The Fourier Transform of a function $f(x)$ can be defined as:

$$\begin{aligned} \color{blue}{\widehat{f(x)} = {1 \over \sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx} f(x)\,dx} \\ \end{aligned}$$

So,

$$\begin{aligned} \color{teal}{\widehat{f'(x)}} &\color{teal}{= {1 \over \sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx} f'(x)\,dx} & \text{substitue } f(x) \text{ by } f'(x) \\ \end{aligned}$$

Using integration by parts, let

$$\begin{aligned} dv &= f'(x)\,dx \\ v &= f(x) \\ u &= e^{-ikx} \\ du &= -ike^{-ikx}\,dx \\ \end{aligned}$$

Now,

$$\begin{aligned} \color{teal}{{1 \over \sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx} f'(x)\,dx} &= {1 \over \sqrt{2\pi}} \int_{-\infty}^{\infty} u\,dv \\ &= {1 \over \sqrt{2\pi}} \left( uv \bigg|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} v\,du \right) \\ &= \color{blue}{{1 \over \sqrt{2\pi}}} \left( e^{-ikx} f(x) \bigg|_{-\infty}^{\infty} + ik \color{blue}{\int_{-\infty}^{\infty} e^{-ikx}f(x)\,dx} \right) \\ \end{aligned}$$

Suppose $f(x) \rightarrow 0$ as $x \rightarrow \pm \infty$,

$$\begin{aligned} \color{teal}{{1 \over \sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx} f'(x)\,dx} &= ik \color{blue}{{1 \over \sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx}f(x)\,dx} \\ \therefore \widehat{f'(x)} &= ik \cdot \widehat{f(x)} \\ \end{aligned}$$

Telescoping,

$$\begin{aligned} \widehat{f^{(n)}} &= (ik)^{n} \cdot \widehat{f} \\ \end{aligned}$$

$\Box$

Source: Computational Methods for Data Analysis.

# posted by rot13(Unafba Pune) @ 8:12 PM