Fourier Transform in a different form

Question

This is from Mathworld Wolfram and I am unfamiliar with this notation.

"In general, the Fourier transform pair may be defined using two arbitrary constants a and b as $$F(w)=\sqrt{\frac{|b|}{(2\pi)^{1-a}}}\int^∞_{-∞}f(t)e^{ibwt}dt$$.........(15) $$f(t)=\sqrt{\frac{|b|}{(2\pi)^{1+a}}}\int^∞_{-∞}F(w)e^{-ibwt}dw$$..............(16)"

I am unsure about this cause I didn't see such form of Fourier transform. Where do the b and a come from?

Edited: Is there a way to derive if from the conventional Fourier transform equation?

Answer 1

From a physics point of view, the $b$ factor is just a change of units for either $b$ or $\omega$, or both. If $b=1$ then we require that the inverse Fourier transform of the Fourier transform of $f(t)$ is equal to $f(t)$. In the $i\omega t$ notation, we must choose a $\frac{1}{2\pi}$ coefficient. We can put it in the direct transform $(a=-1)$, in the inverse transform $(a=1)$, or we can split it between the two equations, each one with a $\sqrt{\frac{1}{2\pi}}$ factor$(a=0)$. If we multiply the Fourier transform by any non-zero constant, we must multiply the inverse transform by one over that constant.

Answer 2

These constants allow to cope with different conventions.

Here is an adapted extract of Mathematica's Help for Fourier Transform function:

Some common choices for $a, b$ are:

• $a=0, \ b=-2 \pi$ (the most usual, in particular in signal and image processing),

• $a=1, \ b=-1$ (current in mathematics and systems engineering),

• $a=0, \ b=1$ (in modern physics),

• $a=-1, \ b=1$ (in classical physics).