Duality of the continuous-time Fourier transform - derivation and notation Suppose we have the Fourier transform pair $x(t)$ and $X(\omega)$ such that
 $$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t$$

The duality property states that $X(t)$ and $2\pi x(-\omega)$ constitute a Fourier transform pair. I was trying to prove this statement when I ran into the following problem: the Fourier transform of a signal is sometimes denoted by $X(j\omega)$, such that $$X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t$$

Using this notation, how do we even state the duality property? Note that evaluating $X(t)$ in this case would effectively replace $j\omega$ (as opposed to just $\omega$ in the previous case) by $t$ which is clearly wrong. Would we have to write $X(jt)$ for consistency? 

Returning to the question of deriving the duality property and using the notation I used in the beginning, what is the flaw in the following approach?

Since $$X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \mathrm{d}t$$
we let $t$ be dummy variable $u$, and evaluate the function $X(\cdot)$ at $t$. We then have $$X(t) = \int_{-\infty}^{\infty} x(u) e^{-j ut} \mathrm{d}u$$
With the change of variables $\omega = -u$ we get $$X(t) = -\int_{-\infty}^{\infty} x(-\omega) e^{j\omega t} \mathrm{d}\omega$$ $$X(t) = -\frac{1}{2\pi}\int_{-\infty}^{\infty} 2\pi x(-\omega) e^{j\omega t} \mathrm{d}\omega$$

Which is almost the correct result except for the additional minus sign. Where did I go wrong? Thank you in advance.

I would also like to add that I feel uncomfortable about simply changing $t$ and $\omega$ in the functions' argument because they have different dimensions.
You've answered your first question yourself: if you use the notation $X(j\omega)$ then you'll need to use $X(jt)$, because you simply replace $\omega$ by $t$.

The sign error in your derivation comes from the fact that you forgot to change the sign of the limits of the integral. After substituting $\omega=-u$ you should get

$$X(t) = -\int_{\infty}^{-\infty} x(-\omega) e^{j\omega t} \mathrm{d}\omega = \int_{-\infty}^{\infty} x(-\omega) e^{j\omega t} \mathrm{d}\omega$$