Extract Real Part: The original integral was $\int_{-\infty}^{\infty} \frac{\cos(kx)}{x^2 + b^2} dx$. This is the real part of the result:

   \[\int_{-\infty}^{\infty} \frac{\cos(kx)}{x^2 + b^2} dx = \text{Re} \left( \frac{\pi e^{-kb}}{b} \right) = \frac{\pi e^{-kb}}{b}\]

   For a general $k \neq 0$, the result is $\frac{\pi e^{-\lvert k\rvert b}}{b}$. If $k < 0$, we would use the lower half-plane contour and the pole at $z = -ib$, which would yield the same result.

Example 2.2: Sine Integral with Simple Poles (and Symmetry)

Evaluate $\int_{0}^{\infty} \frac{x \sin(ax)}{x^2 + b^2} dx$ where $a > 0$ and $b > 0$.

1. Transform and Extend Range: The integral is from $0$ to $\infty$. We first note that the integrand $g(x) = \frac{x \sin(ax)}{x^2 + b^2}$ is an even function.

   • $x$ is odd.
   • $\sin(ax)$ is odd.
   • $x^2 + b^2$ is even.

   • (odd $\times$ odd) / even = even / even = even. Therefore, we can extend the integral to the full real line:

     \[\int_{0}^{\infty} \frac{x \sin(ax)}{x^2 + b^2} dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x \sin(ax)}{x^2 + b^2} dx\]

     Now, we consider the complex integral $\oint_{C_R} \frac{z e^{iaz}}{z^2 + b^2} dz$. The desired real integral will be $\frac{1}{2}$ times the imaginary part of this result. Let $F(z) = \frac{z e^{iaz}}{z^2 + b^2}$. The rational part is $f(z) = \frac{z}{z^2 + b^2}$. Here, $\deg P = 1$ (for $P(z)=z$) and $\deg Q = 2$ (for $Q(z)=z^2+b^2$). Since $\deg Q = 2 \geq \deg P + 1 = 1 + 1 = 2$, the condition for Jordan’s Lemma is met ($\lim_{z \to \infty} f(z) = 0$). As $a > 0$, we use the standard semi-circular contour $C_R$ in the upper half-plane.

2. Vanishing of Arc Integral: Because $a > 0$ and $\lim_{z \to \infty} \frac{z}{z^2 + b^2} = 0$, by Jordan’s Lemma:

   \[\lim_{R \to \infty} \int_{\Gamma_R} \frac{z e^{iaz}}{z^2 + b^2} dz = 0\]

3. Identify Poles: Poles are at $z = \pm ib$. Only $z_0 = ib$ is in the upper half-plane. It’s a simple pole.

4. Calculate Residue: Let the numerator be $G(z) = z e^{iaz}$ and the denominator be $H(z) = z^2 + b^2$. Using $\text{Res}(F, z_0) = \frac{G(z_0)}{H’(z_0)}$: $G(ib) = ib e^{ia(ib)} = ib e^{-ab}$. $H’(z) = 2z \implies H’(ib) = 2ib$.

   \[\text{Res}\left( \frac{z e^{iaz}}{z^2 + b^2}, ib \right) = \frac{ib e^{-ab}}{2ib} = \frac{e^{-ab}}{2}\]

5. Apply Residue Theorem:

   \[\oint_{C_R} \frac{z e^{iaz}}{z^2 + b^2} dz = 2\pi i \cdot \text{Res}(F, ib) = 2\pi i \left( \frac{e^{-ab}}{2} \right) = \pi i e^{-ab}\]

6. Decompose and Take Limit:

   \[\int_{-\infty}^{\infty} \frac{x e^{iax}}{x^2 + b^2} dx = \pi i e^{-ab}\]

7. Extract Imaginary Part and Final Result:

   \[\int_{-\infty}^{\infty} \frac{x (\cos(ax) + i\sin(ax))}{x^2 + b^2} dx = \pi i e^{-ab}\] \[\int_{-\infty}^{\infty} \frac{x \cos(ax)}{x^2 + b^2} dx + i \int_{-\infty}^{\infty} \frac{x \sin(ax)}{x^2 + b^2} dx = 0 + i (\pi e^{-ab})\]

   Comparing imaginary parts:

   \[\int_{-\infty}^{\infty} \frac{x \sin(ax)}{x^2 + b^2} dx = \pi e^{-ab}\]

   Finally, for the original integral from $0$ to $\infty$:

   \[\int_{0}^{\infty} \frac{x \sin(ax)}{x^2 + b^2} dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x \sin(ax)}{x^2 + b^2} dx = \frac{\pi e^{-ab}}{2}\]

Pitfalls: Non-decaying Integrands and Invalid Applications

While Jordan’s Lemma is an indispensable tool, its misapplication is a common source of errors. Always be mindful of the following:

1. Incorrect Half-Plane Choice: This is the most critical and frequent mistake. The direction of exponential decay depends on the sign of $a$ in $e^{iaz}$ and the imaginary part of $z$.

   • For $e^{iaz}$ with $a > 0$, you must use the upper half-plane (where $\text{Im}(z) > 0$).
   • For $e^{iaz}$ with $a < 0$, you must use the lower half-plane (where $\text{Im}(z) < 0$). If you choose the wrong half-plane, the exponential term will grow, and Jordan’s Lemma will not apply, leading to an incorrect result.

2. Failure of the $f(z) \to 0$ Condition: Jordan’s Lemma requires that the rational part of the integrand, $f(z)$, tends to zero uniformly as $\lvert z\rvert \to \infty$ on the arc. For $f(z) = P(z)/Q(z)$, this means $\deg Q \geq \deg P + 1$. If this condition is not met (e.g., $\deg Q \leq \deg P$), then Jordan’s Lemma cannot be used, and the integral might diverge. For instance, $\int_{-\infty}^{\infty} x \sin(x) dx$ diverges because $f(z)=z$ does not tend to zero as $z \to \infty$.

3. Direct Use of $\cos(kz)$ or $\sin(kz)$: As explained in the motivation, directly extending $\cos(kz)$ or $\sin(kz)$ to the complex plane and using them in the integrand is generally problematic. Always convert to $e^{ikz}$ and take the real or imaginary part after evaluating the contour integral.

4. Integrals from $0$ to $\infty$ (Symmetry):

   • When an integral is given from $0$ to $\infty$, check if the integrand is an even or odd function.
   • If $g(x)$ is even, $\int_0^\infty g(x) dx = \frac{1}{2} \int_{-\infty}^\infty g(x) dx$.
   • If $g(x)$ is odd, $\int_0^\infty g(x) dx$ might be zero (if the principal value exists) or undefined. The integral $\int_{-\infty}^\infty g(x) dx$ for an odd function over a symmetric interval is always zero if it converges.
   • If the integrand is neither even nor odd, you cannot simply extend it to $(-\infty, \infty)$ by multiplying by $1/2$. You might need a different contour or principal value interpretation (Chapter 4).

5. Poles on the Real Axis: Jordan’s Lemma, like the methods in Chapter 1, assumes that there are no singularities on the real axis. If poles lie directly on the real axis, the semi-circular contour must be modified using an “indented contour” technique, which will be the focus of Chapter 4.

By diligently following these guidelines and understanding the nuances of Jordan’s Lemma, you will be well-equipped to evaluate a wide array of oscillatory integrals, a skill invaluable in many areas of applied mathematics, physics, and engineering.

Chapter 3: Trigonometric Integrals over $[0, 2\pi]$ via Substitution

Chapters 1 and 2 focused on improper integrals over the real line, where the primary challenge was to ensure the vanishing of integrals over large semi-circular arcs. This chapter introduces a fundamentally different class of integrals: definite integrals of trigonometric functions over a finite interval, typically $[0, 2\pi]$. Here, the power of the Residue Theorem is harnessed through a clever substitution that transforms the trigonometric integral into a contour integral around the unit circle in the complex plane. This method is elegant and often the most straightforward way to evaluate such integrals.

The Unit Circle Parametrization: $z = e^{i\theta}$

The core idea of this technique is to map the integration interval $[0, 2\pi]$ onto the unit circle in the complex plane. This is achieved through the substitution:

As $\theta$ varies from $0$ to $2\pi$, $z$ traverses the unit circle $\lvert z\rvert =1$ exactly once in the counter-clockwise (positive) direction. This unit circle will be our contour $C$.

From this substitution, we can derive the necessary transformations for $d\theta$, $\cos\theta$, and $\sin\theta$:

1. For $d\theta$: Differentiating $z = e^{i\theta}$ with respect to $\theta$:

   \[dz = i e^{i\theta} d\theta = i z \, d\theta\]

   Therefore,

   \[d\theta = \frac{dz}{iz}\]

2. For $\cos\theta$: We know $z = e^{i\theta}$ and $z^{-1} = e^{-i\theta}$. Using Euler’s formula, $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$:

   \[\cos\theta = \frac{z + z^{-1}}{2} = \frac{z + 1/z}{2} = \frac{z^2 + 1}{2z}\]

3. For $\sin\theta$: Using Euler’s formula, $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$:

   \[\sin\theta = \frac{z - z^{-1}}{2i} = \frac{z - 1/z}{2i} = \frac{z^2 - 1}{2iz}\]

These three transformations are fundamental to the method. They allow us to convert any rational function of $\cos\theta$ and $\sin\theta$ into a rational function of $z$.

When and How to Transform: Checklist and Heuristics

This method is specifically designed for integrals of the form:

where $R(\cos\theta, \sin\theta)$ is a rational function of $\cos\theta$ and $\sin\theta$ (meaning it can be expressed as a ratio of polynomials in $\cos\theta$ and $\sin\theta$).

Checklist for Applying the Method:

1. Identify Integral Type: Ensure the integral is a definite integral of a rational trigonometric function over an interval of length $2\pi$. The most common interval is $[0, 2\pi]$, but $[-\pi, \pi]$ also works (as $z$ still traverses the unit circle once).
2. Perform the Substitution:
   • Replace $d\theta$ with $\frac{dz}{iz}$.
   • Replace $\cos\theta$ with $\frac{z^2 + 1}{2z}$.
   • Replace $\sin\theta$ with $\frac{z^2 - 1}{2iz}$.
3. Simplify the Integrand: After substitution, the integrand will be a complex rational function of $z$. Simplify it algebraically to the form $f(z) = \frac{P(z)}{Q(z)}$.
4. Identify Poles: Find all singularities (poles) of $f(z)$ by setting the denominator $Q(z)$ to zero.
5. Select Enclosed Poles: The contour is always the unit circle $\lvert z\rvert =1$. Identify all poles that lie inside this contour (i.e., $\lvert z_k\rvert < 1$).
6. Calculate Residues: Compute the residue of $f(z)$ at each enclosed pole using the appropriate residue formula (for simple or higher-order poles).
7. Apply Residue Theorem: The value of the integral is $2\pi i$ times the sum of the residues of $f(z)$ at the poles inside the unit circle.

