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I've been trying to understand the time shift proof with fourier transforms, however I'm confused as to how some variables can change but not others. My intuitive understanding of the time shift is as follows:

We define $f(y) = f(x-c)$

$$ \mathcal{F}[f(x-c)] (s) = \int^{\infty}_{-\infty} f(y) e^{-isy} dy $$ $$ = \int^{\infty}_{-\infty} f(x-c) e^{-isx} dx $$ The confusion comes when it's chosen that $x = x+c$, and $f(x-c) = f(x)$. I dont't understand why this can be done. Has this something to do with the fact that the function is integrated across the whole real line? Since $y = x-c$, why wouldn't $e^{-isy} = e^{-is(x-c)}$?

I looked at this similar question, however it provided little clarification on why we $e^{-isy}$ to $e^{-is(x+c)}$.

User2001
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    It's not clear to me what the issue is. You have $$\mathcal{F}[f(x-c)] = \int_{\mathbb{R}} f(x-c) e^{-isx} dx \underbrace{=}{\color{red}{x \to x + c}} \int{\mathbb{R}} f(x) e^{-is(x+c)} dx = e^{-isc} \int_{\mathbb{R}} f(x) e^{-isx} dx = e^{-isc} \mathcal{F}[f(x)]$$ – Matthew Cassell Dec 12 '22 at 17:39
  • @MatthewCassell Oh, wow. I've just realised how stupid that was. I misunderstood the answer, didn't realise that's what was happening. Thanks for the help! – User2001 Dec 12 '22 at 17:41
  • @MatthewCassell Actuallly, looking at this a bit longer, I'm still confused. Does the definition of the fourier tranform state that the function will be followed by $e^{-isx}$ always? Like for example, the fourier transform of $f(x)$ is $\int^{\infty}{-\infty} f(x) e^{-isx} dx$, why wouldn't the fourier transform of $f(x-c)$ not be $\int^{\infty}{-\infty} f(x-c) e^{-is(x - c)} dx$? – User2001 Dec 12 '22 at 17:49
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    $\int_{-\infty}^{\infty}f(x-c)e^{-is(x-c)}dx=e^{isc}\int_{-\infty}^{\infty}f(x-c)e^{-isx}dx$. Because the integral range is from $-\infty$ to $\infty$, it's same. – Hypernova Dec 12 '22 at 18:46
  • @Hypernova Just to make sure I understand correctly, is this because $\lim_{x\to \pm \infty} x - c = x$? – User2001 Dec 12 '22 at 19:24
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    @user2001 yes that is correct. – VeDAN Dec 12 '22 at 20:10

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