Integral tranforms and symmetry

1. Jan 27, 2014

Jhenrique

The fourier transform, wrt to angular frequency, needs of a factor (1/2π) for get f(t) or F(ω), actually, this factor is broken in 2 factors (1/√2pi) and each kernel, direct and inverse, receives one factor for keep the symmetry in equation.
$$F(\omega)=\int_{-\infty }^{+\infty }\frac{e^{-i\omega t}}{\sqrt{2\pi}} f(t)dt$$
$$f(t)=\int_{-\infty }^{+\infty }\frac{e^{+i\omega t}}{\sqrt{2\pi}} F(\omega)d\omega$$

Why others transform, by definition, don't have its "conversion factor" "broken" in 2 and distributed in each kernel? Is wrong define the transforms in this way? Prejudice the calculus?

2. Jan 27, 2014

Simon Bridge

You can break the any transformation any way you like - it's just not always useful to do so.