AMath 501

Outline

Basic Vector Manipulations

Dot products and cross products (area of a parallelogram)
Relation between norms and dot products
Scalar triple products (signed volume)
Geometric understanding of the concepts above

Coordinates & Position Vectors

Cartesian Coordinates:
- $r = x \hat{i} + y \hat{j} + z \hat{k}$
Spherical Coordinates:
- $r = r {\hat{e}}_{r}$
Cylindrical Coordinates:
- $r = r {\hat{e}}_{r} + z {\hat{e}}_{z}$

Curvilinear Coordinates

Unit vectors:

{\hat{e}}_{i} = \frac{\partial r / \partial u_{i}}{‖ \partial r / \partial u_{i} ‖}, i = 1, 2, 3

Scale factors:

h_{i} = ‖ \frac{\partial r}{\partial u_{i}} ‖

All scale factors in coordinates are stated above.

Derivatives and Arc Length

Derivative of position vector $d r$ :

d r = \frac{\partial r}{\partial u_{1}} d u_{1} + \frac{\partial r}{\partial u_{2}} d u_{2} + \frac{\partial r}{\partial u_{3}} d u_{3}

This can be written as:

d r = h_{1} d u_{1} {\hat{e}}_{1} + h_{2} d u_{2} {\hat{e}}_{2} + h_{3} d u_{3} {\hat{e}}_{3}

Arc length differential $(d s)^{2}$ :

(d s)^{2} = d r \cdot d r = h_{1}^{2} (d u_{1})^{2} + h_{2}^{2} (d u_{2})^{2} + h_{3}^{2} (d u_{3})^{2}

In orthogonal coordinates.

Arc Length Formula:

L = \int_{a}^{b} | \frac{d r (t)}{d t} | d t

Gradient Operator

$\nabla f$ for a scalar $f$ ,

\nabla := \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z} (in Cartesian)

\nabla := \frac{1}{h_{1}} {\hat{e}}_{1} \frac{\partial}{\partial u_{1}} + \frac{1}{h_{2}} {\hat{e}}_{2} \frac{\partial}{\partial u_{2}} + \frac{1}{h_{3}} {\hat{e}}_{3} \frac{\partial}{\partial u_{3}}

Geometric explanation: Represents the steepest ascent.

Divergence Operator

In Cartesian coordinates:

div F = \nabla \cdot F = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}

In general:

div F = \frac{1}{h_{1} h_{2} h_{3}} [\frac{\partial}{\partial u_{1}} (F_{1} h_{2} h_{3}) + \frac{\partial}{\partial u_{2}} (F_{2} h_{1} h_{3}) + \frac{\partial}{\partial u_{3}} (F_{3} h_{1} h_{2})]

where:

F = F_{1} {\hat{e}}_{1} + F_{2} {\hat{e}}_{2} + F_{3} {\hat{e}}_{3}

Write down div in spherical/cylindrical coordinates.

Physical explanation: Represents the net flux out of a volume per unit volume.

Curl Operator

Curl F = \frac{1}{h_{1} h_{2} h_{3}} | \begin{matrix} h_{1} {\hat{e}}_{1} & h_{2} {\hat{e}}_{2} & h_{3} {\hat{e}}_{3} \\ \frac{\partial}{\partial u_{1}} & \frac{\partial}{\partial u_{2}} & \frac{\partial}{\partial u_{3}} \\ h_{1} F_{1} & h_{2} F_{2} & h_{3} F_{3} \end{matrix} |

Physical Explanation

The circulation around a small area per unit area.

Line Integral

W = \int_{r (a)}^{r (b)} F \cdot d r = \int_{r (a)}^{r (b)} [F_{1} (h_{1} d u_{1}) + F_{2} (h_{2} d u_{2}) + F_{3} (h_{3} d u_{3})]

Parameterizing the line integral:

W = \int_{a}^{b} F \cdot \frac{d r}{d t} d t

= \int_{a}^{b} [F_{1} h_{1} \frac{d u_{1}}{d t} + F_{2} h_{2} \frac{d u_{2}}{d t} + F_{3} h_{3} \frac{d u_{3}}{d t}] d t

Conservative Field

If $F = \nabla Φ$ , where $Φ$ is a scalar.
If $F$ is conservative, then:

W = \int_{r (a)}^{r (b)} F \cdot d r = \int d Φ = Φ (r (b)) - Φ (r (a))

The independence of the path:
If $F$ is conservative, then $curl F = 0$ .
The opposite way is also true.

