I found something interesting when you use duality on the same transform pair over and over... Note: I can not find a way to display a proper fourier symbol, so I went with the $ \displaystyle{\bf{F}} $ as seen below.
(1)$ \displaystyle{\bf{F}} (e^{-at}u(t))=\frac{1}{a+j\omega} $
By Duality of (1):
(2)$ \displaystyle{\bf F} (\frac{1}{a+jt})=2\pi e^{a\omega}u(\omega) $
By Duality of (2) (and interestingly Time Reversal of (1)):
(3)$ \quad\displaystyle{\bf F} (2\pi e^{at}u(-t))=\frac{2\pi}{a-j\omega} $
By Linearity of (3) the $ 2\pi $ divides out of both sides:
(4)$ \quad\displaystyle {\bf F} (e^{at}u(-t))=\frac{1}{a-j\omega} $
By Duality of (4) (and again interestingly Time Reversal of (2)):
(5)$ \quad\displaystyle {\bf F} (\frac{1}{a-jt})=2\pi e^{-a\omega}u(-\omega) $
By Duality of (5) we see that we get back to (1).
If this is done in a more general since, it becomes clear that it is only necessary to take the dual of a Fourier Transform Pair once. After taking the dual once, one might as well use time reversal. Taking the dual four times will always result in the original pair again after the extra $ 2\pi $'s are divided out.
