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From here; $\hat f=\mathcal{F}(f)$, bar = complex conjugate:

  1. Time-shift property: $x(t-b) \Leftrightarrow e^{-j\omega b}{\bf X} (\omega)$, so why is it $+$ (red)?
  2. What at all is happening? Looks like convolution theorem, taking $f$ and $\psi$'s time-domain multiplication into frequency domain, except don't we need a second integral for conv.?
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That's an application of Parseval's (Plancherel's) Theorem:

$$\int_{-\infty}^{\infty}f(t)g^*(t)dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)G^*(\omega)d\omega\tag{1}$$

where $F(\omega)$ and $G(\omega)$ are the Fourier transforms of $f(t)$ and $g(t)$, respectively. I've used $^*$ to denote complex conjugation.

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    $\begingroup$ Of course, your explanation is much more plausible than the one I posted. Though I came (almost) to the same conclusion, just in a more complicated manner. I will delete mine though as yours is more helpful. $\endgroup$ – Florian Oct 19 '20 at 9:18
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    $\begingroup$ Very fair, @Florian $\endgroup$ – Laurent Duval Oct 19 '20 at 10:20
  • $\begingroup$ @Florian: Your answer went in the right direction but using Parseval's theorem is probably more obvious. $\endgroup$ – Matt L. Oct 19 '20 at 10:34
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    $\begingroup$ @OverLordGoldDragon: You need the complex conjugate, hence the positive sign. $\endgroup$ – Matt L. Oct 19 '20 at 21:24
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    $\begingroup$ @OverLordGoldDragon: I've changed the link, because the wikipedia article on Parseval's theorem indeed doesn't mention Eq. (1) of my answer. In mathematics, that formula is more often called Plancherel theorem, whereas in signal processing we usually call it (generalized) Parseval's theorem. $\endgroup$ – Matt L. Oct 20 '20 at 8:59
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Matt isn't wrong, but there's a more satisfactory answer; it is convolution theorem:

$$ \begin{align} \int_{-\infty}^{\infty}f(t)\psi^*(t-b)dt &= \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega) \Psi^*(\omega) e^{j\omega b}d\omega \\ &= \frac{1}{2\pi} \mathcal{F}^{-1}( F(\omega) \Psi^*(\omega) ) \end{align} $$

and the $+b$ timeshift is per the complex conjugate. Note that without the $b$ in $\psi()$, Parseval / Plancherel's theorem is more directly applicable, so here they're equivalent, but not necessarily with same assumptions/interpretations. Convolution is 'correct' in context, and enables $\text{FFT}$.

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