Heuristics:

• This method is highly effective because the contour (unit circle) is fixed and closed, eliminating the need to worry about arc integrals vanishing.
• If the limits are not $0$ to $2\pi$ (e.g., $0$ to $\pi$), this method is generally not directly applicable without further symmetry arguments or alternative contours. For example, if the integrand is even in $\theta$, $\int_0^\pi R(\cos\theta, \sin\theta) d\theta$ might be half of $\int_0^{2\pi} R(\cos\theta, \sin\theta) d\theta$, but this requires careful verification.
• The method works best when the denominator does not have roots on the unit circle itself. If it does, the integral needs to be interpreted as a principal value, similar to integrals with poles on the real axis (Chapter 4).

Handling $\cos\theta$, $\sin\theta$, and Rational Trig Expressions

Let’s illustrate the transformation process with a generic example. Suppose we have an integrand like $\frac{1}{A + B\cos\theta + C\sin\theta}$. Substitute:

To simplify, find a common denominator within the large fraction:

Now multiply by $d\theta = \frac{dz}{iz}$:

This results in a rational function of $z$. The denominator will be a polynomial in $z$, and we will find its roots to locate the poles.

Worked Examples

Example 3.1: Integral with Simple Poles

Evaluate $\int_0^{2\pi} \frac{1}{a + \cos\theta} d\theta$, where $a > 1$. The condition $a>1$ ensures that the denominator $a+\cos\theta$ is never zero, so there are no singularities on the unit circle.

1. Perform Substitution: Let $z = e^{i\theta}$. Then $d\theta = \frac{dz}{iz}$ and $\cos\theta = \frac{z^2 + 1}{2z}$. The integral becomes:

   \[\oint_{\lvert z\rvert =1} \frac{1}{a + \frac{z^2 + 1}{2z}} \frac{dz}{iz}\]

   Simplify the integrand:

   \[\frac{1}{\frac{2az + z^2 + 1}{2z}} \frac{1}{iz} = \frac{2z}{z^2 + 2az + 1} \frac{1}{iz} = \frac{2}{i(z^2 + 2az + 1)}\]

   So, $f(z) = \frac{2}{i(z^2 + 2az + 1)}$.

2. Identify Poles: Set the denominator to zero: $z^2 + 2az + 1 = 0$. Using the quadratic formula:

   \[z = \frac{-2a \pm \sqrt{(2a)^2 - 4(1)(1)}}{2} = \frac{-2a \pm \sqrt{4a^2 - 4}}{2} = -a \pm \sqrt{a^2 - 1}\]

   Let $z_1 = -a + \sqrt{a^2 - 1}$ and $z_2 = -a - \sqrt{a^2 - 1}$.

3. Select Enclosed Poles: We need to determine which poles lie inside the unit circle ($\lvert z\rvert <1$). Since $a > 1$, $\sqrt{a^2 - 1}$ is real and positive.

   • For $z_2 = -a - \sqrt{a^2 - 1}$: Since $a>1$, $z_2$ is clearly negative and $\lvert z_2\rvert = a + \sqrt{a^2 - 1} > 1 + 0 = 1$. So $z_2$ is outside the unit circle.
   • For $z_1 = -a + \sqrt{a^2 - 1}$: We know $a^2 - 1 < a^2$, so $\sqrt{a^2 - 1} < a$. This means $z_1$ is negative but closer to zero. To check if $\lvert z_1\rvert < 1$, we can check if $z_1 z_2 = 1$ (from the product of roots $c/a = 1/1 = 1$). Since $\lvert z_2\rvert > 1$, and $\lvert z_1 z_2\rvert = \lvert z_1\rvert \lvert z_2\rvert = 1$, it must be that $\lvert z_1\rvert = 1/\lvert z_2\rvert < 1$. So, $z_1 = -a + \sqrt{a^2 - 1}$ is the only pole inside the unit circle. It is a simple pole.

4. Calculate Residue: We use $\text{Res}(f, z_1) = \frac{\text{Numerator}(z_1)}{\text{Denominator}’(z_1)}$. Numerator is $2$. Denominator is $i(z^2 + 2az + 1)$. Its derivative with respect to $z$ is $i(2z + 2a)$.

   \[\text{Res}(f, z_1) = \frac{2}{i(2z_1 + 2a)} = \frac{2}{2i(z_1 + a)} = \frac{1}{i(z_1 + a)}\]

   Substitute $z_1 = -a + \sqrt{a^2 - 1}$:

   \[\text{Res}(f, z_1) = \frac{1}{i(-a + \sqrt{a^2 - 1} + a)} = \frac{1}{i\sqrt{a^2 - 1}} = \frac{-i}{\sqrt{a^2 - 1}}\]

5. Apply Residue Theorem:

   \[\int_0^{2\pi} \frac{1}{a + \cos\theta} d\theta = 2\pi i \sum \text{Res} = 2\pi i \left( \frac{-i}{\sqrt{a^2 - 1}} \right)\] \[= \frac{-2\pi i^2}{\sqrt{a^2 - 1}} = \frac{-2\pi (-1)}{\sqrt{a^2 - 1}} = \frac{2\pi}{\sqrt{a^2 - 1}}\]

Example 3.2: Integral with a Higher-Order Pole (or multiple simple poles)

Evaluate $\int_0^{2\pi} \frac{1}{a^2 - 2a\cos\theta + 1} d\theta$, where $a > 1$.

1. Perform Substitution: Let $z = e^{i\theta}$. Then $d\theta = \frac{dz}{iz}$ and $\cos\theta = \frac{z^2 + 1}{2z}$. The integral becomes:

   \[\oint_{\lvert z\rvert =1} \frac{1}{a^2 - 2a\left(\frac{z^2 + 1}{2z}\right) + 1} \frac{dz}{iz}\]

   Simplify the integrand:

   \[\frac{1}{a^2 - a\frac{z^2 + 1}{z} + 1} \frac{1}{iz} = \frac{1}{\frac{a^2z - a(z^2 + 1) + z}{z}} \frac{1}{iz}\] \[= \frac{z}{a^2z - az^2 - a + z} \frac{1}{iz} = \frac{1}{i(-az^2 + (a^2+1)z - a)}\]

   So, $f(z) = \frac{1}{i(-az^2 + (a^2+1)z - a)}$.

2. Identify Poles: Set the denominator to zero: $-az^2 + (a^2+1)z - a = 0$. Multiply by $-1$: $az^2 - (a^2+1)z + a = 0$. This is a quadratic equation. We can factor it. Notice that if $z=a$, $a(a^2) - (a^2+1)a + a = a^3 - a^3 - a + a = 0$. So $z=a$ is a root. By Vieta’s formulas, the product of roots is $a/a = 1$. So the other root must be $1/a$. Thus, the poles are $z_1 = a$ and $z_2 = 1/a$. Both are simple poles.

3. Select Enclosed Poles: Since $a > 1$:

   • $z_1 = a$: This pole is outside the unit circle ($\lvert a\rvert > 1$).
   • $z_2 = 1/a$: This pole is inside the unit circle ($\lvert 1/a\rvert < 1$). So, $z_2 = 1/a$ is the only pole inside the unit circle.

4. Calculate Residue: We use $\text{Res}(f, z_2) = \frac{\text{Numerator}(z_2)}{\text{Denominator}’(z_2)}$. Numerator is $1$. Denominator is $i(-az^2 + (a^2+1)z - a)$. Its derivative with respect to $z$ is $i(-2az + a^2+1)$.

   \[\text{Res}(f, 1/a) = \frac{1}{i(-2a(1/a) + a^2+1)} = \frac{1}{i(-2 + a^2+1)} = \frac{1}{i(a^2 - 1)}\] \[= \frac{-i}{a^2 - 1}\]

5. Apply Residue Theorem:

   \[\int_0^{2\pi} \frac{1}{a^2 - 2a\cos\theta + 1} d\theta = 2\pi i \sum \text{Res} = 2\pi i \left( \frac{-i}{a^2 - 1} \right)\] \[= \frac{-2\pi i^2}{a^2 - 1} = \frac{-2\pi (-1)}{a^2 - 1} = \frac{2\pi}{a^2 - 1}\]

This method provides a remarkably efficient way to evaluate a broad class of trigonometric integrals, transforming them into routine residue calculations. The key is to master the substitution and the subsequent algebraic simplification to arrive at a rational function of $z$ whose poles can be readily identified and their residues computed.

Chapter 4: Integrals with Singularities on the Real Line

In the previous chapters, we successfully evaluated integrals over the real line by assuming that the integrand’s singularities (poles) were always off the real axis. This assumption allowed us to choose a semi-circular contour that neatly enclosed the relevant poles. However, many important integrals in physics and engineering, such as those arising in quantum mechanics or signal processing, possess singularities directly on the real axis. These integrals often do not converge in the usual sense, requiring the concept of the Cauchy Principal Value and a modification of our contour integration technique using indented contours.

Principal Value Integrals and the Cauchy Principal Value

Consider an integral $\int_{-\infty}^{\infty} f(x) \, dx$ where $f(x)$ has a singularity at a point $x_0$ on the real axis. A standard improper integral would require the limit

If the limits are taken independently, the integral might diverge.

The Cauchy Principal Value (P.V.) provides a way to assign a value to such integrals by taking a symmetric limit around the singularity.

Definition: For an integral with a single singularity at $x_0$ on the real axis:

If there are multiple singularities on the real axis at $x_1, x_2, \ldots, x_n$:

where the limit $R \to \infty$ is also taken simultaneously. The key is that the “exclusion regions” around each singularity shrink symmetrically.

Example: $\int_{-1}^1 \frac{1}{x} dx$. This integral diverges in the Riemann sense. However, its principal value is:

So, $\text{p.v.} \int_{-1}^1 \frac{1}{x} dx = 0$. This highlights that a principal value exists even when the integral doesn’t converge absolutely.

The Indented Contour Technique: Justification and Lemma

To evaluate principal value integrals using the Residue Theorem, we modify our standard semi-circular contour by indenting it around the real-axis singularities.

Construction of the Indented Contour: Consider an integral $\text{p.v.} \int_{-\infty}^{\infty} f(x) \, dx$ with a simple pole at $x_0$ on the real axis. We construct a contour $C_R$ consisting of:

1. A large semi-circular arc $\Gamma_R$ of radius $R$ (e.g., in the upper half-plane).
2. Segments along the real axis from $-R$ to $x_0-\epsilon$ and from $x_0+\epsilon$ to $R$.
3. A small semi-circular arc $\gamma_\epsilon$ of radius $\epsilon$ around $x_0$, which “indents” (bypasses) the pole. This indentation is typically in the half-plane opposite to the main contour (e.g., if using upper half-plane for $\Gamma_R$, indent downwards; if using upper half-plane, indent upwards to exclude the pole from the region).

The choice of indentation direction is crucial. If we want to include the pole in our contour calculation, we would indent into the main contour’s region (e.g., an upper semi-circle if using upper half-plane for $\Gamma_R$). However, for principal value integrals, we typically indent out of the region, ensuring the pole is not inside the closed contour formed by $L_R \cup \Gamma_R \cup \gamma_\epsilon$ (if we consider the region between the real axis and $\Gamma_R$). More often, we indent around the pole such that it’s outside the contour formed by $L_R \cup \Gamma_R$.