Surface Integral

\int_{S} \int F \cdot \hat{n} d s on a surface S

We have to find $\hat{n}$ and $d s$ .
In general:
The surface $S$ can be expressed in terms of $u_{1}$ and $u_{2}$ .

\begin{aligned} d S = \hat{n} d s & = (\frac{\partial r}{\partial u_{1}} \times \frac{\partial r}{\partial u_{2}}) d u_{1} d u_{2} \\ = \frac{\frac{\partial r}{\partial u_{1}} \times \frac{\partial r}{\partial u_{2}}}{‖ \frac{\partial r}{\partial u_{1}} \times \frac{\partial r}{\partial u_{2}} ‖} \cdot ‖ \frac{\partial r}{\partial u_{1}} \times \frac{\partial r}{\partial u_{2}} ‖ d u_{1} d u_{2} \\ = \hat{n} d s \end{aligned}

If the surface is written as $f (x, y, z) = 0$ , then:

\hat{n} = \frac{\nabla f}{‖ \nabla f ‖}, which is easier than: \hat{n} = \frac{\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y}}{‖ \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} ‖}

Surface Representation

If we use $x, y$ to write the surface as $z = h (x, y)$ , then:

d s = \sqrt{1 + {(\frac{\partial h}{\partial x})}^{2} + {(\frac{\partial h}{\partial y})}^{2}} d x d y

Since:

r = x \hat{i} + y \hat{j} + h (x, y) \hat{k}

and:

\frac{\partial r}{\partial x} = \hat{i} + \frac{\partial h}{\partial x} \hat{k}, \frac{\partial r}{\partial y} = \hat{j} + \frac{\partial h}{\partial y} \hat{k}

\hat{n} d s = \frac{\nabla f}{\sqrt{{(\frac{\partial f}{\partial x})}^{2} + {(\frac{\partial f}{\partial y})}^{2} + {(\frac{\partial f}{\partial z})}^{2}}} \cdot \sqrt{1 + {(\frac{\partial h}{\partial x})}^{2} + {(\frac{\partial h}{\partial y})}^{2}} d x d y

If $f (x, y, z) = z - h (x, y) = 0$ , then:

\hat{n} d s = \nabla f d x d y

Surface Area

A = \iint d s

Divergence Theorem

Compute the net flux through the surface of a volume by using the volume integral:

\iint_{\partial V} F \cdot \hat{n} d s = ∭_{V} div F d V

Where:

d V = h_{1} h_{2} h_{3} d u_{1} d u_{2} d u_{3}

Stokes' Formula/Theorem

\int_{S} \int curl F \cdot \hat{n} d s = \oint_{C} F \cdot d r,

The circulation around an area $S$ , where $C$ is the boundary of $S$ .
RHS (Right-Hand Side) is easier to compute.

This expands to:

\nabla \times F = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}) \hat{i} - (\frac{\partial F_{3}}{\partial x} - \frac{\partial F_{1}}{\partial z}) \hat{j} + (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) \hat{k}

In 2D:

curl F = \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}

Definition of Holomorphic Functions / Analytic Functions

A holomorphic function (or analytic function) is a complex function $f (z)$ that is differentiable at every point in an open subset of the complex plane $C$ . Differentiability in the complex sense means that the limit defining the derivative:

f' (z) = lim_{h \to 0} \frac{f (z + h) - f (z)}{h}

exists and is the same no matter how $h$ approaches $0$ in the complex plane.

Cauchy-Riemann Relations:

For $f (z) = u (x, y) + i v (x, y)$ (where $z = x + i y$ ), the function is holomorphic if:

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} .

Branch Cuts / Points:

Some functions (like $\ln z$ or $\sqrt{z}$ ) are not holomorphic everywhere because they involve multivalued branches. For example:

$\ln z$ : Has a branch cut along the negative real axis.
$\sqrt{z}$ : Often has a branch cut along the negative real axis to ensure single-valuedness.

Example

Holomorphic Function: $f (z) = z^{2} + i$

Here, $u (x, y) = x^{2} - y^{2}$ and $v (x, y) = 2 x y + 1$ . These satisfy the Cauchy-Riemann equations:

\frac{\partial u}{\partial x} = 2 x, \frac{\partial v}{\partial y} = 2 x

and

\frac{\partial u}{\partial y} = - 2 y, \frac{\partial v}{\partial x} = - 2 y .

