# Integrals of Complex Functions

• xspook
In summary, the problem involves finding the integral from negative infinity to positive infinity of the given functions using the Cauchy Integral Formula and Theorem. The general approach is to integrate about the boundary of a disk, where the values contained in the disk are equal to the values on the boundary. The Cauchy Integral Theorem states that for any closed path, no work is done, making it a useful tool in solving these types of problems. Factoring the denominators into a form of 1/(z-a) is necessary in solving these integrals.
xspook

## Homework Statement

What is the integral from negative infinity to positive infinity of the following functions?

a) f(z) = $\frac{e^{-i5z}}{z^{2}+1}$

b) f(z) = $\frac{e^{-i5z}}{z^{2}-1}$

c) f(z) = $\frac{1}{π}$$\frac{a}{z^{2}+a^{2}}$

d) f(z) = $e^{\frac{-(z-ia)^{2}}{2}}$

e) f(z) = $\frac{sinz}{z}$

## Homework Equations

All my professor showed was an example, which would be (c) above

$\frac{1}{π}$$\int\frac{a}{z^{2}+a^{2}}$dz
$\frac{1}{π}$$\int\frac{1}{2ai}$($\frac{1}{z-ia}$-$\frac{1}{z+ia}$)dz
$\frac{1}{2iπ}$$\int\frac{1}{z-ia}$-$\frac{1}{z+ia}$ dz=1
dz$\Rightarrow$$Re^{iθ}idθ$

limits of integration are now from 0 to π
$\int\frac{1}{R^{2}e^{2iθ}+a^{2}}$$Re^{iθ}iθdθ$
$R^{2}$$\rightarrow$∞
$\oint$f(z)dz=0
f(z) is analytic
f(z) = $\frac{1}{2πi}$

and this is where the class time expired and he said "you can figure out the rest"

## The Attempt at a Solution

I have no clue where to start with these. The class is Mathematical Methods of Physics and this year there weren't enough students in the class so they offered it as an independent study. The professor that "observes" our work didn't tell us how to do these or where to start. I'm just looking for guidance on what to do, that an idiot can understand. Often I read threads on here and the way some of you convey your knowledge is out of my league. I would appreciate any help! Thanks

So the general approach is to use the Cauchy Integral Formula and Cauchy Integral Theorem. The theorem has a wiki page, as does the formula: http://en.wikipedia.org/wiki/Cauchy's_integral_formula.

You are basically integrating about the boundary of a disk, where the values contained in the disk are equal to the values on the boundary. This should seem familiar, as it is the basis for Stokes' Theorem, and consequently Green's Theorem.

If you look at the Cauchy Integral Theorem and its proof, you can see ∫f(z)dz = 0. This is to say that for any closed path, no work is done.

My suggestion would be to compute ∫(1/z) dz on the unit circle, draw a graph of this (it will be a circle of radius 1 with a hole at R = 0), and examine its physical significance. From here, compare with the Cauchy Integral Formula and Theorem, and understand why this works.

Your reasoning behind factoring these denominators into a form of 1/(z-a) is necessary because it is just a translation of the solution to ∫(1/z) dz.

## 1. What is an integral of a complex function?

An integral of a complex function is a mathematical operation that calculates the area between the curve of the function and the horizontal axis in the complex plane. It is a generalization of the concept of integration for real-valued functions.

## 2. How is the integral of a complex function different from the integral of a real-valued function?

The integral of a complex function involves integrating over a complex plane, which has two dimensions (real and imaginary) as opposed to just one dimension (real) in the case of real-valued functions. This means that the integral of a complex function takes into account both the real and imaginary components of the function.

## 3. What are some common techniques used to solve integrals of complex functions?

Some common techniques used to solve integrals of complex functions include the Cauchy-Riemann equations, contour integration, and the residue theorem. These techniques involve manipulating the complex function into a more manageable form and using properties of complex numbers and functions to evaluate the integral.

## 4. What is the significance of integrals of complex functions in mathematics and science?

Integrals of complex functions have many applications in mathematics and science, including in the fields of physics, engineering, and finance. They are used to solve problems involving complex systems, such as electrical circuits, fluid dynamics, and quantum mechanics. They also have important theoretical implications in the study of complex analysis and the behavior of functions in the complex plane.

## 5. Are there any limitations or challenges to solving integrals of complex functions?

One of the main challenges in solving integrals of complex functions is that the techniques used for real-valued integrals may not be applicable or may require additional modifications. Additionally, complex functions can have infinitely many paths of integration, so choosing the correct path can be a difficult task. Moreover, some complex functions may not have a closed-form solution, making it necessary to use numerical methods to approximate the integral.

Replies
14
Views
2K
Replies
13
Views
1K
Replies
8
Views
715
Replies
19
Views
1K
Replies
17
Views
2K
Replies
1
Views
939
Replies
3
Views
937
Replies
1
Views
1K
Replies
2
Views
1K
Replies
6
Views
1K