Fourier transform

Proof: Fourier Transform of the n-th Derivative

Let's start by expressing $g(t)$ in terms of its Fourier components $G(\omega )$:

$$g(t)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}G(\omega )e^{i\omega t}d\omega$$

We can differentiate both sides of this equation with respect to $t$ $n$ times:

$$\frac{d^{n}g(t)}{d{t}^n}=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(i\omega {)}^{n}G(\omega )e^{i\omega t}d\omega$$

We can now rewrite the right-hand side of this equation in terms of the Fourier Transform of $\frac{d^n}{d{t}^n}g(t)$ by grouping all of the $(i\omega {)}^n$ terms:

$$\frac{d^{n}g(t)}{d{t}^n}=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}[(i\omega {)}^{n}G(\omega )]e^{i\omega t}d\omega$$

Therefore, we have shown that the Fourier Transform of the $n$-th derivative of a function $g(t)$ is given by $(i\omega {)}^n$ times the Fourier Transform of $g(t)$:

$${\left(\frac{d^{n}g(t)}{d{t}^n}\right)}^{\mathcal{F}}=(i\omega {)}^{n}G(\omega )$$