Non-Holomorphic Function: $f (z) = | z |^{2} = x^{2} + y^{2}$

Here, $u (x, y) = x^{2} + y^{2}$ and $v (x, y) = 0$ . The Cauchy-Riemann equations fail:

\frac{\partial u}{\partial x} = 2 x \neq \frac{\partial v}{\partial y} = 0.

Cauchy Theorem

If $f (z)$ is holomorphic (analytic) in a simply connected domain $D$ and $γ$ is a closed contour in $D$ , then:

\int_{γ} f (z) d z = 0.

Key Idea:
The integral of an analytic function over a closed loop is zero if the function is analytic everywhere inside and on the contour.

Example:

Let $f (z) = z^{2}$ , and $γ$ is the circle $| z | = 1$ :

\int_{γ} z^{2} d z = 0

since $f (z)$ is holomorphic everywhere.

Cauchy Integral Formula

If $f (z)$ is holomorphic in a simply connected domain $D$ and $γ$ is a simple closed contour in $D$ , then for any point $a$ inside $γ$ :

f (a) = \frac{1}{2 π i} \int_{γ} \frac{f (z)}{z - a} d z .

Key Idea:
The value of a holomorphic function inside a contour can be computed using the values of the function on the contour.

Example:

Let $f (z) = z^{2}$ , and $γ$ is the circle $| z | = 2$ . Compute $f (0)$ :

f (0) = \frac{1}{2 π i} \int_{γ} \frac{z^{2}}{z} d z = \frac{1}{2 π i} \int_{γ} z d z .

Since $\int_{γ} z d z = 0$ , $f (0) = 0$ .

Residue

Intuition: Circulation Around the Singularity

The residue measures how much the function “circulates” around the singularity. If you imagine walking along the contour $C$ , the residue tells you the strength and direction of this “circulation” (positive for counterclockwise, negative for clockwise).

In $f (z) = \frac{1}{z - 1}$ , the residue is $1$ , meaning a unit of circulation counterclockwise around $z = 1$ .

Key Points About Residues and Singularities

Residue Explains Local Behavior Near a Singularity
The residue at a singularity $z = z_{0}$ describes the “circulation” of $f (z)$ near that singularity.
For example, in $f (z) = \frac{1}{(z + 2) (z - 1)}$ :

$z = - 2$ and $z = 1$ are the two singularities.
Each singularity has its own residue, calculated by isolating the behavior of $f (z)$ near that singularity (locally).

Residues Are Independent of Distance From the Singularity
Residues depend only on the singularity itself, not on the distance from it. The contour $C$ used to calculate the residue can have any radius, as long as it encloses only that singularity (and no others).

This is because the integral $\oint_{C} f (z) d z$ over any closed contour enclosing the singularity gives the same result, thanks to the Cauchy Integral Formula.

Residue Measures Local Circulation
Residue tells us how $f (z)$ “circulates” locally around the singularity. This is like isolating the behavior of $f (z)$ at that point.

Each residue at $z = - 2$ and $z = 1$ is independent of the other singularity.

Why Distance Doesn’t Affect Residue
The residue at a singularity depends only on the coefficient of $\frac{1}{z - z_{0}}$ in the Laurent series expansion. Since Laurent series expansions focus on the singularity and ignore the behavior elsewhere, the residue is unaffected by the distance from the singularity.

Residue Theorem

Let $f (z)$ be holomorphic in a domain $D$ , except for isolated singularities. If $γ$ is a simple closed contour in $D$ that encloses these singularities, then:

\int_{γ} f (z) d z = 2 π i \sum Residues of f (z) inside γ .

Key Idea:
The integral around a closed contour is determined by the sum of residues of the function’s singularities inside the contour.

Single Singular Point at z = 0

In your specific example, the contour $C$ is the unit circle $| z | = 1$ , and the function $f (z)$ has only one singularity inside the contour, at $z = 0$ . For this special case:

\oint_{C} f (z) d z = 2 π i \cdot Res (f (z), 0) .

The residue of $f (z)$ at $z = 0$ is the coefficient of $\frac{1}{z}$ in the Laurent series expansion of $f (z)$ around $z = 0$ . That coefficient is denoted as $a_{- 1}$ .

So, in this case:

\oint_{C} f (z) d z = 2 π i \cdot Res (f (z), 0) = 2 π i \cdot a_{- 1} .

