How to use inverse Fourier transform to get the original signal

Question

Suppose I have a signal $u(x)$ where $x\in[0,2\pi]$ and the signal has following form in Fourier space,

$\hat{u}(\kappa)=\begin{cases} \frac{1+i\tan{(\varphi_u(\kappa))}}{\sqrt{1+\tan^2{(\varphi_u(\kappa))}}}\kappa^{\alpha/2},\ \ 0<\kappa\le\kappa_c \\ 0+i0,\ \ \ else \end{cases}$

where $\varphi_u(\kappa)$ is a random number for each $\kappa$, and $\alpha$ is a constant.

How can I get $u(x)$ using inverse Fourier transform?

Answer 1

The expectation value $E[u](x)$ can be obtained by inverse Fourier transform of the expectation value of $\hat{u}(k)$, since the Fourier transform is a linear operation; If I interpret the statement in the OP that the phase $\phi$ is a "random number" as saying that $\phi$ is uniformly distributed, then the integral $$\frac{1}{\pi}\int_0^\pi \frac{1+i\tan\phi}{\sqrt{1+\tan^2\phi}}\,d\phi=\frac{2}{\pi}$$ gives $E[u](x)$ as the inverse Fourier transform of $(2/\pi)\kappa^{\alpha/2}$, which is an incomplete Gamma function.

_{I should note that the statement in the OP that both $u(x)$ and $\hat{u}(k)$ have compact support is inconsistent.}