$\hat{f}(x)=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f(x)e^{-i\omega x}dx}$ and $\hat{f}(x)=\int_{-\infty}^{\infty}{f(x)e^{-2\pi ixy}}dx$

Question

Fourier transform - difference between $$\hat{f}(x)=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f(x)e^{-i\omega x}dx}$$ and $$\hat{f}(x)=\int_{-\infty}^{\infty}{f(x)e^{-2\pi ix\omega}}dx$$

Can you explain this difference please? thanks!

Answer 1

In the first case $\hat{f}(\omega+2k\pi)=\hat{f}(\omega)$ for all $k\in\mathbb Z$; in the second case we arrive at $\hat{f}(\omega+k)=\hat{f}(\omega)$, instead. In other words, you are changing periodicity of the Fourier transform.

Answer 2

The Fourier transform \begin{equation} \mathscr{F}f(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t)e^{-i\omega t} dt \end{equation} was defined with the coefficient $\frac{1}{2\pi}$ in front of the integral. It came from the coefficients of the Fourier series, when Fourier integral was developed. In practice, however, the expressions in calculations have to be as simple as possible and there is depending on the application also other definitions available. The most common is \begin{eqnarray} \mathscr{F}f(\omega) = \int_{-\infty}^\infty f(x)e^{-i\omega x} dx \ , \end{eqnarray} that is applied in electrical engineering, for example. The definition \begin{eqnarray} \mathscr{F}f(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) e^{-i\omega t} dt \end{eqnarray} is applied in physics, where also unitarity of $\mathscr{F}$ is of interest. Mathematicians often use the definition \begin{eqnarray} \mathscr{F}f(\textrm{w}) = \int_{-\infty}^\infty f(x)e^{-2\pi i \textrm{w} x} dx \end{eqnarray} because it is unitary and doesn't have an additional coefficient. It is written for frequency $f$ instead of angular frequency $\omega$. This is also the difference between the definitions.

Answer 3

Both are used as the definition of the Fourier transform. I will explain why one has two definitions (and actually more than that). The Fourier transform is actually a map to functions on the dual group. But since $\mathbb{R}$ is self dual ($\mathbb{R} \cong \hat{\mathbb{R}}$), it maps into functions on $\mathbb{R}$. So when one writes

$$ \hat{f}(\omega)=\frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty}{f(x)e^{-i\omega x}dx} $$

with $\omega$ a real number, one has identified $\mathbb{R}$ with its dual through an isomorphism. Depending on the isomorphism, one gets a different "definition" of the Fourier transform. The factor $\frac{1}{\sqrt{2 \pi}}$ is there to make the Haar measure self-dual, that is the Haar measure on the dual group in the Fourier inversion theorem is equal to the Haar measure on the group itself.