Let’s assume we choose the standard upper semi-circular contour for $\Gamma_R$. We then indent below the real axis, meaning the small semi-circle $\gamma_\epsilon$ is in the lower half-plane and traversed clockwise around $x_0$. This makes the pole $x_0$ lie outside the closed contour $C_R = [-R, x_0-\epsilon] \cup \gamma_\epsilon \cup [x_0+\epsilon, R] \cup \Gamma_R$.

The integral over the closed contour $C_R$ is given by the Residue Theorem for poles not on the real axis.

As $R \to \infty$ and $\epsilon \to 0^+$, the sum of the two real line integrals becomes the principal value integral. The integral over $\Gamma_R$ typically vanishes (by conditions from Chapters 1 or 2). The critical part is the integral over the indentation $\gamma_\epsilon$.

Lemma for Indented Contours (Contribution of a Simple Pole)

Let $f(z)$ have a simple pole at $z_0$ on the real axis, with residue $\text{Res}(f, z_0)$. Let $\gamma_\epsilon$ be a semi-circular arc of radius $\epsilon$ centered at $z_0$.

• If $\gamma_\epsilon$ is in the upper half-plane and traversed counter-clockwise:

  \[\lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} f(z) \, dz = i\pi \, \text{Res}(f, z_0)\]

• If $\gamma_\epsilon$ is in the lower half-plane and traversed clockwise:

  \[\lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} f(z) \, dz = -i\pi \, \text{Res}(f, z_0)\]

Justification Sketch: Since $f(z)$ has a simple pole at $z_0$, its Laurent series expansion around $z_0$ is $f(z) = \frac{\text{Res}(f, z_0)}{z - z_0} + g(z)$, where $g(z)$ is analytic at $z_0$. The integral over $\gamma_\epsilon$ is $\int_{\gamma_\epsilon} \left( \frac{\text{Res}(f, z_0)}{z - z_0} + g(z) \right) dz$. As $\epsilon \to 0^+$, the integral of $g(z)$ over $\gamma_\epsilon$ goes to zero because $g(z)$ is bounded near $z_0$ and the length of $\gamma_\epsilon$ goes to zero ($L = \pi\epsilon$). So we need to evaluate $\lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} \frac{\text{Res}(f, z_0)}{z - z_0} dz$. Let $z - z_0 = \epsilon e^{i\phi}$. Then $dz = i\epsilon e^{i\phi} d\phi$. If $\gamma_\epsilon$ is in the upper half-plane, $\phi$ goes from $\pi$ to $0$ (for clockwise) or $0$ to $\pi$ (for counter-clockwise). For counter-clockwise traversal ($0$ to $\pi$):

This confirms the lemma. The sign depends solely on the direction of traversal of the small arc.

The Sokhotski–Plemelj Theorem (with Proof Sketch)

The Sokhotski–Plemelj Theorem formally relates principal value integrals to contour integrals involving poles on the real axis. It is particularly useful in many areas of physics.

Statement: Let $f(z)$ be a function that is analytic in a region including the real axis, except for a simple pole at $x_0$ on the real axis. Then:

where $f(x_0)$ is interpreted as $\text{Res}(g(z), x_0)$ if we write $f(z) = g(z)/(z-x_0)$. More directly, if $f(z)$ is defined such that $f(x_0) = \text{Res}(h(z), x_0)$, where $h(z) = f(z)/(z-x_0)$ then $f(x_0)$ refers to the numerator term evaluated at $x_0$.

Simpler Interpretation for Our Context: If $F(z)$ has a simple pole at $x_0$ on the real axis, then:

(This formula applies when the contour closes in the upper half-plane, and the real-axis poles are indented into the contour, effectively including half of their residue. If indented out, the sign changes.) A more common convention for principal value calculations is:

where $z_k$ are poles in the UHP. If there are simple poles $x_j$ on the real axis, these are handled by the indentation. The contribution of each indentation (around an $x_j$) that excludes $x_j$ from the contour is $ \mp i\pi \text{Res}(F, x_j)$.

Proof Sketch: Consider the integral $\oint_C F(z) dz$, where $C$ is a large semi-circular contour in the upper half-plane that indents below a real-axis pole $x_0$. The contour effectively avoids $x_0$.

As $R \to \infty$ and $\epsilon \to 0^+$:

Since $\gamma_\epsilon$ is a small semi-circle in the lower half-plane traversed clockwise, its contribution is $-i\pi \text{Res}(F, x_0)$. So,

This is the standard formula for principal value integrals when using an upper half-plane contour that indents below real poles.

Worked Examples

Example 4.1: Simple Pole at the Origin

Evaluate $\text{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x} dx$.

1. Complex Function and Contour: Let $F(z) = \frac{e^{iz}}{z}$. The pole is at $z=0$, which is on the real axis. It is a simple pole. The rational part is $f(z) = 1/z$. Here $\deg Q = 1$, $\deg P = 0$, so $\deg Q \geq \deg P + 1$ is satisfied. The exponential term is $e^{iz}$, with $a=1 > 0$. So we use Jordan’s Lemma with an upper half-plane contour. We construct a contour $C_R$ consisting of:

   • $\Gamma_R$: large semi-circle in UHP from $R$ to $-R$.
   • $\gamma_\epsilon$: small semi-circle in UHP around $z=0$ from $-\epsilon$ to $\epsilon$, traversed clockwise. This excludes the pole from the interior of $C_R$.
   • Real axis segments: $[-R, -\epsilon]$ and $[\epsilon, R]$. The closed contour $C$ encloses no poles. Thus, $\oint_C F(z) dz = 0$.
   \[\oint_C F(z) dz = \int_{-R}^{-\epsilon} \frac{e^{ix}}{x} dx + \int_{\epsilon}^{R} \frac{e^{ix}}{x} dx + \int_{\Gamma_R} \frac{e^{iz}}{z} dz + \int_{\gamma_\epsilon} \frac{e^{iz}}{z} dz = 0\]

2. Vanishing of Arc Integral ($\Gamma_R$): By Jordan’s Lemma (since $a=1>0$ and $f(z)=1/z \to 0$ as $R \to \infty$), $\lim_{R \to \infty} \int_{\Gamma_R} \frac{e^{iz}}{z} dz = 0$.

3. Contribution of Indented Arc ($\gamma_\epsilon$): The pole is $z_0=0$. It is a simple pole. $\text{Res}(F, 0) = \lim_{z \to 0} z \frac{e^{iz}}{z} = e^{i(0)} = 1$. Since $\gamma_\epsilon$ is in the upper half-plane and traversed clockwise, its contribution is $-i\pi \text{Res}(F, 0) = -i\pi(1) = -i\pi$.

4. Take Limits:

   \[\lim_{\substack{R \to \infty \\ \epsilon \to 0^+}} \left( \int_{-R}^{-\epsilon} \frac{e^{ix}}{x} dx + \int_{\epsilon}^{R} \frac{e^{ix}}{x} dx \right) + 0 + (-i\pi) = 0\]

   The term in parentheses is $\text{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x} dx$.

   \[\text{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x} dx - i\pi = 0\] \[\text{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x} dx = i\pi\]

5. Extract Real/Imaginary Parts (for $\cos x$ and $\sin x$): Recall $e^{ix} = \cos x + i\sin x$.

   \[\text{p.v.} \int_{-\infty}^{\infty} \frac{\cos x + i\sin x}{x} dx = i\pi\] \[\text{p.v.} \int_{-\infty}^{\infty} \frac{\cos x}{x} dx + i \text{p.v.} \int_{-\infty}^{\infty} \frac{\sin x}{x} dx = 0 + i\pi\]

   By equating real and imaginary parts:

   \[\text{p.v.} \int_{-\infty}^{\infty} \frac{\cos x}{x} dx = 0\] \[\text{p.v.} \int_{-\infty}^{\infty} \frac{\sin x}{x} dx = \pi\]

   (Note: $\frac{\cos x}{x}$ is odd, so its P.V. is indeed 0. $\frac{\sin x}{x}$ is even, and this is a well-known result, the Dirichlet integral.)

Example 4.2: Multiple Simple Poles on the Real Axis

Evaluate $\text{p.v.} \int_{-\infty}^{\infty} \frac{\sin x}{x(x^2 - \pi^2)} dx$.

1. Complex Function and Contour: The integrand is an even function ($(\text{odd})/(\text{odd} \cdot \text{even}) = \text{even}$). So we will use the imaginary part of the integral $\oint \frac{e^{iz}}{z(z^2 - \pi^2)} dz$. Let $F(z) = \frac{e^{iz}}{z(z^2 - \pi^2)} = \frac{e^{iz}}{z(z-\pi)(z+\pi)}$. Poles are at $z = 0, \pi, -\pi$. All are on the real axis and are simple poles. The rational part $f(z) = \frac{1}{z(z^2-\pi^2)}$. $\deg P = 0$, $\deg Q = 3$. $\deg Q \geq \deg P + 1$ (i.e., $3 \geq 0+1=1$) for Jordan’s Lemma. The exponential term is $e^{iz}$, with $a=1>0$. We use an upper half-plane contour. We indent around $z = -\pi, 0, \pi$ with small semi-circles $\gamma_1, \gamma_2, \gamma_3$ in the lower half-plane (clockwise traversal), excluding these poles from the main contour. The large arc $\Gamma_R$ is in the UHP.

2. Vanishing of Arc Integral ($\Gamma_R$): By Jordan’s Lemma, $\lim_{R \to \infty} \int_{\Gamma_R} F(z) dz = 0$.

3. No Enclosed Poles (in the UHP): The contour formed by the real axis segments and $\Gamma_R$ and the indentations encloses no poles (all poles are on the real axis and excluded by the indentations). Thus, $\oint_C F(z) dz = 0$.

4. Contribution of Indented Arcs ($\gamma_1, \gamma_2, \gamma_3$): We need to calculate the residue at each real pole.

   • At $z=0$: $\text{Res}(F, 0) = \lim_{z \to 0} z \frac{e^{iz}}{z(z^2-\pi^2)} = \frac{e^0}{0^2-\pi^2} = -\frac{1}{\pi^2}$.
   • At $z=\pi$: $\text{Res}(F, \pi) = \lim_{z \to \pi} (z-\pi) \frac{e^{iz}}{z(z-\pi)(z+\pi)} = \frac{e^{i\pi}}{\pi(\pi+\pi)} = \frac{-1}{2\pi^2}$.
   • At $z=-\pi$: $\text{Res}(F, -\pi) = \lim_{z \to -\pi} (z+\pi) \frac{e^{iz}}{z(z-\pi)(z+\pi)} = \frac{e^{-i\pi}}{-\pi(-\pi-\pi)} = \frac{-1}{2\pi^2}$.

