Fourier shift theorem: What happens when the shift is very large (e.g., multiplying a signal by a shifting bandpass filter)?

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I'm trying to apply the Fourier transform shift theorem and the convolution theorem to a problem. I thought I was using them logically, but I have a case where my logic has to break.

I have a given signal, which I multiply by a rectangular window function (the window starts off centered on the origin) and then I Fourier transform (FT) the result. The FT of the window function is a sinc function, so my end result is a convolution of the FT of my signal and the sinc function (Gibbs ringing --- I'm okay with this so far).

I'd then like to move the window to a different position along my signal (not centered on the origin) and FT that region instead. From the FT shift theorem, the FT of the shifted window is still a sinc function, but with an added phase offset. $$\mathcal{F}\{f(t-t_0)\}=F(\omega)e^{-j\omega t_0}$$ So I multiply my signal by this shifted window and take the FT. The end result is a convolution of the FT of my signal and the phase-offset sinc function.

My thought was that the phase offset should have no affect on the magnitude of my result. Therefore, the magnitude of my end result will always be the same, regardless of the position of the window function. This clearly cannot be the case. For example, if I shift the window so far in time that I only FT a region of noise. My above logic would say that the magnitude of the result is the same as if my window were centered (this cannot be true).

Where does my thinking fall apart?