Fix a rapidly decreasing function $\psi \in \mathcal{S}(\mathbb{R})$ with the properties that $\int_\mathbb{R} \psi = 0$, $\mathrm{Re}(\psi(\cdot))$ is an even function, and $\mathrm{Im}(\psi(\cdot))$ is an odd function.
Given any bounded integrable function $x \colon \mathbb{R} \to \mathbb{R}$, define \begin{align*} \hat{x}(\xi) \ :=& \ \int_\mathbb{R} x(\tau)e^{-2\pi i\xi\tau} \, d\tau \\ \mathring{x}(\xi,t) \ :=& \ \int_\mathbb{R} x(\tau)\psi(\xi(\tau-t)) \, d\tau \end{align*} for all $\xi,t \in \mathbb{R}$. (Obviously $\hat{x}$ is the Fourier transform; $\mathring{x}$ is a version of the wavelet transform.)
It is known that
$$ \hspace{20mm} \int_\mathbb{R} x(t)^2 \, dt \ = \ \int_\mathbb{R} |\hat{x}(\xi)|^2 \, d\xi \ = \ C_\psi\!\iint_{\mathbb{R}^2} |\mathring{x}(\xi,t)^2\xi| \, d\xi dt \hspace{18mm} (1) $$
where $C_\psi$ is a constant that depends only on $\psi$ and not on $x$, namely $$ C_\psi \ = \ \left( \int_\mathbb{R} \frac{\hat{\psi}(r)^2}{|r|} \, dr \right)^{\!-1}\,. $$ It is also known that the first equality in $(1)$ can be extended to the 3rd moment by
$$\hspace{26mm} \int_\mathbb{R} x(t)^3 \, dt \ = \ \iint_{\mathbb{R}^2} \hat{x}(\xi_1)\hat{x}(\xi_2)\overline{\hat{x}(\xi_1+\xi_2)} \, d\xi_1d\xi_2 \hspace{24mm} (2) $$
where the bar denotes complex conjugate. This naturally leads to the question of whether the second equality in $(1)$ can be extended to the 3rd moment; in case it helps, we can assume that $x$ is a test function:
Does there exist a function $f_\psi \colon \mathbb{R}^2 \to \mathbb{C}$ such that for every real-valued test function $x \in C_c^\infty(\mathbb{R},\mathbb{R})$, $$ \int_\mathbb{R} x(t)^3 \, dt \ = \ \iiint_{\mathbb{R}^3} \mathring{x}(\xi_1,t)\mathring{x}(\xi_2,t)\overline{\mathring{x}(\xi_1+\xi_2,t)} \, f_\psi(\xi_1,\xi_2) d\xi_1d\xi_2dt \, ? $$ (If not, then [a more open-ended question] are there reasonable conditions on $\psi$ under which one can find a function $f_\psi$ such that the above equality "approximately" holds?)
A little bit of numerical experimentation by a friend and myself seems to suggest that the function $$ f_\psi(\xi_1,\xi_2) \ = \ D_\psi\mathbf{1}_{[0,\infty)\times[0,\infty)\!}(\xi_1,\xi_2)(\xi_1+\xi_2) $$ might work for some constant $D_\psi \in \mathbb{C}$.
[Remark. In all the above formulae (except possibly the unknown formula in the question itself), one only needs to consider positive $\xi$-values:
- $\int_\mathbb{R} |\hat{x}|^2 = 2\int_{[0,\infty)} |\hat{x}|^2$;
- $\int_{\mathbb{R}^2} \mathcal{B}x = 6\int_{[0,\infty)\times[0,\infty)} \mathcal{B}x\ $ where $\mathcal{B}x$ is the integrand in the RHS of $(2)$;
- due to our assumptions on the symmetry of $\psi$, $\int_\mathbb{R} |\mathring{x}(\xi,t)^2\xi| \, d\xi = 2\int_{[0,\infty)} |\mathring{x}(\xi,t)|^2\xi \, d\xi$;
this is why my conjectured form of $f_\psi$ is restricted to nonnegative $\xi$-values.]
UPDATE: I must admit, it is starting to look unlikely to me that a function $f_\psi$ with the desired property exists. I think I can show that \begin{align*} \mathrm{RHS} &= \iint_{\mathbb{R}^2} \frac{-f_\psi(\xi_1,\xi_2)}{\xi_1\xi_2(\xi_1\!+\xi_2)} \iint_{\mathbb{R}^2} \hat{\psi}(\tfrac{\omega_1}{\xi_1})\hat{\psi}(\tfrac{\omega_2}{\xi_2})\hat{\psi}(\tfrac{{\color{blue}\omega_{{\color{blue}1}}}{\color{blue}+}{\color{blue}\omega_{{\color{blue}2}}}}{{\color{blue}\xi_{{\color{blue}1}}}{\color{blue}+}{\color{blue}\xi_{{\color{blue}2}}}}) \hat{x}(\omega_1)\hat{x}(\omega_2)\hat{x}(\omega_1\!+\omega_2) \, d\omega_1d\omega_2 \, d\xi_1d\xi_2 \\ \mathrm{LHS} &= \iint_{\mathbb{R}^2} \ \ \frac{D_\psi}{\xi_1\xi_2} \ \iint_{\mathbb{R}^2} \hat{\psi}(\tfrac{\omega_1}{\xi_1})\hat{\psi}(\tfrac{\omega_2}{\xi_2})\hat{\psi}(\tfrac{{\color{blue}\omega_{{\color{blue}1}}}}{{\color{blue}\xi_{{\color{blue}1}}}} {\color{blue}+} \tfrac{{\color{blue}\omega_{{\color{blue}2}}}}{{\color{blue}\xi_{{\color{blue}2}}}}) \hat{x}(\omega_1)\hat{x}(\omega_2)\hat{x}(\omega_1\!+\omega_2) \, d\omega_1d\omega_2 \, d\xi_1d\xi_2 \end{align*} where $D_\psi$ is some constant.
After the inner double integral, the only difference between the two formulae is what is highlighted in blue. If it were not for this difference, the conjectured solution $f_\psi=\mathrm{cst}\!\cdot\!(\xi_1+\xi_2)$ would work.
However, I think I can also show that the conjectured solution is guaranteed to give $0$ rather than $\int_\mathbb{R} x(t)^3 \, dt$.