This is why $a_{- 1}$ alone represents the residue for the integral in this case—there is only one singularity inside the contour, so the residue theorem reduces to this single contribution.

Steps to Find Residues:

For a simple pole at $z = a$ , residue:

Res (f, a) = lim_{z \to a} (z - a) f (z) .

For higher-order poles or Laurent expansions, use:

Res (f, a) = coefficient of \frac{1}{z - a} in Laurent series of f (z) .

Example:

Let $f (z) = \frac{1}{z^{2} + 1}$ , and $γ$ is the circle $| z | = 2$ . The singularities are at $z = i$ and $z = - i$ . Residues:

Res (f, i) = lim_{z \to i} (z - i) \frac{1}{z^{2} + 1} = \frac{1}{2 i}, Res (f, - i) = \frac{- 1}{2 i} .

Thus:

\int_{γ} \frac{1}{z^{2} + 1} d z = 2 π i (\frac{1}{2 i} - \frac{1}{2 i}) = 0.

Calculation

Res (f, i) = lim_{z \to i} (z - i) \frac{1}{z^{2} + 1} = (z - i) \cdot \frac{1}{(z - i) (z + i)} = lim_{z \to i} \frac{1}{z + i} = \frac{1}{2 i}

Singularities

A singularity of a complex function is a point $z_{0}$ where the function is not holomorphic (analytic). There are three main types of singularities, which are classified based on the behavior of $f (z)$ near $z_{0}$ .

Classification of Singularities

Removable Singularity

Definition: A singularity $z_{0}$ is removable if $f (z)$ can be redefined at $z_{0}$ to make the function holomorphic.
Test: If $lim_{z \to z_{0}} f (z)$ exists and is finite, the singularity is removable.
Example: $f (z) = \frac{\sin z}{z}$ .
At $z = 0$ , $f (z)$ is undefined, but:

lim_{z \to 0} \frac{\sin z}{z} = 1.

So, $z = 0$ is a removable singularity. Redefine $f (0) = 1$ to make $f (z)$ analytic everywhere.

Pole

Definition: A singularity $z_{0}$ is a pole if $| f (z) | \to \infty$ as $z \to z_{0}$ , and $f (z)$ can be written as: $f (z) = \frac{g (z)}{(z - z_{0})^{n}}$ , where $g (z)$ is analytic and $g (z_{0}) \neq 0$ , and $n$ is a positive integer (the order of the pole).

Types:

Simple Pole: If $n = 1$ .
Higher-Order Pole: If $n > 1$ .

Example: $f (z) = \frac{1}{z - 1}$ .
Here, $z = 1$ is a simple pole since $| f (z) | \to \infty$ as $z \to 1$ . Another example: $f (z) = \frac{1}{(z - 1)^{2}}$ .
Here, $z = 1$ is a second-order pole.

Essential Singularity

Definition: A singularity $z_{0}$ is essential if $f (z)$ has no Laurent series expansion that terminates (like poles) or can’t be redefined (like removable singularities). Around $z_{0}$ , the behavior is unpredictable (the function takes on all possible values near $z_{0}$ ).
Example: $f (z) = e^{1 / z}$ .
At $z = 0$ , $f (z)$ has an essential singularity because:

e^{1 / z} = \sum_{n = 0}^{\infty} \frac{1}{n! z^{n}},

which is infinitely oscillatory as $z \to 0$ .

Laurent Expansions

A Laurent series is a representation of a complex function $f (z)$ as an infinite series that includes both positive and negative powers of $(z - z_{0})$ :

f (z) = \sum_{n = - \infty}^{\infty} c_{n} (z - z_{0})^{n} .

Here:

$z_{0}$ : The center of the series.
$c_{n}$ : Coefficients of the series, given by:

c_{n} = \frac{1}{2 π i} \int_{γ} \frac{f (z)}{(z - z_{0})^{n + 1}} d z,

where $γ$ is a closed contour around $z_{0}$ .

Taylor series works only if the function is analytic (holomorphic) everywhere within a region.
Laurent series works for functions with singularities by expanding into two parts:
- A regular part (positive powers of z ).
- A principal part (negative powers of z ).

Laurent Series Centered at z = 0

If the expansion is centered at $z_{0} = 0$ , then $(z - z_{0}) = z$ , and the Laurent series becomes:

f (z) = \sum_{n = - \infty}^{\infty} c_{n} z^{n} .

