Have computed Fourier transformation of the $U(0,1)$ PDF (aka characteristic function) as
\begin{equation} PDF(x) = 1 (0<x<1) \end{equation}
\begin{equation} CF = \frac{e^{i\omega} - 1}{i\omega} \end{equation}
How I could get back PDF from CF? I'm struggling with inverse Fourier transform. I know the definition of inverse transform, I just cannot get back the square bump of PDF.
Can be done by computing the residues. Let $F(w) = e^{i a w}/w, a > 0$. If $C_\epsilon$ and $C_R$ are half-circles with radii $\epsilon$ and $R$ in the upper half-plane, then $$\lim_{\substack {\epsilon \to 0 \\ R \to \infty}} \left( \int_{-R}^{-\epsilon} F(w) dw + \int_\epsilon^R F(w) dw \right) = \operatorname{v.\!p.} \int_{-\infty}^\infty F(w) dw, \\ \lim_{\epsilon \to 0} \int_{C_\epsilon} F(w) dw = -i \pi \operatorname{Res}_{w = 0} F(w) = -i \pi, \\ \lim_{R \to \infty} \int_{C_R} F(w) dw = 0 \Rightarrow \\ \operatorname{v.\!p.} \int_{-\infty}^\infty F(w) dw = i \pi.$$ In the same way, the principal value integral for negative $a$ is $-i \pi$. Now $$\frac 1 {2 \pi} \int_{-\infty}^\infty \frac {(e^{i w} - 1) e^{-i x w}} {i w} dw = \frac 1 {2 \pi i} i \pi (\operatorname{sgn} (1 - x) - \operatorname{sgn} (-x))= [0 < x < 1].$$