Derivative property of the Fourier transform

傅里叶变换与导数性质

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$f (t)$ has the property:

The domain is $(∞,−∞)(\infty, -\infty)$
The integral of $f (t)$ is limited
$∫−∞∞f(t)dt=A\int ^{\infty} _{-\infty}f(t)dt= A$
and
$\lim_{t \to \infty} f(t)=0}$
The Fourier transform of $f (t)$

$\rightarrow F(\omega)$
is
$F(ω)=∫−∞∞f(t)exp(−jωt)dtF(\omega)=\int ^{\infty} _{-\infty}f(t)exp(-j\omega t)dt$
What is about the Fourier transform of the first derivative of $f (t)$ ?
Let’s use the definition
$f′(t)→F(ω)f^{\prime}(t) \rightarrow F(\omega)$
$F(f′(t))=∫−∞∞f′(t)exp(−jωt)dt\mathcal{F}\left (f^{\prime}(t) \right)=\int ^{\infty} _{-\infty}f^{\prime}(t) exp(-j\omega t)dt$
Integral by part
$F(ω)=f(t)exp(−jωt)∣−∞∞−∫−∞∞f(t)(exp(−jωt))′dtF(\omega)= f(t)exp(-j\omega t) \bigg |^{\infty}_{-\infty}-\int ^{\infty} _{-\infty}f(t) \left ( exp(-j\omega t) \right) ^{\prime} dt$
$f(t)exp(-j\omega t) \bigg |^{\infty}_{-\infty} + j\omega \int ^{\infty} _{-\infty}f(t) exp(-j\omega t) dt$
$f(t)exp(-j\omega t) \bigg |^{\infty}_{-\infty} + j\omega \int ^{\infty} _{-\infty}f(t) exp(-j\omega t) dt$
$j\omega \int ^{\infty} _{-\infty}f(t) exp(-j\omega t) dt$
Therefore,
$F(fn(t))=(jω)n∫−∞∞f(t)exp(−jωt)dt\mathcal{F}(f^n(t))= (j\omega)^n \int ^{\infty} _{-\infty}f(t) exp(-j\omega t) dt$