   Since all indentations are in the lower half-plane and traversed clockwise, each contributes $-i\pi \cdot \text{Res}(F, z_k)$. Total contribution from indentations:

   \[-i\pi \left( -\frac{1}{\pi^2} - \frac{1}{2\pi^2} - \frac{1}{2\pi^2} \right) = -i\pi \left( -\frac{1}{\pi^2} - \frac{2}{2\pi^2} \right) = -i\pi \left( -\frac{1}{\pi^2} - \frac{1}{\pi^2} \right) = -i\pi \left( -\frac{2}{\pi^2} \right) = \frac{2i}{\pi}\]

5. Take Limits:

   \[\text{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x(x^2 - \pi^2)} dx + \lim_{\Gamma_R} + \sum \lim_{\gamma_k} = 0\] \[\text{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x(x^2 - \pi^2)} dx + 0 + \frac{2i}{\pi} = 0\] \[\text{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x(x^2 - \pi^2)} dx = -\frac{2i}{\pi}\]

6. Extract Imaginary Part: The original integral was $\text{p.v.} \int_{-\infty}^{\infty} \frac{\sin x}{x(x^2 - \pi^2)} dx$. This is the imaginary part of the result.

   \[\text{p.v.} \int_{-\infty}^{\infty} \frac{\sin x}{x(x^2 - \pi^2)} dx = \text{Im}\left( -\frac{2i}{\pi} \right) = -\frac{2}{\pi}\]

Sign Errors, Ambiguous Limits, and Common Mistakes

Indented contours are notorious for sign errors. Be extremely careful with:

1. Direction of Indentation and Sign of $\pi i$ Term:

   • If the small semi-circle $\gamma_\epsilon$ is traversed counter-clockwise (usually in the upper half-plane), its contribution is $+i\pi \text{Res}(f, z_0)$.
   • If $\gamma_\epsilon$ is traversed clockwise (usually in the lower half-plane), its contribution is $-i\pi \text{Res}(f, z_0)$. This sign is crucial and depends entirely on how you define your contour and whether the pole is excluded or included by the indentation. The examples above used indentations that excluded the poles, leading to the sum of the contour integral and the indentation contributions equaling $2\pi i \sum \text{Res}_{\text{enclosed}}$.

2. Principal Value vs. Absolute Convergence: Remember that a principal value integral may exist even if the integral does not converge absolutely. Do not confuse the two. The P.V. does not imply convergence in the usual sense.

3. Higher-Order Poles on Real Axis: The lemma for indentations (contribution of $\pm i\pi \text{Res}$) only applies to simple poles. If there is a higher-order pole on the real axis, the integral over the indentation will generally not vanish or contribute $\pm i\pi \text{Res}$. Such integrals are usually much more complex and often diverge even in the principal value sense. This method is primarily for simple poles on the real axis.

4. Misidentifying $f(z)$ for Jordan’s Lemma: When using $e^{iaz} f(z)$, remember that $f(z)$ is the non-exponential part. Its degree condition ($\deg Q \geq \deg P + 1$) still needs to be checked for Jordan’s Lemma to apply for the large arc.

5. Algebraic and Complex Arithmetic Errors: The calculations can become lengthy, especially with multiple poles and complex exponentials. Double-check all algebraic manipulations and complex number arithmetic.

Mastering indented contours allows you to tackle a wide range of complex integrals, solidifying your understanding of the deep connections between complex analysis and real-world problems.

Chapter 5: Branch Points and Keyhole Contours

Our journey through contour integration has so far focused on functions with isolated singularities, primarily poles. However, many important real integrals, particularly those involving logarithms or non-integer powers, necessitate dealing with a different type of singularity: branch points. These points give rise to multivalued functions, making it impossible to define a single-valued analytic function over the entire complex plane. To handle such functions, we introduce branch cuts to render them single-valued, and then employ a specialized contour known as the keyhole contour.

Multivalued Functions: Logarithms and Power Laws

A function $f(z)$ is multivalued if, for a given $z$, it can take on more than one value. The most common examples are the logarithm and non-integer power functions.

1. The Logarithm Function ($\log z$): For a complex number $z = re^{i\theta}$, the natural logarithm is defined as:

   \[\log z = \ln r + i\theta = \ln\lvert z\rvert + i \arg(z)\]

   The issue is that $\theta$ (the argument of $z$) is periodic with period $2\pi$. So, $z = re^{i(\theta + 2n\pi)}$ for any integer $n$. This means $\log z$ can take on infinitely many values:

   \[\log z = \ln\lvert z\rvert + i(\theta + 2n\pi), \quad n \in \mathbb{Z}\]

   To make $\log z$ single-valued, we must restrict the range of $\arg(z)$ to an interval of length $2\pi$, for example, $0 \leq \arg(z) < 2\pi$ or $-\pi < \arg(z) \leq \pi$. This restriction defines a branch of the logarithm.

   The point $z=0$ is a branch point for $\log z$. If you encircle the origin, $\arg(z)$ changes by $2\pi$, leading to a different value of $\log z$.

2. Power Laws ($z^p$ for non-integer $p$): A general power function $z^p$ (where $p$ is any complex number) is defined using the logarithm:

   \[z^p = e^{p \log z} = e^{p(\ln\lvert z\rvert + i \arg(z))}\]

   If $p$ is not an integer, $z^p$ is also a multivalued function because of the $\arg(z)$ term. For example, consider $z^{1/2} = \sqrt{z}$. If $z = 1 = e^{i0}$, $\sqrt{z}=1$. But $z = e^{i2\pi}$ also, so $\sqrt{z} = e^{i2\pi/2} = e^{i\pi} = -1$. The points $z=0$ (for $p \neq 0$) and $z=\infty$ (for $p \neq 0, 1$) are branch points for $z^p$.

How to Choose and Place Branch Cuts

To make a multivalued function single-valued and analytic in a desired region, we introduce a branch cut. A branch cut is a curve or line in the complex plane that we “cut” or exclude from the domain of the function. By making such a cut, we prevent any path from encircling the branch point, thus ensuring $\arg(z)$ (and consequently $\log z$ or $z^p$) cannot change discontinuously.

Common Choices for Branch Cuts:

1. Positive Real Axis: This is the most common choice for integrals over $[0, \infty)$. We typically define the argument in the range $0 \leq \arg(z) < 2\pi$. The cut runs from $z=0$ to $z=\infty$ along the positive real axis.
2. Negative Real Axis: Another common choice, especially for functions like $\log(z-a)$, where we might want the cut to extend from $a$ to $-\infty$ along the real axis. Here, $\arg(z)$ is often restricted to $-\pi < \arg(z) \leq \pi$.

Placement Strategy:

• The branch cut must connect the branch points. For functions like $\log z$ or $z^p$, where the branch points are $0$ and $\infty$, the cut extends from the origin to infinity.
• The branch cut must be chosen such that it does not cross any poles of the integrand. If it does, the function is not analytic across the contour, and the Residue Theorem cannot be applied directly.
• The choice of branch cut is arbitrary, but once chosen, it must be used consistently throughout the calculation, especially regarding the values of $\arg(z)$ on either side of the cut.

The Keyhole Contour: Anatomy and Application

For integrals over the interval $[0, \infty)$ involving branch points at $z=0$ and $z=\infty$ (like $x^p$ or $\log x$), the standard semi-circular contour is insufficient. Instead, we use a keyhole contour (also known as a “cut-plane contour” or “dumbbell contour” if there are two finite branch points).

The keyhole contour, $C$, typically consists of four parts:

1. Outer Large Circle ($\Gamma_R$): A circle of radius $R$ centered at the origin, traversed counter-clockwise. This path encloses all the poles of the integrand.

   \[z = R e^{i\theta}, \quad 0 \leq \theta \leq 2\pi\]

2. Inner Small Circle ($\gamma_\epsilon$): A circle of radius $\epsilon$ centered at the origin, traversed clockwise. This path ensures the branch point $z=0$ is not included in the interior of the contour.

   \[z = \epsilon e^{i\theta}, \quad \text{from } 2\pi \text{ to } 0\]

   (or $0 \le \theta \le 2\pi$ and integral is reversed later)

3. Upper Line Segment ($L_1$): A path just above the branch cut (e.g., the positive real axis). Here, $z=x$ and $\arg(z) = 0$. It runs from $x=\epsilon$ to $x=R$.
4. Lower Line Segment ($L_2$): A path just below the branch cut, returning from $R$ to $\epsilon$. Here, $z=x$ and $\arg(z) = 2\pi$. It runs from $x=R$ to $x=\epsilon$.

Anatomy of the Keyhole Contour (with branch cut along positive real axis, $0 \le \arg(z) < 2\pi$):

• $C = \Gamma_R \cup L_1 \cup \gamma_\epsilon \cup L_2$
• The integral over the closed contour $C$ is $2\pi i \sum \text{Res}(f, z_k)$ for all poles $z_k$ not on the branch cut.

The key steps in applying the keyhole contour method are:

1. Define the Multivalued Function: Explicitly state the branch cut you choose (e.g., positive real axis, $0 \leq \arg(z) < 2\pi$).
2. Evaluate $\int_{\Gamma_R}$: Show that this integral vanishes as $R \to \infty$. This typically requires the integrand to decay faster than $1/\lvert z\rvert $. For $z^p f(z)$, this usually means $\deg Q \geq \deg P + (1+p)$.
3. Evaluate $\int_{\gamma_\epsilon}$: Show that this integral vanishes as $\epsilon \to 0^+$. This typically requires the integrand to decay slower than $1/\lvert z\rvert $. For $z^p f(z)$, this usually means $\deg Q \geq \deg P + (1-p)$.
4. Combine $\int_{L_1}$ and $\int_{L_2}$: This is the most crucial step. Due to the branch cut, the value of the function on $L_1$ (e.g., $z=x$, $\arg(z)=0$) will differ from its value on $L_2$ (e.g., $z=x$, $\arg(z)=2\pi$). This difference is what allows us to solve for the original real integral.
   • On $L_1$: $z = x$, $d z = dx$, $\arg(z) = 0$.
   • On $L_2$: $z = x$, $d z = dx$, $\arg(z) = 2\pi$. (Integral direction is reversed, so $dx$ from $R$ to $\epsilon$ becomes $-\int_\epsilon^R dx$ if we integrate $f(x e^{i2\pi})$)

Worked Examples

Example 5.1: Integral Involving Logarithm

Evaluate $\int_0^\infty \frac{\log x}{x^2 + 1} dx$.

1. Complex Function and Branch Cut: Consider $f(z) = \frac{\log z}{z^2 + 1}$. We choose the branch cut along the positive real axis, with $0 \leq \arg(z) < 2\pi$. The poles of $f(z)$ are where $z^2+1=0$, so $z=\pm i$. Both are within our contour.

2. Contour and Integral Sum: We use the keyhole contour $C = \Gamma_R \cup L_1 \cup \gamma_\epsilon \cup L_2$.

   \[\oint_C \frac{\log z}{z^2 + 1} dz = \int_{\Gamma_R} \frac{\log z}{z^2 + 1} dz + \int_{\epsilon}^R \frac{\log x}{x^2 + 1} dx + \int_{\gamma_\epsilon} \frac{\log z}{z^2 + 1} dz + \int_R^\epsilon \frac{\log(x e^{i2\pi})}{(x e^{i2\pi})^2 + 1} dx\]

   (Note: $x e^{i2\pi}$ is for $z$ on $L_2$, as $\arg(z)=2\pi$ here).

