Using this equality:
$ \mathcal{F} \left(\frac{d^nf(x)}{dx^n}\right) = (i 2\pi u)^n F(u) $
I remember my professor saying that if $n=-1$ we would be able to get $ \int f(x) dx $, is this true? If so, how precise it is? Will it work with any kind of $f(x)$?
IIRC he also said we could use any $n$, even non integers, if possible, does it have any practical use?
If $f=g'$ then $\mathcal F (f)=\mathcal F (g')=(2\pi i u) \mathcal F (g)$. Hence $\mathcal F (g)=\frac 1 {2\pi i u} \mathcal F (f)=(2\pi i u )^{-1}\mathcal F (f)$. This is the Fourier transfom of $g=\int f$. This is valid under the condition $\int_{-\infty} ^{\infty} f(x)\, dx =0$ (in addition to the obvious integrability conditions).