For the usual rectangular pulse of amplitude $A$ and duration $\tau$, from $-\tau/2$ to $\tau/2$, I need to find its Fourier Transform. Using the formula $$X(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-i\omega t}\,dt$$ I get $$X(\omega)= \tau A \displaystyle\frac{\sin{(\omega\tau/2)}}{\omega\tau/2}$$
However, if I want to find $X(\omega)$ using the Fourier Transform properties, I get something else.
I try it by writing $x(t)$ as $A\left[u{\left(t+\frac{\tau}{2}\right)}-u{\left(t-\frac{\tau}{2}\right)}\right]$ and I use the Fourier Transform of the unit step function, which is $\pi\delta(\omega)+1/(i\omega)$. and the property $f(t-t_0)\longrightarrow e^{-i\omega t_0}F(\omega)$, which leads to $$A\,u{\left(t+\frac{\tau}{2}\right)}\longrightarrow Ae^{i\omega\tau/2}\left(\pi\delta(\omega)+\frac{1}{i\omega} \right)$$ $$A\,u{\left(t-\frac{\tau}{2}\right)}\longrightarrow Ae^{-i\omega\tau/2}\left(\pi\delta(\omega)+\frac{1}{i\omega} \right)$$ So, the Fourier Transform of $x(t)$ that I get is $$Ae^{i\omega\tau/2}\left(\pi\delta(\omega)+\frac{1}{i\omega} \right) - Ae^{-i\omega\tau/2}\left(\pi\delta(\omega)+\frac{1}{i\omega} \right) = A\left(\pi\delta(\omega)+\frac{1}{i\omega} \right)\left(e^{i\omega\tau/2} - e^{-i\omega\tau/2} \right)=$$ $$=2iA\left(\pi\delta(\omega)+\frac{1}{i\omega} \right)\sin{(\omega\tau/2)}$$ which is different from what I get by using the definition of the Fourier Transform.
What am I doing wrong when using the property?