Fourier transform

Proof: Fourier Transform of the n-th Derivative

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

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

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

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

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

$$\frac{d^n g(t)}{dt^n} = \frac{1}{2\pi} \int_{-\infty}^{\infty} \left[ (i\omega)^n G(\omega) \right] e^{i\omega t} d\omega$$

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 $g(t)$:

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