We write the Laurent series as:

f (z) = \dots + \frac{a_{- 2}}{z^{2}} + \frac{a_{- 1}}{z} + a_{0} + a_{1} z + a_{2} z^{2} + \dots,

where:

$c_{n} = a_{n}$ are the coefficients for $z^{n}$ ,
$c_{- n} = a_{- n}$ are the coefficients for $\frac{1}{z^{n}}$ .

So this is just notation. Writing it as $\frac{a_{- n}}{z^{n}}$ or $c_{n} z^{n}$ is equivalent, as long as you recognize that the coefficient of $\frac{1}{z^{n}}$ is $c_{- n}$ .

The coefficient $a_{- 1}$ specifically refers to the term $\frac{1}{z}$ , which corresponds to the residue. This notation emphasizes its role in the residue theorem:

Residue at z = 0 = a_{- 1} .

So, while the general Laurent series looks like:

f (z) = \sum_{n = - \infty}^{\infty} c_{n} (z - z_{0})^{n},

in practice, we rewrite it as:

f (z) = \dots + \frac{a_{- 2}}{z^{2}} + \frac{a_{- 1}}{z} + a_{0} + a_{1} z + \dots,

because this form explicitly separates positive and negative powers of $z$ , and it is easier to identify $a_{- 1}$ , the residue.

How to Rewrite

To decide whether to rewrite terms, follow these steps:

Step 1: Identify Singularities

Find all the singularities of the function. These divide the complex plane into regions of convergence.

Step 2: Determine the Region of Interest

Ask: What region am I analyzing? For example:

$| z | < R$ : Inside a circle of radius $R$ .
$R_{1} < | z | < R_{2}$ : An annular region.
$| z | > R$ : Outside a circle of radius $R$ .

Step 3: Check Convergence Behavior

If $| z | < R$ :
- Use $z$ directly for expansion (e.g., $\frac{1}{1 - z} = 1 + z + z^{2} + \dots$ ).
If $| z | > R$ :
- Rewrite terms to involve $\frac{1}{z}$ , ensuring $| \frac{1}{z} | < 1$ (e.g., $\frac{1}{1 - z} = - \frac{1}{z} (1 + \frac{1}{z} + \dots)$ ).
If $R_{1} < | z | < R_{2}$ :
- Carefully choose expansions to match the dominant behavior in the annular region.

General Rule for Laurent Series:

Focus on $| w | < 1$ :
Identify the variable $w$ such that $| w | < 1$ . This ensures convergence of the geometric series.
If the original function satisfies $| w | < 1$ , no rewriting is needed.
- Example: $| z | < 1$ for $\frac{1}{1 + z}$ .
If $| w | \geq 1$ , rewrite the function to redefine $w$ so $| w | < 1$ .
- Example: $| z | > 1$ for $\frac{1}{1 + z}$ , rewrite using $\frac{1}{z}$ .

Example 1:

Laurent Series for $f (z) = \frac{1}{z (z - 1)}$ .
Expand $f (z) = \frac{1}{z (z - 1)}$ around $z_{0} = 0$ :

f (z) = \frac{1}{z} \cdot \frac{1}{z - 1} .

Using the geometric series expansion $\frac{1}{1 - w} = \sum_{n = 0}^{\infty} w^{n}$ (valid for $| w | < 1$ ), rewrite $\frac{1}{z - 1}$ :

\frac{1}{z - 1} = - \frac{1}{1 - z} = - \sum_{n = 0}^{\infty} z^{n} .

Thus:

f (z) = \frac{1}{z} (- \sum_{n = 0}^{\infty} z^{n}) = - \sum_{n = 0}^{\infty} \frac{1}{z^{n - 1}} .

The Laurent series is:

f (z) = - \frac{1}{z} - 1 - z - z^{2} - z^{3} - \dots .

This series converges for $| z | > 1$ .

Complex Method for Solving Real Integrals

General Procedure

Transform the Real Integral into a Complex Integral:
Replace the real variable $x$ with the complex variable $z$ , and rewrite the real integral:

I = \int_{- \infty}^{\infty} f (x) d x = \int_{γ} f (z) d z,

where $γ$ is a contour in the complex plane.

Close the Contour:
Add a semicircular arc in the complex plane to make the integral over a closed contour. Ensure that the contribution from the arc goes to $0$ as the radius tends to infinity:

\int_{γ} f (z) d z = \int_{- \infty}^{\infty} f (x) d x + \int_{arc} f (z) d z .

