Let $x(t) = e^{-|t|}$ and $X(\omega) = \mathcal F\{\mathrm e^{-|t|}\}$ be the Fourier Transform of $x(t)$, which is given by $$X(\omega) = \frac {2} {1+\omega^2}\tag{1}$$ Then, the scaling property states that: $$\mathcal F \{x(at)\} = \frac {1} {|a|} X\left(\frac {\omega} {a}\right)\tag{2}$$ Now, let me have $a = j$, where $j = \sqrt{-1}$. $$\mathcal F \{x(jt)\} = \frac {1} {|j|} X\left(\frac {\omega} {j}\right) = 1 \frac {2} {1+\left(\frac {\omega} {j}\right)^2} = \frac {2} {1-\omega^2}\tag{3}$$ But, if we wouldn't use the scaling property, we would find following result: $$\mathcal F \{x(jt)\} = \mathcal F \{e^{-|jt|}\} = \mathcal F \{e^{-|t|}\} = \frac {2} {1+\omega^2}\tag{4}$$
Obviously, $\frac {2} {1-\omega^2} \neq \frac {2} {1+\omega^2}$.
The fact that $x(jt)=x(t)$ seems to cause the problem.
Am I doing a mistake somewhere?
$t$ of the Fourier transform is only defined over the real numbers. Multiplying $t$ by $j$ breaks this.