# Fourier Transform of Derivative

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 $F(s)$?

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 $$i \omega \, F(\omega)$$

• A very good answer indeed. +1 Nov 9, 2014 at 18:59
• @leonbloy Why exactly can we move the derivative inside the integral (apply Leibniz rule)? Mar 5, 2017 at 12:17
• @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 Mar 16, 2017 at 18:43
• @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 $g(t)$ Sep 10, 2018 at 2:16
• I hate to be that guy, but doesn't your answer rely on the fact that the Fourier inversion formula holds for $f$? Which means you are assuming extra conditions on $f$? Like you are saying that $F$ is integrable (which it doesn't have to be in general). Sep 16, 2021 at 16:58

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

Why?

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 $\mathbb{R}$.)

• Why the first term must vanish? I think we need the additional condition $\lim_{t\to\infty}f(t)=0$ to guarantee this. Feb 27, 2016 at 3:51
• Since $f \in L^1 \cap C^1$ f is continous and integrable, and must tend to zero when t tends to infinity...? Jun 29, 2016 at 17:48
• 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."
– Zach
Jan 12, 2017 at 19:22
• Where did we ever assume that $f$ is absolutely integrable, or was that an assumption appended as a bandage? Dec 6, 2017 at 2:50
• It seems that absolute integrability can NOT imply that $f$ vanishes at the infinity. See Did's answer here: math.stackexchange.com/questions/108191/… Sep 18, 2018 at 2:53