Proving Fourier transform pair with derivatives using duality

Question

I want to use duality to prove the Fourier transform pair $t^nx(t) \overset{\mathscr{F}}{\longleftrightarrow} j^n\frac{d^nX(\omega)}{d\omega^n}$ but I am struggling.

I learned that if $x(t) \overset{\mathscr{F}}{\leftrightarrow} X(\omega)$ then $X(t) \overset{\mathscr{F}}{\leftrightarrow} 2\pi x(-\omega)$, however I am not sure if I can apply it here. My take, going from the derivative first:

$$g(t) = \frac{d^n}{dt^n}x(t)\overset{\mathscr{F}}{\longleftrightarrow} (j\omega)^nX(\omega) = G(\omega)$$

My knowledge is still shaky, so I try to apply duality with pure symbolic manipulation, so I should find that $G(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi g(-\omega)$, which should get us back to the first expression I want to prove:

$$G(t) = (jt)^n X(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi \frac{d^n}{dt^n}x(-\omega) = 2\pi g(-\omega)$$ Obviously, there are a few problems.

1. I want $t^nx(t)$ but I have $(jt)^n X(t)$. What should I do with this $j$?
2. The $2\pi$ should not appear, but the duality formula requires so. Why?
3. Why the negative sign on the $\omega$?

I want to really understand the duality property but for some reason, it just won't click...

Thank you for your time.

Answer 1

Let's start with $n=1$. We know that

$$\frac{dx(t)}{dt}\overset{\mathscr{F}}{\longleftrightarrow} j\omega X(\omega)\tag{1}$$

where $X(\omega)$ is the Fourier transform of $x(t)$.

From the duality property we obtain

$$jtX(t)\overset{\mathscr{F}}{\longleftrightarrow}2\pi\frac{dx(-\omega)}{d(-\omega)}\tag{2}$$

Now the problem is that $x(\omega)$ is not the Fourier transform of $X(t)$. If we use $\hat{X}(\omega)$ to denote the Fourier transform of $X(t)$, we have, again from duality,

$$\hat{X}(\omega)=2\pi x(-\omega)\tag{3}$$

Plugging $(3)$ into $(2)$ gives

$$jtX(t)\overset{\mathscr{F}}{\longleftrightarrow}\frac{d\hat{X}(\omega)}{d(-\omega)}\tag{4}$$

Multiplying both sides of $(4)$ by $-j$ results in

$$tX(t)\overset{\mathscr{F}}{\longleftrightarrow}j\frac{d\hat{X}(\omega)}{d\omega}\tag{5}$$

Finally, applying $(5)$ $n$ times gives the desired relationship

$$t^nX(t)\overset{\mathscr{F}}{\longleftrightarrow}j^n\frac{d^n\hat{X}(\omega)}{d\omega^n}\tag{6}$$