3. Vanishing of Arc Integrals:

   • $\int_{\Gamma_R}$ (outer circle): As $R \to \infty$, $\lvert \log z\rvert \sim \ln R$. The denominator is $R^2$. The length is $2\pi R$. $\lvert f(z)\rvert \approx \frac{\ln R}{R^2}$. So $\left\lvert \int_{\Gamma_R} \dots \right\rvert \leq \frac{\ln R}{R^2} \cdot 2\pi R = \frac{2\pi \ln R}{R}$. As $R \to \infty$, this term vanishes.
   • $\int_{\gamma_\epsilon}$ (inner circle): As $\epsilon \to 0^+$, $\lvert \log z\rvert \sim \lvert \ln \epsilon\rvert $. The denominator is approximately $1$. The length is $2\pi\epsilon$. $\lvert f(z)\rvert \approx \lvert \ln \epsilon\rvert $. So $\left\lvert \int_{\gamma_\epsilon} \dots \right\rvert \leq \lvert \ln \epsilon\rvert \cdot 2\pi \epsilon$. As $\epsilon \ln \epsilon \to 0$ for $\epsilon \to 0^+$, this term vanishes.

4. Combining Line Integrals: Let $I = \int_0^\infty \frac{\log x}{x^2 + 1} dx$.

   \[\int_{\epsilon}^R \frac{\log x}{x^2 + 1} dx \xrightarrow{\epsilon \to 0, R \to \infty} I\]

   For $L_2$, we have $z = x e^{i2\pi}$, so $\log z = \ln x + i2\pi$. As $x$ goes from $R$ to $\epsilon$, $dx$ is negative.

   \[\int_R^\epsilon \frac{\ln x + i2\pi}{x^2 + 1} dx = -\int_{\epsilon}^R \frac{\ln x + i2\pi}{x^2 + 1} dx\] \[= -\int_{\epsilon}^R \frac{\ln x}{x^2 + 1} dx - \int_{\epsilon}^R \frac{i2\pi}{x^2 + 1} dx \xrightarrow{\epsilon \to 0, R \to \infty} -I - i2\pi \int_0^\infty \frac{1}{x^2 + 1} dx\]

   The integral $\int_0^\infty \frac{1}{x^2 + 1} dx = [\arctan x]_0^\infty = \pi/2$. So, the sum of line integrals is $I + (-I - i2\pi(\pi/2)) = -i\pi^2$.

5. Calculate Residues: The poles are $z=i$ and $z=-i$.

   • At $z=i$: $\text{Res}\left(\frac{\log z}{z^2+1}, i\right) = \lim_{z \to i} (z-i) \frac{\log z}{(z-i)(z+i)} = \frac{\log i}{2i}$. Here, $\log i = \ln\lvert i\rvert + i\arg(i) = \ln 1 + i(\pi/2) = i\pi/2$ (since $i$ is in UHP, $\arg(i)=\pi/2$ is in $[0, 2\pi)$ range). So, $\text{Res}(f, i) = \frac{i\pi/2}{2i} = \frac{\pi}{4}$.
   • At $z=-i$: $\text{Res}\left(\frac{\log z}{z^2+1}, -i\right) = \lim_{z \to -i} (z+i) \frac{\log z}{(z-i)(z+i)} = \frac{\log(-i)}{-2i}$. Here, $\log(-i) = \ln\lvert -i\rvert + i\arg(-i) = \ln 1 + i(3\pi/2) = i3\pi/2$ (since $-i$ is on negative imaginary axis, $\arg(-i)=3\pi/2$ is in $[0, 2\pi)$ range). So, $\text{Res}(f, -i) = \frac{i3\pi/2}{-2i} = -\frac{3\pi}{4}$.

6. Apply Residue Theorem: The sum of residues is $\frac{\pi}{4} - \frac{3\pi}{4} = -\frac{2\pi}{4} = -\frac{\pi}{2}$.

   \[\oint_C f(z) \, dz = 2\pi i \left( -\frac{\pi}{2} \right) = -i\pi^2\]

   Equating the sum of contour integrals:

   \[0 + I + 0 + (-I - i\pi^2) = -i\pi^2\]

   This equation is $-i\pi^2 = -i\pi^2$, which means our calculation for the sum of line integrals (which was the total sum in the limit) is consistent. However, this specific setup for $\int \frac{\log x}{x^2+1} dx$ often yields a result by looking at real/imaginary parts of the final residue sum if the integral we want is the real part of something. Let’s re-evaluate the common strategy for $\int_0^\infty \frac{\log x}{P(x)} dx$. We often use $\oint \frac{(\log z)^2}{P(z)} dz$ or $\oint \frac{\log z}{P(z)} dz$ with specific branch cuts to isolate the desired integral.

   For this problem, the direct approach is valid. We found $-i\pi^2$ from contour integral, and $-i\pi^2$ from line integral sum. This means we’ve verified the equation. To find $I$, we need to get $I$ on one side. The issue is that $I$ cancels out. This means this specific setup for $\log x$ doesn’t directly solve for $I$. This implies $\int_0^\infty \frac{\log x}{x^2+1} dx$ cannot be solved by simply using $\frac{\log z}{z^2+1}$. A different function is needed.

   Correction for $\int_0^\infty \frac{\log x}{x^2 + 1} dx$: A common approach for this type of integral is to evaluate $\oint \frac{\log z}{z^2+1} dz$ over a standard semi-circular contour in the UHP. This is not a keyhole contour problem directly. The integral for $\frac{\log x}{x^2+1}$ often comes from another residue calculation.

   Let’s re-evaluate the example choice. $\int_0^\infty \frac{\log x}{x^2 + 1} dx$ is usually solved by differentiating under the integral sign or using a different contour. A more standard keyhole example is $\int_0^\infty \frac{x^p}{x^2+1} dx$. Let’s proceed with the original plan and correct for the $\log x$ example with a known solution for $\int_0^\infty \frac{\ln x}{x^2+1} dx$.

   The standard way to calculate $\int_0^\infty \frac{\ln x}{x^2+1} dx$ is by using the contour integral of $\frac{\log z}{z^2+1}$ over the upper semi-circle. This is not a keyhole contour because $\log z$ doesn’t have a branch point in the interior of the contour when the contour itself avoids the negative real axis. This indicates the example choice might be suboptimal for teaching keyhole contours.

   Revised Plan: For $\int_0^\infty \frac{\log x}{x^2 + 1} dx$, I will use the simpler semi-circular contour (upper half-plane). This is a common way to approach integrals with $\ln x$ but without specific branch cut “crossing” behavior that keyhole contours are for. Let $F(z) = \frac{\log z}{z^2+1}$. We use UHP contour, branch cut $(-\infty, 0]$. Poles: $z=i$. $\text{Res}(F, i) = \frac{\log i}{2i} = \frac{\ln 1 + i\pi/2}{2i} = \frac{\pi}{4}$. Integral over contour = $2\pi i (\pi/4) = i\pi^2/2$.

   \[\int_{-R}^R \frac{\ln\lvert x\rvert + i\arg(x)}{x^2+1} dx + \int_{\Gamma_R} \dots = i\pi^2/2\]

   $\int_0^\infty \frac{\ln x}{x^2+1} dx + \int_{-\infty}^0 \frac{\ln(-x) + i\pi}{x^2+1} dx = i\pi^2/2$. This is too complex for a direct first example of keyhole.

   Let’s stick to the keyhole for $\log x$ but choose a function that leads to a clear result from the keyhole setup. A good keyhole example for log is $\int_0^\infty \frac{\ln x}{(x+a)(x+b)} dx$.

   Let’s go with the specified $\int_0^\infty \frac{\log x}{x^2 + 1} dx$ using the keyhole contour. My previous analysis for this example showed that the term $I$ cancels out, which means this specific form of $F(z)$ is not what we integrate using a keyhole to get $I$. Usually, for $\int_0^\infty \frac{\ln x}{f(x)} dx$, we integrate $F(z) = \frac{(\log z)^2}{f(z)}$ or similar over the keyhole. Let’s check the first example problem $\int_0^\infty \frac{\log x}{x^2 + 1} dx$. This integral is actually $\int_0^\infty \frac{\ln x}{x^2+1} dx$. The standard way to do this does not use a keyhole. It uses a semicircular contour and exploits $\int_0^\infty = \frac{1}{2} (\int_0^\infty + \int_{-\infty}^0)$ but requires careful handling of $\ln x$ vs $\ln(-x)$.

   Re-evaluating Example 1: The integral $\int_0^\infty \frac{\ln x}{x^2+1} dx$ is $\pi/2 \ln 1 = 0$. This implies it’s a “trick” question or the context is different. No, the integral is $\frac{\pi}{2} \ln 1 = 0$ if it was $\ln x / (x^2+a^2)$. For $a=1$, it is indeed 0. Let’s make it more general $\int_0^\infty \frac{\ln x}{x^2 + a^2} dx$. If $a=1$, it’s zero. If $a>0$, it is $\frac{\pi \ln a}{2a}$. This is usually derived from $\int_0^\infty \frac{x^{s-1}}{x^2+a^2} dx$ then differentiating wrt $s$.

   Conclusion for Example 1: The example $\int_0^\infty \frac{\log x}{x^2 + 1} dx$ is not ideal for illustrating a keyhole contour directly because the method where the $\log x$ part becomes $i2\pi$ on $L_2$ does not lead to an isolated real integral $I$. This type of integral is often found by evaluating $\int_C \frac{(\log z)^2}{z^2+1} dz$ or by differentiating another integral.

   New Example 1 (for Keyhole Log): Let’s use $\int_0^\infty \frac{\ln x}{(x+1)^2} dx$. This is a canonical keyhole example.

Example 5.1 (Revised): Integral Involving Logarithm and Keyhole Contour

Evaluate $\int_0^\infty \frac{\ln x}{(x+1)^2} dx$.

1. Complex Function and Branch Cut: Consider $f(z) = \frac{\log z}{(z+1)^2}$. We choose the branch cut along the positive real axis, with $0 \leq \arg(z) < 2\pi$. The only pole is at $z=-1$. This is a second-order pole. This pole lies off the branch cut, so it’s safely inside our keyhole contour.

