$2\pi$ in the Definition of Fourier Transform

Question

Most textbooks I read define Fourier transform of a function $f \in L^2(\mathbb R)$ as

$$ \hat f (\xi) := \int_\mathbb R f(x) e^{-2\pi i x \xi} dx. $$

However, in class my teacher defines it without the $2\pi$. My question is whether this $2\pi$ is necessary and what role does it play here? Thank you!

Answer 1

This is a factor for norming. See the wikipedia article on Fourier Transform for a list of common constants. The one I see more often uses $$\hat f(\xi) := (2\pi)^{-\frac n2} \int_{\mathbb R^n} f(x) e^{-ix\cdot \xi} \mathrm dx \\ \check f(x) = (2\pi)^{-\frac n2} \int_{\mathbb R^n} f(\xi) e^{ix\cdot \xi} \mathrm d\xi$$ Where $\cdot$ denotes the inner product. Both of these are isometric isomorphisms on $L^2(\mathbb R^n)$.
Possible advantages of the two choices are only in constants in the convolution theorem (my version is $\widehat{f\ast g}(\xi) = (2\pi)^n \hat f(\xi) \hat g(\xi)$) and easier (constant-less) partial integration and differentiation for my version: $$\frac{\mathrm d}{\mathrm d\xi} \hat f(\xi) = i\xi \widehat{f'}(\xi)$$

Answer 2

There are many possible conventions, but you can't get around the fact that a $2\pi$ will appear somewhere. In order to have a way of comparing different conventions easily, I wrote up restrictions on the various coefficients in the definitions and comparison between different choices.