Apply the Residue Theorem:
Compute the integral over the closed contour $γ$ using the Residue Theorem:

\int_{γ} f (z) d z = 2 π i \sum Res (f (z), z_{k}),

where $z_{k}$ are the singularities of $f (z)$ inside the contour.

Extract the Real Integral:
The desired integral $I$ is related to the contour integral minus any contributions from the semicircular arc or other portions.

Example

Evaluate:

I = \int_{- \infty}^{\infty} \frac{1}{x^{2} + 1} d x .

Step 1: Rewrite as a Complex Integral:
Replace $x$ with $z$ :

I = \int_{- \infty}^{\infty} \frac{1}{z^{2} + 1} d z .

Step 2: Close the Contour:
Use a semicircular contour in the upper half-plane with radius $R$ . The integral becomes:

\int_{γ} \frac{1}{z^{2} + 1} d z = \int_{- \infty}^{\infty} \frac{1}{z^{2} + 1} d z + \int_{arc} \frac{1}{z^{2} + 1} d z .

As $R \to \infty$ , $\frac{1}{z^{2} + 1} \to 0$ on the arc, so its contribution vanishes.

Step 3: Apply the Residue Theorem:
The singularities of $f (z) = \frac{1}{z^{2} + 1}$ are at $z = i$ and $z = - i$ . For the upper half-plane, only $z = i$ is inside the contour. The residue at $z = i$ is:

Res (\frac{1}{z^{2} + 1}, i) = lim_{z \to i} (z - i) \frac{1}{z^{2} + 1} = \frac{1}{2 i} .

Thus:

\int_{γ} \frac{1}{z^{2} + 1} d z = 2 π i \cdot \frac{1}{2 i} = π .

Step 4: Extract the Real Integral:
Since the arc contribution vanishes:

I = \int_{- \infty}^{\infty} \frac{1}{x^{2} + 1} d x = π .

Jordan’s Lemma

Jordan’s Lemma is a result in complex analysis used to evaluate integrals of the form:

I = \int_{- \infty}^{\infty} e^{i m z} f (z) d z,

where:

$m > 0$ (important for convergence),
$f (z)$ is analytic in the upper half-plane,
$| f (z) | \to 0$ as $| z | \to \infty$ in the upper half-plane.

The key idea is that when closing the contour in the upper half-plane, the contribution of the integral over the semicircular arc vanishes as the radius $R \to \infty$ , provided $f (z)$ decays sufficiently fast.

Application

Jordan’s Lemma is especially useful when evaluating real integrals using contour integration, where the integrand includes oscillatory terms $e^{i m z}$ (for large $m$ ).

Steps to Apply Jordan’s Lemma

Write the real integral in terms of z :

I = \int_{- \infty}^{\infty} e^{i m x} f (x) d x .

Extend the integral to the complex plane, turning it into a contour integral:

\int_{γ} e^{i m z} f (z) d z,

where $γ$ includes the real line and a semicircular arc in the upper half-plane.
3. Show that the contribution from the semicircular arc vanishes as $R \to \infty$ , using Jordan’s Lemma.
4. Apply the Residue Theorem to compute the integral over the closed contour $γ$ .

Example

Evaluate:

I = \int_{- \infty}^{\infty} \frac{e^{i m x}}{x^{2} + 1} d x, m > 0.

Step 1: Rewrite as a Complex Integral:
Replace $x$ with $z$ :

I = \int_{- \infty}^{\infty} \frac{e^{i m z}}{z^{2} + 1} d z .

Step 2: Close the Contour:
Use a semicircular contour in the upper half-plane. The integral becomes:

\int_{γ} \frac{e^{i m z}}{z^{2} + 1} d z = \int_{- \infty}^{\infty} \frac{e^{i m z}}{z^{2} + 1} d z + \int_{arc} \frac{e^{i m z}}{z^{2} + 1} d z .

Step 3: Apply Jordan’s Lemma:
On the semicircular arc $| z | = R$ , $z = R e^{i θ}$ , and:

e^{i m z} = e^{i m R \cos θ} e^{- m R \sin θ} .

The term $e^{- m R \sin θ} \to 0$ as $R \to \infty$ , because $\sin θ > 0$ in the upper half-plane. Thus:

\int_{arc} \frac{e^{i m z}}{z^{2} + 1} d z \to 0 as R \to \infty .

