The Fourier transform of a function $f(t)$ is defined as: $$F(\omega)=\int_{-\infty}^{\infty} f(t)e^{-2\pi i\omega t} dt$$ Express the Fourier transform of $g(t)=f(at)$ in terms of $F,a$ and $\omega$.

Now, for $a>0$ and $a<0$, I know that the Fourier transform is $g(t)=f(at)$ $$G(\omega)=\frac{1}{|a|}F\left(\frac{\omega}{a}\right)$$

Since the question does not specify anything about $a$, I was wondering what the Fourier transform of $g(t)=f(at)$ would be if $a$ is complex, $a\in \mathbb{C}$, i.e. $$a=\text{Re}(a)+i\text{ Im}(a)$$

This is what I have tried so far, I'm not sure if this is the right approach $$G(\omega)=\int_{-\infty}^{\infty} f(at)e^{-2\pi i\omega t} dt$$ Now using the substitution $t'=at=(\text{Re}(a)+i\text{ Im}(a))t$, we have $$G(\omega)=\frac{1}{a} \int_{-\infty}^{\infty} f(t')e^{-2\pi i(\omega/a) t'} dt'$$ $$\frac{1}{a}=\frac{1}{\text{Re}(a)+i\text{ Im}(a)}=\frac{\text{Re}(a)-i\text{ Im}(a)}{\text{Re}(a)^2+\text{ Im}(a)^2}$$ Looking at it, I don't think that this will work.


First notice a small detail. When you do the transformation $ t = a t' $, it converts to $ dt= | a | d t' $, right? Infinitesimals only make sense as sort of real numbers, remember this.

Now, with that correction, write $x = \frac{\omega }{a } $ in the formula $$ G ( \omega ) = \frac{1}{|a|} \int_{-\infty }^{\infty } f(t' ) e^{-2 \pi i x t' } = \frac{1}{|a| } F(x) \, . $$

The substitution $x = \frac{\omega }{a } $ is only to make clear that your right hand side is in fact the Fourier transform of $f$. You should try and train yourself to make these natural observations even before writing up the substitution.

  • $\begingroup$ I just realised that you also have to transform the limits too i.e. $t=\infty$ transforms to $t'=\infty+i\infty$ and $t=-\infty$ transforms to $t'=-\infty-i\infty$. How would this change the result? $\endgroup$ – Hogg Oct 3 '16 at 15:10

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