I was reading proofs on the scaling property of fourier transform:
http://www.thefouriertransform.com/transform/properties.php
http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html
I notice this line:
$$\mathcal{F}[g(ct)] = \int_{-\infty}^{\infty} g(ct) \ e^{-i\omega t} dt \tag{1}$$
I have a few issues with this line:
$1)$ Is there a missing factor of $\frac{1}{\sqrt{2\pi}}$ on the right hand side? Because shouldn't a Fourier transform be (derived from the inversion theorem):
$$\mathcal{F}[g(t)] = \frac{1}{\sqrt{2\pi}} \int_{- \infty}^{\infty} g(t) \ e^{-i \omega t} dt \tag{2}$$
$2)$ I believe that the correct expression should be:
$$\mathcal{F}[g(ct)] = c\int_{- \infty}^{\infty} g(ct) \ e^{-i \omega (ct)}dt$$
Proof of (2)
We let $x = ct$ and hence $dx = c dt$. Hence:
$$\mathcal{F}[g(ct)] = \mathcal{F}[g(x)] = \frac{1}{\sqrt{2\pi}} \int_{- \infty}^{\infty} g(x) \ e^{-i \omega x} dx $$
Now we can reverse the substitution and we obtain:
$$\frac{1}{\sqrt{2\pi}} \int_{- \infty}^{\infty} g(x) \ e^{-i \omega x} dx = c \ \int_{- \infty}^{\infty} g(ct) \ e^{-i \omega (ct)} dt \tag{3}$$
in which $(1) \neq (3)$. Is there something wrong with my proof?
(1) Engineers use the non-unitrary angular-frequency version of Fourier transform. It is not wrong, just a different convention (there are three nonequivalent conventions). Their Fourier inversion formula would have the $2\pi$ but not the Fourier transform formula.
(2) No. For $g_c\colon t\mapsto g(ct)$, the Fourier transform itself has no way of knowing the $c$ is there (assuming you are using the nonunitary version): $$ \mathcal{F}[g_c](\omega):=\int_\mathbb{R} g_c(t)e^{i\omega t}\,\mathrm{d}t =\int_\mathbb{R} g(ct)e^{i\omega t}\,\mathrm{d}t $$ So the substitution $\tau=ct$ then gives $$ \mathcal{F}[g_c](\omega) =\int_\mathbb{R} g(\tau)e^{i\omega \tau/c}\,\mathrm{d}(\tau/c) $$ et cetera.