Step 4: Apply the Residue Theorem:
The singularities of $f (z) = \frac{e^{i m z}}{z^{2} + 1}$ are at $z = i$ and $z = - i$ . In the upper half-plane, only $z = i$ is inside the contour. The residue at $z = i$ is:

Res (\frac{e^{i m z}}{z^{2} + 1}, i) = lim_{z \to i} (z - i) \frac{e^{i m z}}{z^{2} + 1} = \frac{e^{- m}}{2 i} .

Using the Residue Theorem:

\int_{γ} \frac{e^{i m z}}{z^{2} + 1} d z = 2 π i \cdot \frac{e^{- m}}{2 i} = π e^{- m} .

Step 5: Extract the Real Integral:
Since the arc contribution vanishes:

I = \int_{- \infty}^{\infty} \frac{e^{i m x}}{x^{2} + 1} d x = π e^{- m} .

Improper Integrals and Principal Value

Improper Integrals

An improper integral is an integral where:

The limits of integration are infinite, e.g.:

\int_{- \infty}^{\infty} f (x) d x .

The integrand has a singularity within the integration range, e.g.:

\int_{- 1}^{1} \frac{1}{x} d x .

Such integrals may not converge in the usual sense. However, by using Principal Value (P.V.) techniques and complex analysis tools like the Residue Theorem, we can define and compute them.

Principal Value (P.V.)

The Principal Value of an integral is a method to assign a finite value to certain divergent improper integrals.

For an integral with a singularity at $x = c$ (e.g., $\frac{1}{x}$ ):

P.V. \int_{a}^{b} f (x) d x = lim_{ϵ \to 0} (\int_{a}^{c - ϵ} f (x) d x + \int_{c + ϵ}^{b} f (x) d x) .

For an integral over infinite limits:

P.V. \int_{- \infty}^{\infty} f (x) d x = lim_{R \to \infty} \int_{- R}^{R} f (x) d x .

Residue Theorem for Improper Integrals

Improper integrals can often be evaluated using the Residue Theorem, especially for integrals of the form:

\int_{- \infty}^{\infty} f (x) d x .

The idea is to extend the real integral to a closed contour in the complex plane and use the residues of $f (z)$ inside the contour:

\int_{- \infty}^{\infty} f (x) d x = 2 π i \sum Res (f (z), inside the contour) .

Example 1

Evaluate:

P.V. \int_{- 1}^{1} \frac{1}{x} d x .

This integral diverges because $\frac{1}{x}$ has a singularity at $x = 0$ .
The Principal Value is defined as:

P.V. \int_{- 1}^{1} \frac{1}{x} d x = lim_{ϵ \to 0} (\int_{- 1}^{- ϵ} \frac{1}{x} d x + \int_{ϵ}^{1} \frac{1}{x} d x) .

Compute each integral:

\int_{- 1}^{- ϵ} \frac{1}{x} d x = \ln | ϵ | - \ln | 1 | = \ln | ϵ |,

\int_{ϵ}^{1} \frac{1}{x} d x = \ln | 1 | - \ln | ϵ | = - \ln | ϵ | .

Add them:

P.V. \int_{- 1}^{1} \frac{1}{x} d x = \ln | ϵ | - \ln | ϵ | = 0.

Thus:

P.V. \int_{- 1}^{1} \frac{1}{x} d x = 0.

Example 2

Evaluate:

I = \int_{- \infty}^{\infty} \frac{1}{x^{2} + 1} d x .

We solved this earlier using the residue theorem:

Extend the integral to the complex plane:

\int_{- \infty}^{\infty} \frac{1}{x^{2} + 1} d x = \int_{γ} \frac{1}{z^{2} + 1} d z,

where $γ$ is a closed contour in the upper half-plane.

Find the singularities of $f (z) = \frac{1}{z^{2} + 1}$ :

z = i, z = - i .

Apply the Residue Theorem:
The residue at $z = i$ is:

Res (\frac{1}{z^{2} + 1}, i) = \frac{1}{2 i} .

Compute the integral:

\int_{γ} \frac{1}{z^{2} + 1} d z = 2 π i \cdot \frac{1}{2 i} = π .

Thus:

\int_{- \infty}^{\infty} \frac{1}{x^{2} + 1} d x = π .

Fourier Transform and Inverse Transform

The Fourier transform is a powerful mathematical tool used to analyze and represent a time-domain signal $f (t)$ in the frequency domain $F (ω)$ . It is defined as:

F (ω) = \int_{- \infty}^{\infty} e^{- i ω t} f (t) d t,

where:

$F (ω)$ is the frequency-domain representation of $f (t)$ ,
$ω$ is the angular frequency.

