Consider a function $f(t)$ with Fourier Transform $F(s)$. So $$F(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f(t) \ dt$$

What is the Fourier Transform of $f'(t)$? Call it $G(s)$.So $$G(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f'(t) \ dt$$

Would we consider $\frac{d}{ds} F(s)$ and try and write $G(s)$ in terms of $F(s)$?


The Fourier transform of the derivative is (see, for instance, Wikipedia) $$ \mathcal{F}(f')(\xi)=2\pi i\xi\cdot\mathcal{F}(f)(\xi). $$


Use integration by parts: $$ \begin{align*} u&=e^{-2\pi i\xi t} & dv&=f'(t)\,dt\\ du&=-2\pi i\xi e^{-2\pi i\xi t}\,dt & v&=f(t) \end{align*} $$ This yields $$ \begin{align*} \mathcal{F}(f')(\xi)&=\int_{-\infty}^{\infty}e^{-2\pi i\xi t}f'(t)\,dt\\ &=e^{-2\pi i\xi t}f(t)\bigr\vert_{t=-\infty}^{\infty}-\int_{-\infty}^{\infty}-2\pi i\xi e^{-2\pi i \xi t}f(t)\,dt\\ &=2\pi i\xi\cdot\mathcal{F}(f)(\xi) \end{align*} $$ (The first term must vanish, as we assume $f$ is absolutely integrable on $\mathbb{R}$.)

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    $\begingroup$ Why the first term must vanish? I think we need the additional condition $\lim_{t\to\infty}f(t)=0$ to guarantee this. $\endgroup$ – Xiang Yu Feb 27 '16 at 3:51
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    $\begingroup$ Since $f \in L^1 \cap C^1$ f is continous and integrable, and must tend to zero when t tends to infinity...? $\endgroup$ – user202542 Jun 29 '16 at 17:48
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    $\begingroup$ The limit need not exist, although if it exists it must be zero. There are smooth, i.e., $C^\infty$, $L^1$ functions that do not tend to zero as $x \to \infty$. For an example, just make smooth "spikes" of height 1 at each integer $n$, such that the spike at $n$ has width $2^{-n}$. The limit must be zero, however, if you replace "continuous" with "uniformly continuous." $\endgroup$ – Zach Jan 12 '17 at 19:22
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    $\begingroup$ Where did we ever assume that $f$ is absolutely integrable, or was that an assumption appended as a bandage? $\endgroup$ – Shamisen Expert Dec 6 '17 at 2:50
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    $\begingroup$ It seems that absolute integrability can NOT imply that $f$ vanishes at the infinity. See Did's answer here: math.stackexchange.com/questions/108191/… $\endgroup$ – Sam Wong Sep 18 '18 at 2:53

A simpler way, using the anti-transform:

$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega$$

$$f'(t) = \frac{d}{dt}\!\left( \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, e^{i \omega t} d\omega \right)= \frac{1}{2\pi} \int_{-\infty}^{\infty} i \omega \, F(\omega) \, e^{i \omega t} d\omega$$

Hence the Fourier transform of $f'(t)$ is $ i \omega \, F(\omega)$

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    $\begingroup$ A very good answer indeed. +1 $\endgroup$ – Swapnil Tripathi Nov 9 '14 at 18:59
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    $\begingroup$ @leonbloy Why exactly can we move the derivative inside the integral (apply Leibniz rule)? $\endgroup$ – Konstantin Mar 5 '17 at 12:17
  • $\begingroup$ @Konstantin why not? As you mentioned, we've used Leibniz rule here as the function inside the integral has a continuous partial derivative w.r.t. the variable t with respect to we're differentiating $\endgroup$ – Arkya Chatterjee Mar 16 '17 at 18:43
  • $\begingroup$ Could you elaborate on how that second inequality tells you that the Fourier transform of $f'(t)$ is $i\omega F(\omega)$? $\endgroup$ – Atsina Sep 9 '18 at 17:19
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    $\begingroup$ @Atsina In general, if you have $g(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega) \, e^{i \omega t} d\omega $, then you know that $G(\omega)$ is the Fourier transform of $g(t)$ $\endgroup$ – leonbloy Sep 10 '18 at 2:16

protected by Zev Chonoles Dec 8 '15 at 2:59

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