I'm having a hard time with Fourier Transform's Duality Property.
The Duality Property states that, if $$\mathcal{F}\left\{x(t)\right\} = X(\nu),$$ then $$\mathcal{F}\left\{X(t)\right\} = 2\pi x(-\nu).$$
How does this property interacts with other properties of the Fourier Transform? For example, knowing that $$\mathcal{F}\left\{\frac{\text{d}}{\text{d} t} x(t) \right\} = i\nu X(\nu),$$ applying the Duality Property, one finds that $$\mathcal{F}\left\{i t x(t) \right\} = 2\pi \frac{\text{d}}{\text{d}(-\nu)} X(-\nu).$$ The correct property, as stated in Wikipedia, is that $$\mathcal{F}\left\{i t x(t) \right\} = -\frac{\text{d}}{\text{d}\nu} X(\nu).$$
I can't find the error, nor can I see where the $2\pi$ came from. I couldn't interchange between the two results, proving that they are the same.
Thanks.
Best Answer
You've applied the duality property wrongly. What you should get is
$$\mathcal{F}\{it X(t)\}=-2\pi\frac{dx(-\nu)}{d\nu}\tag{1}$$
Note that $X(t)$ is the Fourier transform of $x(\nu)$, not the other way around, as you supposed in your equation. Let
$$\tilde{X}(\nu)=\mathcal{F}\{X(t)\}\tag{2}$$
From the duality property we know that
$$\tilde{X}(\nu)=2\pi x(-\nu)\tag{3}$$
Plugging $(3)$ into $(1)$ gives
$$\mathcal{F}\{it X(t)\}=-\frac{d\tilde{X}(\nu)}{d\nu}\tag{4}$$
which is equivalent to the last equation in your question.