# Trouble Understanding the Fourier Integral

• ColdFusion85
In summary, the Fourier Integral Formula involves integrating a function f(\xi) multiplied by cosine and sine functions. The example given in the book may not be helpful, but the integral can be simplified by considering f(x) as a constant within certain bounds. The notation f(\xi) refers to the value that the function f takes when x is substituted with \xi.
ColdFusion85
[SOLVED] Trouble Understanding the Fourier Integral

I am having a bit of trouble understanding how to use the Fourier Integral Formula.

The integral is $$\int_{0}^{\infty} [A_{\omega}cos(\omega x) + B_{\omega}sin(\omega x)]d\omega$$

Where,

$$A_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)cos(\omega\xi)d\xi$$
and
$$B_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)sin(\omega\xi)d\xi$$

The example in the book doesn't help because f(x) is 1 for -1<=x<=1, and 0 for |x|>1. This doesn't clear up my confusion as to what the integral is supposed to look like with $$f(\xi)$$ inserted. What is $$f(\xi)$$? Is it the function given, f(x)? Like if I had f(x) = sin(x) on [-2, 2], would A_w look like:

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega\xi)d\xi$$,

with sin(x) being treated as a constant that can be brought outside the integral, or am I supposed to make A_w be

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega x)dx$$?

This is where I am confused.

ColdFusion85 said:
I am having a bit of trouble understanding how to use the Fourier Integral Formula.

The integral is $$\int_{0}^{\infty} [A_{\omega}cos(\omega x) + B_{\omega}sin(\omega x)]d\omega$$

Where,

$$A_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)cos(\omega\xi)d\xi$$
and
$$B_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)sin(\omega\xi)d\xi$$

The example in the book doesn't help because f(x) is 1 for -1<=x<=1, and 0 for |x|>1. This doesn't clear up my confusion as to what the integral is supposed to look like with $$f(\xi)$$ inserted. What is $$f(\xi)$$? Is it the function given, f(x)? Like if I had f(x) = sin(x) on [-2, 2], would A_w look like:

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega\xi)d\xi$$,

with sin(x) being treated as a constant that can be brought outside the integral, or am I supposed to make A_w be

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega x)dx$$?

This is where I am confused.

Your book gives: f(x) = 1 for values between -1 and 1. Let's just say you wanted to integrate this function:

eg. $$\int_{-\infty}^{\infty}f(x)dx$$
This integral would reduce to: $$\int_{-1}^{1}dx$$ since it is zero outside of x=-1 and x=1 and is unity otherwise.

Note: You could also integrate this function as:
$$\int_{-\infty}^{\infty}f(\xi)d\xi$$
and you would get the same thing.

The $\xi$ is a dummy variable.

For example, imagine I wanted to define a quantity that was the area of an arbitrary rectangle. Let's call this number $$R_{AB}$$ where A and B are the lengths of the rectangle respectively.

Well I could simply define $$R_{AB} = R \times B$$ or I could also write it as:

$$R_{AB} = \int_0^A \int_0^B d\xi_1 d\xi_2$$ Does that make sense? Its kinda like a loop.

Last edited:

If you were to integrate:

$$\int_{-\infty}^{\infty} f(\xi)cos(\omega \xi) d\xi$$

Lets let f(x) = 1 for x= 0 to 1 and f(x)=0 otherwise. Thus, your integral would become:

$$A = \int_{a}^{b} (1)cos(\omega \xi) d\xi$$
$$A = \frac{\sin b \omega }{\omega }-\frac{\sin a \omega }{\omega}$$

ColdFusion85 said:
What is $$f(\xi)$$? Is it the function given, f(x)? Like if I had f(x) = sin(x) on [-2, 2], would A_w look like:

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega\xi)d\xi$$,

with sin(x) being treated as a constant that can be brought outside the integral, or am I supposed to make A_w be

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega x)dx$$?

This is where I am confused.

Apparently you are confused over basic definitions. The notation $f(\xi)$ refers to the value that the function f takes when you substitute $x = \xi$, just like f(0) means: the value of f in x = 0, and f(12.4872) means: the value of f in x = 12.8472.
So if you put that into the formula you gave, you get
$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(\xi)cos(\omega\xi)d\xi$$.

Yeah that's what I was having trouble understanding. Pretty simple actually, but it was late at night and my brain wasn't working. I don't know why I didn't see that in the first place. Thanks.

## 1. What is the Fourier integral and why is it important in science?

The Fourier integral is a mathematical tool used to decompose a complex function into simpler components. It is important in science because many natural phenomena and signals can be represented as a combination of simpler functions, making it easier to analyze and understand them.

## 2. How does the Fourier integral work?

The Fourier integral works by representing a function as a sum of sines and cosines of different frequencies. This allows us to break down complex signals into simpler components and analyze them separately.

## 3. What are some applications of the Fourier integral in science?

The Fourier integral has many applications in science, including signal processing, image and sound compression, and studying the behavior of waves and vibrations in various systems. It is also used in fields such as physics, engineering, and economics.

## 4. What are the limitations of the Fourier integral?

While the Fourier integral is a powerful tool, it has some limitations. It can only be applied to functions that are continuous and have a finite number of discontinuities. It also assumes that the function is periodic, which may not always be the case in real-world scenarios.

## 5. How can I improve my understanding of the Fourier integral?

To improve your understanding of the Fourier integral, it is important to have a solid understanding of basic calculus and trigonometry. Practice with different examples and applications, and seek out resources such as textbooks, online tutorials, and lectures. You can also consult with a math or physics tutor for personalized help and guidance.

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