The answer was in
dndtn{f(t)}Fourier⟶(i2πf)n.F(f)
when n=1.
ddt{f(t)}Fourier⟶i2π.f.F(f)
The factor f justifies windowing a function of interest in the frequency domain (or equivalently the time domain) and repeat that window for a periodic variation and then apply the transform. It is also the reason why in the frequency domain we always obtain results in per unit time (per second), over one time period, f=1T.
And first thing first. Although in retrospect, nothing stopped us from making up periodic intervals just because we can obtain the appropriate answer. And the bull drives on.