# What does it mean to Fourier transform a function, i.e. hat(f)(omega) = mathcalF_(omega)[f(x)]?

Jul 24, 2015

The Fourier transform is a way of converting one function to another function (details to follow). It is useful for solving differential equations. I'm not an expert in this, but I believe it's also useful for signal processing, which is at the heart of communication technology (see Wikipedia's page )

#### Explanation:

Given a "sufficiently nice" function $f \left(x\right)$ (details on what "sufficiently nice" means will not be delved into), defined on the entire real line $\mathbb{R}$, the Fourier transform of $f \left(x\right)$, denoted $\hat{f} \left(\omega\right)$, is defined by the formula:

$\hat{f} \left(\omega\right) = \setminus {\int}_{- \setminus \infty}^{\setminus \infty} f \left(x\right) {e}^{i \setminus \omega x} \setminus \mathrm{dx}$, where $i$ is the imaginary unit (${i}^{2} = - 1$).

Note 1: Sometimes people put an extra $- 2 \pi$ in the power of $e$. Alternatively, sometimes people put an extra $\frac{1}{\sqrt{2 \pi}}$ in front of the integral.

Note 2: With respect to the given improper integral, the $\omega$ is a constant. However, since we can (hopefully) do this for many values of $\omega$, we ultimately think of $\omega$ as the variable for the "new function" $\hat{f}$. Typically $\omega$ is a real variable, though $\hat{f} \left(\omega\right)$ is a complex number (in general).

Note 3: Technically speaking, then, the Fourier transform is the (more general kind of) function that maps $f$ to $\hat{f}$. Sometimes this is written as $m a t h c a l \left\{F\right\} \left(f\right) = \hat{f}$ (in other words, $m a t h c a l \left\{F\right\}$ is the symbol for this function that maps $f$ to $\hat{f}$).

As an example, let $f \left(x\right) = {e}^{- | x |}$. Then, not worrying about convergence issues, and using Euler's formula ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$, we get:

$\setminus \hat{f} \left(\omega\right) = \setminus {\int}_{- \setminus \infty}^{0} {e}^{x} {e}^{i \omega x} \setminus \mathrm{dx} + \setminus {\int}_{0}^{\setminus \infty} {e}^{- x} {e}^{i \omega x} \setminus \mathrm{dx}$

$= \setminus {\int}_{- \setminus \infty}^{0} \left({e}^{x} \cos \left(\omega x\right) + i {e}^{x} \sin \left(\omega x\right)\right) \setminus \mathrm{dx} + \setminus {\int}_{0}^{\setminus \infty} \left({e}^{- x} \cos \left(\omega x\right) + i {e}^{- x} \sin \left(\omega x\right)\right) \setminus \mathrm{dx}$.

Now (via Wolfram Alpha or tables of integrals),

$\setminus \int \left({e}^{x} \cos \left(\omega x\right) + i {e}^{x} \sin \left(\omega x\right)\right) \setminus \mathrm{dx} = \frac{i}{i - \omega} {e}^{x} \cos \left(\omega x\right) - \frac{1}{i - \omega} {e}^{x} \sin \left(\omega x\right) + C$

and

$\setminus \int \left({e}^{- x} \cos \left(\omega x\right) + i {e}^{- x} \sin \left(\omega x\right)\right) \setminus \mathrm{dx} = - \frac{1}{1 - i \omega} {e}^{- x} \cos \left(\omega x\right) - \frac{i}{1 - i \omega} {e}^{- x} \sin \left(\omega x\right) + C$

Since $\cos \left(0\right) = 1$ and $\sin \left(0\right) = 0$ and ${e}^{x} \to 0$ as $x \to - \setminus \infty$, this leads to

$\hat{f} \left(\omega\right) = \frac{i}{i - \omega} + \frac{1}{1 - i \omega}$

$= \frac{i \left(- i - \omega\right)}{\left(i - \omega\right) \left(- i - \omega\right)} + \frac{1 + i \omega}{\left(1 - i \omega\right) \left(1 + i \omega\right)}$

$= \frac{1 - i \omega}{1 + {\omega}^{2}} + \frac{1 + i \omega}{1 + {\omega}^{2}} = \frac{2}{1 + {\omega}^{2}}$