The Fourier transform of a function under a derivative is:
$$ \mathcal{F}\Bigg({df (t) \over dt}\Bigg) = {1\over \sqrt{2 \pi}}\int _{-\infty}^{\infty}{df (t) \over dt} e^{-iwt}dt \tag 1$$
Using integrations by parts we get:
$$ \mathcal{F}\Bigg({df (t)\over dt}\Bigg) = {1\over \sqrt{2 \pi}}\Bigg ( f(t)e^{-iwt} \Bigg |_{-\infty}^{\infty} + iw\int _{-\infty}^{\infty}{f(t)} e^{-iwt}dt \Bigg ) \tag 2$$
$$ \mathcal{F}\Bigg({df (t)\over dt}\Bigg) = {f(t)e^{-iwt}\over \sqrt{2 \pi}} \Bigg |_{-\infty}^{\infty} + iwF(w) \tag 3$$
The thing that is actually confusing me is the first term on the right-hand side that we are assuming that it vanishes at the limits:
$$ {f(t)e^{-iwt}\over \sqrt{2 \pi}} \Bigg |_{-\infty}^{\infty} = 0 \tag 4 $$
According to this post, the statement $(4)$ is valid if the function $f(t)$ is continuous and smooth. However, does this term vanish even if $t$ belongs to a bounded domain $[t_0, t_1]$? I am currently studying spectral methods for numerically solving PDE's. There, the domain is bounded, but it is assumed that this term still vanishes.