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I have been trying to understand the formula for the Fourier Transform of a derivative, but I am getting hung up in the integration. Looking at this post.

So far, I understand everything up to the step:

$$ \begin{align*} \mathcal{F}(f')(\xi)&=\int_{-\infty}^{\infty}e^{-2\pi i\xi t}f'(t)\,dt\\ &=e^{-2\pi i\xi t}f(t)\bigr\vert_{t=-\infty}^{\infty}-\int_{-\infty}^{\infty}-2\pi i\xi e^{-2\pi i \xi t}f(t)\,dt\\ & \end{align*} $$

I understand also, that since f(t) must vanish, the first term must vanish. It may be that I am missing something easy, but my problem is in understanding how to compute the integral on the right, and why it evaluates to

$$ \begin{align*} 2\pi i\xi\cdot\mathcal{F}(f)(\xi) \end{align*} $$

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    $\begingroup$ $2\pi i\xi$ is a constant respect to $t$. $\endgroup$ – Nosrati Sep 6 '18 at 17:09
  • $\begingroup$ I just realized that I was basically asking how to do a normal Fourier Transform. Thanks for your response $\endgroup$ – anapollosun Sep 6 '18 at 22:09

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