Different forms of Fourier Transform

Question

Why is that some text books use $\left(\frac{1}{\sqrt{2\pi}}\right)$? And some use $\left(\frac{1}{2\pi}\right)$?

Answer 1

If you are a physicist, you would (generally) like to use $$ \hat f (\xi)=\sqrt{\left(\frac{\hbar}{2\pi}\right)^d} \int_{\mathbb{R}^d} f(x)e^{-i \hbar \, x \cdot \xi}\,dx $$ the advantage is that it takes into account planck's constant, i.e it is unitary, $\lVert f\rVert_2 = \lVert \hat f\rVert_2^2$, and shares the same constant as its inverse, but $\widehat{f * g} = \sqrt{\frac{(2\pi)^d}{\hbar^d}}\hat{f}\hat{g} \not=\hat{f}\hat{g}$. (Generally) Mathmaticians use $$ \hat f (\xi)=\int_{\mathbb{R}^d} f(x)e^{-2\pi i\, x \cdot \xi}\,dx,\qquad \mathcal{F}^{-1}f(x) =\int_{\mathbb{R}^d} f(x)e^{2\pi i\, x \cdot \xi}\,d\xi $$ this is a $*$-algebra morphism from $L^1(G=\mathbb{R}^d)$ to $C_0(G)$, $\widehat{f * g} =\hat{f}\hat{g}$. and it is unitary at the level of $L^2(G)$ it has all the right properties and one does not have to bother with any constant.

On the other hand some people, for reasons unclear, prefer to use $$ \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} f(x)e^{-i \,x \cdot \xi}\,dx $$ this one doesnt have any good property and its inverse doesnt even have the same constant, i.e. $$ \mathcal{F}^{-1}f(x) = \int_{\mathbb{R}^d} f(x)e^{i\,x \cdot \xi}\,d\xi. $$

But as you might notice, all are basically the same transform minus some change of variable of the form $x \mapsto cx$, they all share the same properties but you have to keep track of the constants.