Denote the Fourier transform of a given function $f$ by $$\widehat{f}(w)=\int\limits_{\mathbb{R}} f(t)e^{-2\pi i wt}dt.$$ Let $c\in\mathbb{R}$ and define $f_c$ as the time-shifted $f$, $$f_c(t)=f(t+c).$$ So its Fourier transform is $$\widehat{f_c}(w)=\int\limits_{\mathbb{R}} f(t+c)e^{-2\pi i wt}dt$$ $$=e^{-2\pi i w c}\widehat{f}(w).$$ Then, if $c$ is integer, $$\widehat{f_c}(w) = e^{(-2\pi i c )w }\widehat{f}(w) = (e^{-2\pi i c })^w =1^w\widehat{f}(w) = \widehat{f}(w),$$ contradicting the Fourier inversion theorem! What is wrong with my argument?
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$\int\limits_{\mathbb{R}} f(t+c)e^{-2\pi i wt}dt= \int\limits_{\mathbb{R}} f(u)e^{-2\pi i w(u-c)}du = e^{2\pi i w c}\widehat{f}(w)$. And it is not true that $e^{ab} = (e^a)^b$ for $a,b$ non positive. What is true is $e^r = e^s$ iff $r - s = 2i \pi k$. – reuns Mar 06 '19 at 23:17
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As reuns describes in the question comment, and it's also stated in the answer by user21820 in Why $e^{i(π/3)} \ne -1$?, the basic issue is that
It is not at all true that $(x^a)^b = x^{ab}$ for complex numbers $x,a,b$ in general, under any reasonable definition of complex exponentiation.
The answer provides a quite good explanation of why this is the case, including providing several counter-examples, and the specific general cases where you can state that $(x^a)^b = x^{ab}$ does hold. In your case, you are assuming this in
$e^{(-2\pi i c )w }\widehat{f}(w) = (e^{-2\pi i c })^w$
John Omielan
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