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As seen in https://math.stackexchange.com/a/430885/634773, we can find the Fourier Transform of the derivative of a function through the anti-transform.

But if we do integration by parts, shouldn't it yield the same answer?

However, unlike what we see in https://math.stackexchange.com/a/430863/634773, the first term doesn't necessarily have a limit, or does it?

Sorry if I used bad english.

Joao Victor
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  • Do you see how it works for $f(x) = e^{-x} 1_{x > 0}$ ? The derivative is $f'(x) = \delta(x)-e^{-x} 1_{x > 0}$ (a distribution, not a function), its primitive is $\int_a^x f'(y)dy = C + f(x)$, the Fourier transform of $\delta(x)$ is the boundary term you'll have when integrating by parts $\int_{-\infty}^\infty f(x) e^{- i \omega x}dx=\int_0^\infty e^{-x}e^{-i \omega x}dx$, and $\mathcal{F}f' = i \omega\mathcal{F}f$ – reuns Jun 23 '19 at 03:26
  • Perfect English – gen-ℤ ready to perish Jun 23 '19 at 03:59
  • There are many details being ignored in the posts you reference. My favorite justification was "why not?" – Disintegrating By Parts Jun 23 '19 at 17:15

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One useful set of assumptions is that $f,f'$ are square integrable on $\mathbb{R}$ and $f$ is absolutely continuous. But then their transforms must be interpreted as limits in $L^2$ of the truncated transforms. This simple set of assumptions works because $ff' \in L^1$, which forces $\int ff'dx = f^2/2$ to have limits at $\pm\infty$, and those limits must be $0$ in order for $f^2$ to integral. Then you can integrate by parts to legitimately obtain $\hat{f'}(\xi)=i\xi\hat{f}(\xi)$.

Disintegrating By Parts
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