The Fourier transform is defined as $$ \hat{f}(\xi) = \int^{\infty}_{-\infty} f(x) e^{-2\pi i x \xi} dx $$
And the inverse Fourier transform as:
$$ f(x) = \int^{\infty}_{-\infty} \hat{f}(\xi) e^{+2\pi i x \xi} d\xi $$
On Wikipedia
they make the following substitution $\omega = 2\pi\xi$ to get the following form of the equations from the first 2 equations I have written above:
Fourier transform: $$ \hat{x}_1(\omega) = \hat{x}(\frac{\omega}{2\pi}) = \int^{\infty}_{-\infty} x(t) e^{-i t \omega} dx $$ Inverse Fourier transform $$ x(t) = \dfrac{1}{2\pi} \int^{\infty}_{-\infty} \hat{x}_1(\omega) e^{+ i t \omega} d\omega = \dfrac{1}{2\pi} \int^{\infty}_{-\infty} \hat{x}(\frac{\omega}{2\pi}) e^{+ i t \omega} d\omega $$
Surely one must take note of the $\hat{x}(\frac{\omega}{2\pi})$ in the inverse transform, but everywhere I look I just see $\hat{x}(\omega)$ used in the inverse transform.
Likewise for the alternate form of the Fourier and inverse Fourier transforms with the $\frac{1}{\sqrt{2\pi}}$ factor in front, they are always written in terms of $x(t)$ and $\hat{x}(\omega)$ not $x(\frac{t}{\sqrt{2\pi}})$ and $\hat{x}(\frac{\omega}{\sqrt{2\pi}})$.
Why is this? Why can these factors be seemingly ignored?
The reason is because the bounds of the integral are $-\infty$ and $\infty$.
So we have the following Inverse Fourier transform $$ x(t) = \dfrac{1}{2\pi} \int^{\infty}_{-\infty} \hat{x}_1(\omega) e^{+ i t \omega} d\omega = \dfrac{1}{2\pi} \int^{\infty}_{-\infty} \hat{x}(\frac{\omega}{2\pi}) e^{+ i t \omega} d\omega $$
but since $\int^{\infty}_{-\infty}f(\frac{y}{2\pi})dy = \int^{\infty}_{-\infty}f(y)dy$ for any variable $y$ we can simply exchange the $\hat{x}(\frac{\omega}{2\pi})$ inside the integral for $\hat{x}(\omega)$ like so:
$$ x(t) = \dfrac{1}{2\pi} \int^{\infty}_{-\infty} \hat{x}(\frac{\omega}{2\pi}) e^{+ i t \omega} d\omega = \dfrac{1}{2\pi} \int^{\infty}_{-\infty} \hat{x}(\omega) e^{+ i t \omega} d\omega $$