In our course, the Fourier transform of $f\in L^2(\mathbb{R})$ is defined as, $$\mathcal{F}(f) = \hat{f}(\zeta) = \int e^{-ix\zeta}f(x)dx.$$ Therefore, the inverse fourier transform is defined as, $$\bar{\mathcal{F}}(f) = f(x) =\frac{1}{2\pi} \int e^{ix\zeta} \hat{f}(\zeta)d\zeta.$$
We have computed the Fourier transform of $f=1_{[-1,1]}$, which is, $$\hat{f}(\zeta) = \frac{\sin(\zeta)}{\zeta}.$$
Now the goal is to compute the Fourier transform of the function $g(x) = \frac{\sin(x)}{x}.$ I am not sure whether I am arguing correctly here.
Here is what I have done so far. We know that, $$\hat{f}(\zeta) =g(\zeta)\iff \mathcal{F}(f) = g\\ \implies f = (2\pi)^{-1}\bar{\mathcal{F}}(g)\\ \implies \bar{f} = (2\pi)^{-1}\overline{\overline{\mathcal{F}}(g)}\\ \implies (2\pi)\bar{f}=\hat{g}. $$
Thus we conclude that $\hat{g} = \frac{1}{2\pi}1_{[-1,1]}(x).$ Is this correct reasoning? I apologize for the non-standard notation, but I can't do much about it at this point.