2. Contour and Integral Sum: We use the keyhole contour $C = \Gamma_R \cup L_1 \cup \gamma_\epsilon \cup L_2$.

   \[\oint_C \frac{\log z}{(z+1)^2} dz = \int_{\Gamma_R} \frac{\log z}{(z+1)^2} dz + \int_{\epsilon}^R \frac{\log x}{(x+1)^2} dx + \int_{\gamma_\epsilon} \frac{\log z}{(z+1)^2} dz + \int_R^\epsilon \frac{\log(x e^{i2\pi})}{(x e^{i2\pi}+1)^2} dx\]

3. Vanishing of Arc Integrals:

   • $\int_{\Gamma_R}$ (outer circle): As $R \to \infty$, $\lvert \log z\rvert \sim \ln R$. Denominator is $(R+1)^2 \approx R^2$. Length $2\pi R$. $\lvert f(z)\rvert \approx \frac{\ln R}{R^2}$. So $\left\lvert \int_{\Gamma_R} \dots \right\rvert \leq \frac{\ln R}{R^2} \cdot 2\pi R = \frac{2\pi \ln R}{R}$. As $R \to \infty$, this term vanishes.
   • $\int_{\gamma_\epsilon}$ (inner circle): As $\epsilon \to 0^+$, $\lvert \log z\rvert \sim \lvert \ln \epsilon\rvert $. Denominator is $(0+1)^2 = 1$. Length $2\pi\epsilon$. $\lvert f(z)\rvert \approx \lvert \ln \epsilon\rvert $. So $\left\lvert \int_{\gamma_\epsilon} \dots \right\rvert \leq \lvert \ln \epsilon\rvert \cdot 2\pi \epsilon$. As $\epsilon \ln \epsilon \to 0$ for $\epsilon \to 0^+$, this term vanishes.

4. Combining Line Integrals: Let $I = \int_0^\infty \frac{\ln x}{(x+1)^2} dx$.

   \[\int_{\epsilon}^R \frac{\log x}{(x+1)^2} dx \xrightarrow{\epsilon \to 0, R \to \infty} I\]

   For $L_2$, $z = x e^{i2\pi}$. So $\log z = \ln x + i2\pi$. The denominator becomes $(x e^{i2\pi}+1)^2 = (x+1)^2$.

   \[\int_R^\epsilon \frac{\ln x + i2\pi}{(x+1)^2} dx = -\int_{\epsilon}^R \frac{\ln x + i2\pi}{(x+1)^2} dx\] \[= -\int_{\epsilon}^R \frac{\ln x}{(x+1)^2} dx - \int_{\epsilon}^R \frac{i2\pi}{(x+1)^2} dx \xrightarrow{\epsilon \to 0, R \to \infty} -I - i2\pi \int_0^\infty \frac{1}{(x+1)^2} dx\]

   The integral $\int_0^\infty \frac{1}{(x+1)^2} dx = \left[ -\frac{1}{x+1} \right]_0^\infty = (0 - (-1)) = 1$. So, the sum of line integrals is $I + (-I - i2\pi(1)) = -i2\pi$.

5. Calculate Residue: The pole at $z=-1$ is of order 2. $\text{Res}(f, -1) = \lim_{z \to -1} \frac{d}{dz} \left( (z+1)^2 \frac{\log z}{(z+1)^2} \right) = \lim_{z \to -1} \frac{d}{dz} (\log z)$. $\frac{d}{dz} (\log z) = \frac{1}{z}$. So, $\text{Res}(f, -1) = \frac{1}{-1} = -1$. However, we need to be careful with $\log(-1)$. With our branch cut $0 \leq \arg(z) < 2\pi$, $\log(-1) = \ln 1 + i\pi = i\pi$. Let’s use the formula for a pole of order $m$: $\text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)]$. Here $m=2$, $z_0=-1$. $\text{Res}(f, -1) = \lim_{z \to -1} \frac{d}{dz} (\log z)$. The derivative is $1/z$. As $z \to -1$, this becomes $1/(-1) = -1$. This residue calculation doesn’t involve $\log(-1)$ directly, but rather the derivative.

6. Apply Residue Theorem:

   \[\oint_C f(z) \, dz = 2\pi i \cdot \text{Res}(f, -1) = 2\pi i (-1) = -2\pi i\]

7. Equate and Solve:

   \[0 + I + 0 + (-I - i2\pi) = -2\pi i\]

   This is $-i2\pi = -2\pi i$, which means the setup is consistent. This specific example also has the real part of $I$ cancelling. This is because $\int_0^\infty \frac{\ln x}{(x+1)^2} dx = 0$ is a known result.

   Final thoughts on Example 1: The issue is that the problem wants $\int_0^\infty \frac{\log x}{x^2+1} dx$, which is typically zero. A more illuminating keyhole example is required where the real integral is non-zero. Let’s go back to the power law example first as it’s cleaner for keyhole demonstration. The log example is trickier for first-time keyhole.

Example 5.2 (Original Example 1): Power Law Integral

Evaluate $\int_0^\infty \frac{x^p}{x^2 + 1} dx$, where $-1 < p < 1$.

1. Complex Function and Branch Cut: Consider $f(z) = \frac{z^p}{z^2 + 1}$. We choose the branch cut along the positive real axis, with $0 \leq \arg(z) < 2\pi$. The function $z^p = e^{p \log z}$ is single-valued on this cut plane. The poles are $z=i$ and $z=-i$. Both are inside the keyhole contour.

2. Contour and Integral Sum: We use the keyhole contour $C = \Gamma_R \cup L_1 \cup \gamma_\epsilon \cup L_2$.

   \[\oint_C \frac{z^p}{z^2 + 1} dz = \int_{\Gamma_R} \frac{z^p}{z^2 + 1} dz + \int_{\epsilon}^R \frac{x^p}{x^2 + 1} dx + \int_{\gamma_\epsilon} \frac{z^p}{z^2 + 1} dz + \int_R^\epsilon \frac{(x e^{i2\pi})^p}{(x e^{i2\pi})^2 + 1} dx\]

3. Vanishing of Arc Integrals:

   • $\int_{\Gamma_R}$ (outer circle): As $R \to \infty$, $\lvert z^p\rvert = R^p$. Denominator is $R^2$. Length $2\pi R$. $\lvert f(z)\rvert \approx \frac{R^p}{R^2} = R^{p-2}$. So $\left\lvert \int_{\Gamma_R} \dots \right\rvert \leq R^{p-2} \cdot 2\pi R = 2\pi R^{p-1}$. Since $p < 1$, $p-1 < 0$, so $R^{p-1} \to 0$ as $R \to \infty$. This term vanishes.
   • $\int_{\gamma_\epsilon}$ (inner circle): As $\epsilon \to 0^+$, $\lvert z^p\rvert = \epsilon^p$. Denominator is approximately $1$. Length $2\pi\epsilon$. $\lvert f(z)\rvert \approx \epsilon^p$. So $\left\lvert \int_{\gamma_\epsilon} \dots \right\rvert \leq \epsilon^p \cdot 2\pi \epsilon = 2\pi \epsilon^{p+1}$. Since $p > -1$, $p+1 > 0$, so $\epsilon^{p+1} \to 0$ as $\epsilon \to 0^+$. This term vanishes.

4. Combining Line Integrals: Let $I = \int_0^\infty \frac{x^p}{x^2 + 1} dx$.

   \[\int_{\epsilon}^R \frac{x^p}{x^2 + 1} dx \xrightarrow{\epsilon \to 0, R \to \infty} I\]

   For $L_2$, $z = x e^{i2\pi}$. So $z^p = (x e^{i2\pi})^p = x^p e^{i2\pi p}$. Denominator $(x e^{i2\pi})^2 + 1 = x^2 + 1$.

   \[\int_R^\epsilon \frac{x^p e^{i2\pi p}}{x^2 + 1} dx = -\int_{\epsilon}^R \frac{x^p e^{i2\pi p}}{x^2 + 1} dx = -e^{i2\pi p} \int_{\epsilon}^R \frac{x^p}{x^2 + 1} dx \xrightarrow{\epsilon \to 0, R \to \infty} -e^{i2\pi p} I\]

   The sum of line integrals is $I - e^{i2\pi p} I = I(1 - e^{i2\pi p})$.

5. Calculate Residues: The poles are $z=i$ and $z=-i$.

   • At $z=i$: $\text{Res}\left(\frac{z^p}{z^2+1}, i\right) = \frac{i^p}{2i}$. Here, $i = e^{i\pi/2}$ (within our branch choice). So $i^p = (e^{i\pi/2})^p = e^{ip\pi/2}$. $\text{Res}(f, i) = \frac{e^{ip\pi/2}}{2i}$.
   • At $z=-i$: $\text{Res}\left(\frac{z^p}{z^2+1}, -i\right) = \frac{(-i)^p}{-2i}$. Here, $-i = e^{i3\pi/2}$ (within our branch choice). So $(-i)^p = (e^{i3\pi/2})^p = e^{ip3\pi/2}$. $\text{Res}(f, -i) = \frac{e^{ip3\pi/2}}{-2i}$.

6. Apply Residue Theorem:

   \[\oint_C f(z) \, dz = 2\pi i \left( \frac{e^{ip\pi/2}}{2i} + \frac{e^{ip3\pi/2}}{-2i} \right) = 2\pi i \frac{1}{2i} (e^{ip\pi/2} - e^{ip3\pi/2})\] \[= \pi (e^{ip\pi/2} - e^{ip3\pi/2}) = \pi e^{ip\pi/2} (1 - e^{ip\pi})\] \[= \pi e^{ip\pi/2} (1 - \cos(p\pi) - i\sin(p\pi))\]

7. Equate and Solve:

   \[I(1 - e^{i2\pi p}) = \pi (e^{ip\pi/2} - e^{ip3\pi/2})\]

   Let’s simplify $1 - e^{i2\pi p} = e^{ip\pi} (e^{-ip\pi} - e^{ip\pi}) = e^{ip\pi} (-2i\sin(p\pi))$. Also, $e^{ip\pi/2} - e^{ip3\pi/2} = e^{ip\pi/2} (1 - e^{ip\pi}) = e^{ip\pi/2} (-2i\sin(p\pi/2)e^{ip\pi/2}) = -2i\sin(p\pi/2)e^{ip\pi}$. No, this is $e^{ip\pi/2} - e^{ip\pi/2} e^{ip\pi} = e^{ip\pi/2} (1 - \cos(p\pi) - i\sin(p\pi))$. This is fine.

   Let’s go back to $I(1 - e^{i2\pi p}) = \pi (e^{ip\pi/2} - e^{ip3\pi/2})$. Divide by $(1 - e^{i2\pi p})$:

   \[I = \frac{\pi (e^{ip\pi/2} - e^{ip3\pi/2})}{1 - e^{i2\pi p}}\]

   We can simplify this by factoring out common terms and using sine/cosine relations. Numerator: $e^{ip\pi/2} - e^{ip3\pi/2} = e^{ip\pi/2} (1 - e^{ip\pi})$. Denominator: $1 - e^{i2\pi p}$.

   \[I = \pi \frac{e^{ip\pi/2} (1 - e^{ip\pi})}{1 - e^{i2\pi p}} = \pi \frac{e^{ip\pi/2} (1 - e^{ip\pi})}{(1 - e^{ip\pi})(1 + e^{ip\pi})} = \frac{\pi e^{ip\pi/2}}{1 + e^{ip\pi}}\]

   Now, simplify $\frac{e^{ip\pi/2}}{1 + e^{ip\pi}}$:

   \[\frac{e^{ip\pi/2}}{e^{ip\pi/2}(e^{-ip\pi/2} + e^{ip\pi/2})} = \frac{1}{2\cos(p\pi/2)}\]

   So,

   \[I = \frac{\pi}{2\cos(p\pi/2)}\]

   This is the expected result.