Conditions for Existence

For the Fourier transform to exist, the function $f (t)$ must satisfy the absolute integrability condition:

\int_{- \infty}^{\infty} | f (t) | d t < \infty .

This ensures that $f (t)$ has finite energy and can be transformed into the frequency domain.

Inverse Fourier Transform

The original signal $f (t)$ can be recovered from its Fourier transform $F (ω)$ using the inverse Fourier transform:

f (t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{i ω t} F (ω) d ω .

Properties of the Fourier Transform

Linearity:

F {a f_{1} (t) + b f_{2} (t)} = a F_{1} (ω) + b F_{2} (ω),

where $F$ represents the Fourier transform operator.

Time Shift:
If $g (t) = f (t - t_{0})$ , then:

F {g (t)} = e^{- i ω t_{0}} F (ω) .

Frequency Shift:
If $g (t) = e^{i ω_{0} t} f (t)$ , then:

F {g (t)} = F (ω - ω_{0}) .

Parseval’s Theorem:
The energy of the signal in the time domain is equal to the energy in the frequency domain:

\int_{- \infty}^{\infty} | f (t) |^{2} d t = \frac{1}{2 π} \int_{- \infty}^{\infty} | F (ω) |^{2} d ω .

Example 1: Fourier Transform of a Gaussian

Let $f (t) = e^{- t^{2}}$ . The Fourier transform is:

F (ω) = \int_{- \infty}^{\infty} e^{- i ω t} e^{- t^{2}} d t .

Using a standard result for Gaussian integrals:

F (ω) = \sqrt{π} e^{- ω^{2} / 4} .

Example 2: Fourier Transform of a Rectangular Pulse

Let:

f (t) = {\begin{cases} 1, & | t | \leq T, \\ 0, & | t | > T . \end{cases}

The Fourier transform is:

F (ω) = \int_{- T}^{T} e^{- i ω t} d t .

Solve the integral:

F (ω) = \frac{\sin (ω T)}{ω} .

This result demonstrates how a rectangular time-domain signal transforms into a sinc function in the frequency domain.

One-Sided Fourier Transform

The one-sided Fourier transform is a variation of the Fourier transform where the integration is limited to $t \geq 0$ . It is used when the signal $f (t)$ is causal (i.e., $f (t) = 0$ for $t < 0$ ).

The one-sided Fourier transform is defined as:

F (ω) = \int_{0}^{\infty} e^{- i ω t} f (t) d t .

Use Case: The one-sided Fourier transform is primarily used to deal with time-limited or causal signals, often found in engineering and physics problems.

Laplace Transform

The Laplace transform generalizes the Fourier transform by replacing $e^{- i ω t}$ with $e^{- s t}$ , where s is a complex variable $s = σ + i ω$ .

The Laplace transform of $f (t)$ is defined as:

F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t, where s = σ + i ω .

Key Difference:

The Fourier transform uses $s = i ω$ (purely imaginary).
The Laplace transform allows $s$ to have a real part $σ$ , which accounts for exponential growth or decay.

Convergence

The Laplace transform converges for functions $f (t)$ satisfying:

\int_{0}^{\infty} | f (t) | e^{- σ t} d t < \infty,

which allows it to handle a broader class of functions than the Fourier transform.

Connection Between Fourier and Laplace Transforms

If $F (s)$ is the Laplace transform of $f (t)$ , then the Fourier transform can be derived by setting $s = i ω$ :

F (ω) = F (s) | s = i ω = \int 0^{\infty} e^{- i ω t} f (t) d t .

Example

Compute the Laplace transform of $f (t) = e^{- a t}$ , $a > 0$ .

Using the definition:

F (s) = \int_{0}^{\infty} e^{- s t} e^{- a t} d t = \int_{0}^{\infty} e^{- (s + a) t} d t .

Solve the integral:

F (s) = {[\frac{e^{- (s + a) t}}{- (s + a)}]}_{0}^{\infty} = \frac{1}{s + a}, for Re (s) > - a .

Example: Fourier Transform from Laplace Transform

Given the Laplace transform:

F (s) = \frac{1}{s + a} .

The Fourier transform is obtained by substituting $s = i ω$ :

F (ω) = \frac{1}{i ω + a} .