# Is the Fourier transform of $\mathcal{F}[g(ct)] = \int_{-\infty}^{\infty} g(ct) \ e^{-i\omega t} dt$ wrong?

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.
• Hi thanks for the reply. but why does the Fourier transform have no way of knowing the c is there? In your 2nd line, couldn’t I just take the derivative $d\tau = c dt$? And use $dt$ instead of $d(\tau/c)$? – Akira D.Soul Jun 25 at 8:29
• @AkiraD.Soul You want the formula to be a Fourier Transform itself, vis $$\int_\Bbb R g_{\small 2}(\tau) \exp({i \omega_{\small 2} \tau})\mathrm d \tau$$ where $\omega_{\small 2}=\omega/c$ and $g_{\small 2}(\tau)=\tfrac 1cg(\tau)$ – Graham Kemp Jun 25 at 9:00
• @GrahamKemp Thanks for the reply. I can understand we want to stick to the original fourier transform formula. I just don't really understand the meaning of "the fourier transform has no way of knowing the c is there" In the line:$$\mathcal{F}[g_c](\omega):=\int_\mathbb{R} g_c(t)e^{i\omega t}\,\mathrm{d}t =\int_\mathbb{R} g(ct)e^{i\omega t}\,\mathrm{d}t$$ why doesn't the last term changes to: $$\int_\mathbb{R} g_c(t)e^{i\omega t}\,\mathrm{d}t =\int_\mathbb{R} g(ct)e^{i\omega (ct)}\,d(ct)$$ – Akira D.Soul Jun 25 at 9:14
• Because it is $g_c$, not $g$. – user10354138 Jun 25 at 9:17
• You could do it: \begin{align} \mathcal F_t[g(ct)](\omega) &= \int_\Bbb R g(ct)\exp(i\omega t)~\mathrm d t\\ &= \int_\Bbb R g(ct)\exp(i(\omega/c)(ct))~\tfrac 1c\mathrm d (ct)\\&=\tfrac 1c\mathcal F_{c t}[g(ct)](\omega/c)\\&=\tfrac 1c\mathcal F_\tau[g(\tau)](\omega/c) \end{align} By "not knowing the $c$ is there", user10354138 means the transformation of $g(ct)$ is implicitly of argument $t$, not $ct$, as I indicate by using a subscript. – Graham Kemp Jun 25 at 9:29