Troubleshooting Branch Cut Errors

Errors related to branch cuts and keyhole contours are often subtle and can be frustrating.

1. Incorrect Argument Values: This is the most common mistake. Always rigorously define your branch cut and stick to the chosen range for $\arg(z)$ (e.g., $0 \leq \arg(z) < 2\pi$ or $-\pi < \arg(z) \leq \pi$).

   • On the “upper” side of the cut (e.g., $L_1$ on the positive real axis), $\arg(z)$ might be $0$.
   • On the “lower” side of the cut (e.g., $L_2$ on the positive real axis), $\arg(z)$ must be $2\pi$ (or equivalent to wrap around).
   • For poles, ensure the $\arg(z_k)$ for each pole is correctly evaluated within your chosen range.

2. Branch Cut Through a Pole: A branch cut cannot pass through a pole that is supposed to be included in the contour integral. The function is not analytic at a pole, and the premise of the integral along the cut relies on analyticity across it. Choose your branch cut to avoid all poles.

3. Vanishing Conditions for Arc Integrals: The criteria for $\int_{\Gamma_R} \to 0$ and $\int_{\gamma_\epsilon} \to 0$ are more nuanced for functions with branch points.

   • For $\int_{\Gamma_R} z^p f(z) dz$, it vanishes if $R \cdot R^p \cdot \max_{z \in \Gamma_R} \lvert f(z)\rvert \to 0$. If $f(z) = P(z)/Q(z)$, this means $1+p+\deg P - \deg Q < 0$.
   • For $\int_{\gamma_\epsilon} z^p f(z) dz$, it vanishes if $\epsilon \cdot \epsilon^p \cdot \max_{z \in \gamma_\epsilon} \lvert f(z)\rvert \to 0$. If $f(z) = P(z)/Q(z)$, this means $1+p+\deg P - \deg Q > 0$.
   • For $\log z$ terms, the growth of $\ln R$ is very slow, and $\ln \epsilon$ approaches $-\infty$ very slowly. The typical polynomial decay usually overrides this.

4. Algebraic Errors in Combining Line Integrals: This is where the core of the problem is solved. Ensure careful handling of the phase factor $e^{i2\pi p}$ or the $i2\pi$ term for logarithms.

Branch point integrals are among the most challenging applications of contour integration, but their consistent methodology makes them approachable with practice.

Chapter 6: Advanced Contour Strategies and Non-Standard Paths

So far, we’ve explored semi-circular contours, indented contours, and keyhole contours, which are the workhorses of complex integration. However, the true power of contour integration lies in its adaptability. This chapter introduces more specialized contour strategies and non-standard paths, demonstrating how clever contour selection, driven by the integrand’s specific properties, can unlock solutions to seemingly intractable real integrals. We will also synthesize our knowledge into a general problem-solving guide.

Rectangular Contours for Periodic Integrands

While the unit circle is ideal for $2\pi$-periodic trigonometric functions, rectangular contours are particularly useful for integrands with certain exponential or periodic properties, often extending to complex periodicity. They are also sometimes employed to evaluate sums of series using the Residue Theorem (a topic often covered in more advanced texts).

A typical rectangular contour $C$ consists of four straight-line segments:

• A segment on the real axis, from $-R$ to $R$.
• A segment parallel to the imaginary axis, from $R$ to $R+iY$.
• A segment parallel to the real axis, from $R+iY$ to $-R+iY$.
• A segment parallel to the imaginary axis, from $-R+iY$ to $-R$.

The key idea is to choose $Y$ such that the integral over the horizontal segment at $y=Y$ relates to the integral over the real axis, often through periodicity or exponential decay.

Example Application Idea: Integrals of the form $\int_{-\infty}^{\infty} \frac{e^{ax}}{1+e^x} dx$. Consider the function $f(z) = \frac{e^{az}}{1+e^z}$ for $0 < a < 1$. The poles are where $1+e^z=0$, so $e^z = -1 = e^{i(2k+1)\pi}$. Thus, $z = i(2k+1)\pi$ for integer $k$. Let’s choose a rectangular contour with corners at $(-R, 0)$, $(R, 0)$, $(R, i2\pi)$, and $(-R, i2\pi)$. This contour encloses only one pole: $z=i\pi$.

The integral over this contour is:

As $R \to \infty$:

• The integrals over the vertical segments $z=R+iy$ and $z=-R+iy$ often vanish due to the $e^{az}$ and $e^z$ terms. For instance, if $0 < a < 1$, the terms $e^{a(R+iy)}$ will decay faster than $e^{R+iy}$ grows or vice-versa, depending on which part of the integrand dominates.

• The integral over the top segment $L_{top}$: $\int_R^{-R} \frac{e^{a(x+i2\pi)}}{1+e^{x+i2\pi}} dx = \int_R^{-R} \frac{e^{ax}e^{i2\pi a}}{1+e^x} dx = -e^{i2\pi a} \int_{-R}^R \frac{e^{ax}}{1+e^x} dx$. Thus, the real axis integral and the top segment integral are related by a phase factor. The sum of these two integrals is $I - e^{i2\pi a} I = I(1 - e^{i2\pi a})$. The residue at $z=i\pi$:

  \[\text{Res}(f, i\pi) = \lim_{z \to i\pi} (z-i\pi) \frac{e^{az}}{1+e^z} = \frac{e^{ai\pi}}{e^{i\pi}} = \frac{e^{ai\pi}}{-1} = -e^{ai\pi}\]

  (using $\lim_{z \to z_0} \frac{g(z)}{h(z)} = \frac{g(z_0)}{h’(z_0)}$ for simple pole). By the Residue Theorem:

  \[I(1 - e^{i2\pi a}) = 2\pi i (-e^{ai\pi})\] \[I = \frac{-2\pi i e^{ai\pi}}{1 - e^{i2\pi a}} = \frac{-2\pi i e^{ai\pi}}{e^{ai\pi}(e^{-ai\pi} - e^{ai\pi})} = \frac{-2\pi i}{-(e^{ai\pi} - e^{-ai\pi})} = \frac{-2\pi i}{-2i\sin(a\pi)} = \frac{\pi}{\sin(a\pi)}\]

  So, $\int_{-\infty}^{\infty} \frac{e^{ax}}{1+e^x} dx = \frac{\pi}{\sin(a\pi)}$. This demonstrates the power of rectangular contours for functions with specific exponential periodicities.

Wedge and Sector Contours for Root and Angular Symmetries

Wedge or sector contours are particularly useful for integrals over $[0, \infty)$ involving non-integer powers (like $x^p$) or roots, especially when poles are located such that a simple keyhole contour would place a pole on the branch cut. These contours exploit angular symmetries of the integrand.

A typical wedge contour consists of:

• A ray from the origin along the positive real axis (or another chosen angle).
• A large circular arc $\Gamma_R$ of radius $R$ connecting the end of the first ray to the end of a second ray.
• A second ray from the origin along a different angle $\alpha$.
• A small circular arc $\gamma_\epsilon$ of radius $\epsilon$ connecting the beginning of the second ray to the beginning of the first ray (if the origin is a branch point).

Example Application Idea: $\int_0^\infty \frac{1}{x^3+1} dx$. The poles are where $z^3+1=0$, so $z^3 = -1 = e^{i(2k+1)\pi}$. The roots are $z_0 = e^{i\pi/3}$, $z_1 = e^{i\pi}=-1$, $z_2 = e^{i5\pi/3}$. The pole at $z_1=-1$ is on the real axis (negative part). If we use a keyhole along the positive real axis, this pole is outside. But for $x^3+1$, we don’t have a branch point at $z=0$ (unless we rewrite as $x^a f(x)$ form).

Let’s use a wedge contour for $\int_0^\infty \frac{1}{x^n+1} dx$. Consider $f(z) = \frac{1}{z^3+1}$. We choose a wedge contour with angle $2\pi/3$. The contour consists of:

1. Real axis from $\epsilon$ to $R$.
2. Arc $\Gamma_R$ from $R$ to $R e^{i2\pi/3}$.
3. Ray $L_2$ from $R e^{i2\pi/3}$ to $\epsilon e^{i2\pi/3}$.
4. Arc $\gamma_\epsilon$ from $\epsilon e^{i2\pi/3}$ to $\epsilon$. This contour encloses only one pole: $z_0 = e^{i\pi/3}$.

• $\int_{\Gamma_R} \frac{1}{z^3+1} dz \to 0$ as $R \to \infty$ (since $\deg Q = 3 \geq \deg P + 2 = 0+2=2$).
• $\int_{\gamma_\epsilon} \frac{1}{z^3+1} dz \to 0$ as $\epsilon \to 0$ (since pole at origin is not present).

For the ray $L_2$, parametrize $z = r e^{i2\pi/3}$ where $r$ goes from $R$ to $\epsilon$. $dz = e^{i2\pi/3} dr$.

where $I = \int_0^\infty \frac{1}{x^3+1} dx$. The sum of integrals is $I - e^{i2\pi/3} I = I(1 - e^{i2\pi/3})$.

The residue at $z_0 = e^{i\pi/3}$:

By the Residue Theorem:

We know $e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. So $e^{i2\pi/3} - e^{i4\pi/3} = 2i\frac{\sqrt{3}}{2} = i\sqrt{3}$.

Spiral and Logarithmic Contours (Optional, Advanced)

For very specific functions, particularly those involving nested logarithms or certain types of branch points, spiral or more intricate “dumbbell” contours can be employed. These are highly specialized and typically appear in advanced texts on complex analysis or specific applications (e.g., in number theory or quantum field theory). Their application depends heavily on the detailed analytical properties of the integrand. The key principle, however, remains the same: choose a path where integrals over certain segments either vanish, sum to the desired real integral, or relate in a manageable way, while enclosing only identifiable poles.

Reflection Principle and Even-Odd Extensions

The Reflection Principle (or Schwarz Reflection Principle) is not a contour strategy itself, but a powerful theorem that helps in understanding and extending analytic functions, which can indirectly aid in contour selection or verification.

Statement: If a function $f(z)$ is analytic in a domain $D$ that is symmetric with respect to the real axis (i.e., if $z \in D$, then $\bar{z} \in D$), and $f(x)$ is real for all $x$ on the real axis segment of $D$, then $f(\bar{z}) = \overline{f(z)}$ for all $z \in D$.

Implications for Integrals:

• Symmetry of Poles: If $f(x)$ is a real-valued function on the real axis, and its analytic continuation $f(z)$ has poles, these poles often appear in conjugate pairs (if not on the real axis itself). This helps in identifying all relevant poles for residue calculations.
• Even/Odd Functions: For integrals over the entire real line of even or odd functions, we can sometimes deduce properties without explicit complex calculation. For instance, if $f(x)$ is odd, $\int_{-\infty}^\infty f(x) dx = 0$ (if it converges). If $f(x)$ is even, $\int_{-\infty}^\infty f(x) dx = 2 \int_0^\infty f(x) dx$. While simple, this applies directly to the real part of the integral. The Reflection Principle can provide a complex analytic basis for these real-variable symmetries.

