- #1
Jhenrique
- 676
- 4
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.
[tex]F(\omega)=\int_{-\infty }^{+\infty }\frac{e^{-i\omega t}}{\sqrt{2\pi}} f(t)dt[/tex]
[tex]f(t)=\int_{-\infty }^{+\infty }\frac{e^{+i\omega t}}{\sqrt{2\pi}} F(\omega)d\omega[/tex]
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?
[tex]F(\omega)=\int_{-\infty }^{+\infty }\frac{e^{-i\omega t}}{\sqrt{2\pi}} f(t)dt[/tex]
[tex]f(t)=\int_{-\infty }^{+\infty }\frac{e^{+i\omega t}}{\sqrt{2\pi}} F(\omega)d\omega[/tex]
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?