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 \(g(t)\):

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