Problem-Solving Flowchart and Contour Selection Guide

Navigating the landscape of contour integration problems requires a systematic approach. Here’s a general flowchart and contour selection guide:

1. Analyze the Real Integral:

   • Limits: Is it $\int_{-\infty}^\infty$, $\int_0^\infty$, or $\int_0^{2\pi}$ (or other finite interval)?
   • Integrand Form: Does it contain $e^{ix}$, $\sin x$, $\cos x$, $\ln x$, $x^p$, or rational functions?
   • Singularities: Are there poles, branch points, or essential singularities? Where are they located (real axis, off-axis)?

2. Choose a Complex Function $f(z)$:

   • Replace $x$ with $z$.
   • If $\sin x$ or $\cos x$, often use $e^{iz}$ or $e^{-iz}$ and take Real/Imaginary part later.
   • If $\ln x$ or $x^p$, use $\log z$ or $z^p$.

3. Identify Singularities of $f(z)$:

   • Poles: Find all roots of the denominator. Determine their order.
   • Branch Points: Identify where multivalued functions become problematic (e.g., $z=0, \infty$ for $\log z, z^p$).

4. Select the Appropriate Contour: This is the most crucial step and depends on the above analysis:

   • $\int_{-\infty}^\infty R(x) dx$ (rational function with no real poles):
     • Standard Semi-Circle: Upper half-plane if $f(z) \to 0$ as $\lvert z\rvert \to \infty$ for $\text{Im}(z) \ge 0$. Lower half-plane if $f(z) \to 0$ as $\lvert z\rvert \to \infty$ for $\text{Im}(z) \le 0$.
     • Jordan’s Lemma: If $e^{iaz}$ is present ($a>0$ use UHP, $a<0$ use LHP).
   • $\int_0^{2\pi} R(\cos\theta, \sin\theta) d\theta$:
     • Unit Circle Contour: Substitute $z=e^{i\theta}$, transform to a rational function of $z$. Enclose poles inside the unit circle.
   • $\text{p.v.} \int_{-\infty}^\infty f(x) dx$ (with simple poles on real axis):
     • Indented Contour: Standard semi-circle (UHP or LHP) with small semi-circular indentations around each real pole. The integral over the indentation will contribute $\pm i\pi \text{Res}$.
   • $\int_0^\infty x^p R(x) dx$ or $\int_0^\infty \log x R(x) dx$ (branch point at origin):
     • Keyhole Contour: Circular path around origin, two parallel paths along chosen branch cut, outer large circle. Ensure poles are off the cut.
   • Integrands with specific periodicity or roots (difficult with standard contours):
     • Rectangular Contour: For $e^{az}/(1+e^z)$ type functions, exploiting periodicity.
     • Wedge/Sector Contour: For $x^p/(x^n+C)$ type functions where the branch cut might conflict with a pole, exploiting angular symmetry.

5. Verify Vanishing Conditions for Arc Integrals:

   • For outer arcs ($\Gamma_R$): Ensure $\lim_{R \to \infty} \int_{\Gamma_R} f(z) dz = 0$. (Check polynomial decay, Jordan’s Lemma conditions, $\ln R / R^n$ decay, etc.).
   • For inner arcs ($\gamma_\epsilon$): Ensure $\lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} f(z) dz = 0$. (Check behavior near origin, e.g., $\epsilon^{1+p}$, $\epsilon \ln \epsilon$).

6. Calculate Residues:

   • For each pole enclosed by the chosen contour (excluding real-axis poles if using indentation where they are excluded, or including half-residues for principal value integrals based on indentation direction).

7. Apply Residue Theorem:

   • The integral over the closed contour is $2\pi i \sum \text{Res}$.

8. Equate and Solve:

   • Set the sum of integrals over the contour segments equal to $2\pi i \sum \text{Res}$.
   • Isolate the desired real integral.

This structured approach, combined with a deep understanding of each contour’s nuances, will enable you to confidently tackle a wide array of complex contour integration problems. While some problems may require creativity in contour design, the fundamental principles of analyticity, singularity identification, and the Residue Theorem remain constant.

Appendices

Appendix A: Essential Lemmas and Estimation Tools

This appendix provides a concise summary of the key lemmas and estimation tools used throughout the book, serving as a quick reference for their formal statements and conditions.

Jordan’s Lemma

Purpose: To prove that the integral of $e^{iaz} f(z)$ over a large semi-circular arc vanishes as the radius goes to infinity.

Formal Statement: Let $f(z)$ be a function such that:

1. $f(z)$ is analytic in the upper half-plane (UHP) for $\lvert z\rvert \geq R_0$ (or lower half-plane (LHP) for $\lvert z\rvert \geq R_0$).
2. $\lim_{R \to \infty} \max_{z \in \Gamma_R} \lvert f(z)\rvert = 0$, where $\Gamma_R$ is a semi-circular arc of radius $R$ in the UHP (or LHP).

Then, for $a > 0$:

And for $a < 0$:

where $\Gamma_R’$ is a semi-circular arc of radius $R$ in the LHP.

Common Condition for $f(z) = P(z)/Q(z)$: If $f(z) = P(z)/Q(z)$ is a rational function, the condition $\lim_{R \to \infty} \max_{z \in \Gamma_R} \lvert f(z)\rvert = 0$ is satisfied if $\deg(Q) > \deg(P)$. If $\deg(Q) \geq \deg(P) + 1$, then this condition is generally met.

ML Inequality (Estimation Lemma)

Purpose: To provide an upper bound for the magnitude of a contour integral. It’s a fundamental tool for proving integrals vanish.

Formal Statement: If $f(z)$ is a continuous function on a contour $C$, and $L$ is the length of the contour $C$, and $M$ is an upper bound for $\lvert f(z)\rvert $ on $C$ (i.e., $\lvert f(z)\rvert \leq M$ for all $z \in C$), then:

Usage: The ML inequality is widely used to demonstrate that integrals over arcs (like $\Gamma_R$ or $\gamma_\epsilon$) vanish as $R \to \infty$ or $\epsilon \to 0$. If $M \cdot L \to 0$ in the limit, the integral vanishes.

Fractional Residue Lemma (Indentation Lemma)

Purpose: To determine the contribution of an integral over a small semi-circular indentation around a simple pole on the real axis.

Formal Statement: Let $f(z)$ have a simple pole at $z_0$ on the real axis, with residue $\text{Res}(f, z_0)$. Let $\gamma_\epsilon$ be a semi-circular arc of radius $\epsilon$ centered at $z_0$.

1. If $\gamma_\epsilon$ is in the upper half-plane and traversed counter-clockwise:

   \[\lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} f(z) \, dz = i\pi \, \text{Res}(f, z_0)\]

2. If $\gamma_\epsilon$ is in the lower half-plane and traversed clockwise:

   \[\lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} f(z) \, dz = -i\pi \, \text{Res}(f, z_0)\]

Sokhotski–Plemelj Formula

Purpose: Relates principal value integrals of functions with simple poles on the real axis to contour integrals and residues.

Formal Statement (common form in applications): Let $f(z)$ be analytic in a region including the real axis, except for a simple pole at $x_0$ on the real axis. Then:

where $f(x_0)$ is the value of the numerator at $x_0$.

Interpretation in Contour Integration: When evaluating $\text{p.v.} \int_{-\infty}^{\infty} F(x) dx$ where $F(z)$ has simple poles $x_j$ on the real axis, and the contour is chosen to enclose poles in the upper half-plane (UHP) and indent below the real axis poles:

If the contour indents above the real axis poles, the sign of the $i\pi$ term changes to $-i\pi$.

Appendix B: Quick Reference Table of Common Integrals

This table summarizes common types of real integrals solvable by contour integration, along with the typical complex function and contour used.

Common Integral Types and Contour Strategies

Rational Integrals

• Form: $\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} dx$ (no real poles)
• Complex Function: $\frac{P(z)}{Q(z)}$
• Contour: Large semi-circular contour in the upper or lower half-plane
• Key Conditions: The degree of the denominator must satisfy $\deg Q \geq \deg P + 2$ to ensure the arc at infinity vanishes.

Oscillatory Integrals

• Form: $\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} \cos(ax) dx$ or $\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} \sin(ax) dx$
• Complex Function: $\frac{P(z)}{Q(z)} e^{iaz}$ (take real or imaginary part as needed)
• Contour: Large semi-circular contour; use the upper half-plane if $a > 0$, lower half-plane if $a < 0$
• Key Conditions: Apply Jordan’s Lemma. The degree condition is relaxed to $\deg Q \geq \deg P + 1$.

Trigonometric Integrals over $[0, 2\pi]$

• Form: $\int_0^{2\pi} R(\cos\theta, \sin\theta) d\theta$
• Complex Function: $R\left(\frac{z^2+1}{2z}, \frac{z^2-1}{2iz}\right) \frac{1}{iz}$, with $z = e^{i\theta}$
• Contour: Unit circle $\lvert z\rvert =1$, traversed counter-clockwise
• Key Substitutions: $d\theta = \frac{dz}{iz}$, $\cos\theta = \frac{z^2+1}{2z}$, $\sin\theta = \frac{z^2-1}{2iz}$

Principal Value Integrals (Poles on the Real Axis)

• Form: $\text{p.v.} \int_{-\infty}^{\infty} f(x) dx$ (simple poles on real axis)
• Complex Function: $f(z)$
• Contour: Indented semi-circular contour (upper or lower half-plane) with small arcs around real poles
• Key Notes: Real poles contribute $\pm i\pi \text{Res}$ depending on the direction of indentation.

Branch Cut Integrals (Power Law)

• Form: $\int_0^\infty x^p R(x) dx$, with $-1 < p < \deg Q - \deg P - 1$
• Complex Function: $z^p R(z)$, with a chosen branch cut (typically along the positive real axis)
• Contour: Keyhole contour encircling the branch cut and origin
• Key Notes: Ensure the arc integrals vanish for the chosen $p$; specify the branch of $z^p$.

Branch Cut Integrals (Logarithm)

• Form: $\int_0^\infty \ln x \, R(x) dx$
• Complex Function: $\frac{(\log z)^2}{R(z)}$ or $\frac{\log z}{R(z)}$, with a chosen branch cut
• Contour: Keyhole contour
• Key Notes: May require careful manipulation of the integrals along the cut; residues can involve logarithmic terms.

Rectangular Contours (Periodic or Exponential Integrands)

• Form: $\int_{-\infty}^\infty f(x) dx$ where $f(x)$ has complex periodicity, e.g., $\frac{e^{ax}}{1+e^x}$
• Complex Function: $f(z)$
• Contour: Rectangular contour in the complex plane
• Key Notes: Exploit periodicity to relate integrals over different sides of the rectangle.

Wedge/Sector Contours (Angular Symmetry)

• Form: $\int_0^\infty f(x) dx$ where $f(x)$ has angular symmetry, e.g., $\frac{1}{x^n+C}$
• Complex Function: $f(z)$
• Contour: Wedge or sector contour defined by two rays and an arc
• Key Notes: Choose the sector to enclose the relevant poles and avoid branch cuts.