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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.

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  • $\begingroup$ Let $\varphi_X(\omega)$ be the characteristic function, then the inverse Fourier transform would be $$ \frac1{2\pi}\int_{\mathbb R} e^{i\omega x}\overline{\varphi_X(\omega)}\ \mathsf d\omega, $$ where $\overline{\;\cdot \;}$ denotes complex conjugate. $\endgroup$ – Math1000 Apr 27 '18 at 15:46
  • $\begingroup$ @Math1000 that is the definition, and I know it. Problem is with particular image. I cannot get back this square bump $\endgroup$ – Severin Pappadeux Apr 27 '18 at 16:11
  • $\begingroup$ Using Mathematica to evaluate the integral, I found that $$ \frac i{2\pi} \int_{-\infty }^{\infty } \frac1\omega \left(-1+e^{-i \omega }\right) e^{i x \omega } \, d\omega = \frac12(\mathsf{Sgn}(1-x)+\mathsf{Sgn}(x), $$ where $\mathsf{Sgn}(\cdot)$ denotes the sign function ($\mathsf{Sgn}(t) = -\mathsf 1_{(-\infty,0)}(t) + \mathsf 1_{(0,\infty)}$). This is equivalent to $t\mapsto\mathsf 1_{(0,1)}(t)$, the density of the $\mathsf{Unif}(0,1)$ distribution. $\endgroup$ – Math1000 Apr 27 '18 at 16:29
  • $\begingroup$ @Math1000 Sure, no doubts, I know the answer after all. I also have the book with FT tables. But how to get it is the question... $\endgroup$ – Severin Pappadeux Apr 27 '18 at 16:30
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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 $$\frac 1 {2 \pi} \int_{-\infty}^\infty \frac {(e^{i w} - 1) e^{-i x w}} {i w} dw = \frac 1 {2 \pi i} i \pi (\operatorname{sgn} (1 - x) - \operatorname{sgn} (-x))= [0 < x < 1].$$

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  • $\begingroup$ Well, I'm back to this problem, and thank you for the answer. The only question I have is why $C_\epsilon$ is in upper half-plane? I thought it will live pole out and whole integral would be zero. Shouldn't $C_\epsilon$ be in lower half-plane so whole integration contour include pole at 0? Just a bit confused... $\endgroup$ – Severin Pappadeux Sep 6 '18 at 23:42
  • $\begingroup$ Correct, the contour leaves the pole out, which means that $I_{\text{v.p.}} + I_\epsilon +I_R = 0$, but notice that $I_\epsilon$ doesn't tend to zero: $I_R \to 0, I_\epsilon \to -\pi i$. You can take $C_\epsilon$ in the lower half-plane, then $I_{\text{v.p.}} + I_\epsilon + I_R = 2 \pi i$ and $I_\epsilon \to \pi i$. $\endgroup$ – Maxim Sep 7 '18 at 1:33
  • $\begingroup$ This is an "aha" moment, thanks $\endgroup$ – Severin Pappadeux Sep 7 '18 at 1:41

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