This question on mathoverflow asks if normal distribution can be expressed as product of two i.i.d. variables, which is answered in this paper in more general setting. The paper shows, that if $W_1, W_2, \ldots W_k$ are i.i.d. with distribution $$ W_1 \overset D = \varepsilon \exp \left(C - \sum_{m=0}^\infty \tilde G_{1/k, 1/(2m + 1)}\right) $$ where $\varepsilon$ is Rademacher variable, $C$ is some appropriate constant, $\tilde G_{a, b}$ is gamma distribution shifted to have null expectation with shape $a$ and scale $b$, and $G$'s and $\varepsilon$ being independent, then $W_1 W_2 \ldots W_k \overset D = \mathcal N (0,1)$. It manages to do so with the help of exp-normal distribution, which is $\ln | \mathcal N (0, 1)|$.
I want to find characteristic function of $W_1$. So far I maneged to get charasteristic function of $\ln |W_1|$ (the inside of the $\exp$ function) which is $$\mathrm e^{itD}\sqrt[k]{\prod_{m=0}^\infty\frac 1 {1 + \frac t {2m + 1}}}$$ for some constant $D$.
Is there a way to get the characteristic function of $\exp (\ln | W_1|)$ from that? Or perhaps some different approach is needed?
Update: I tried plugging inverse transform formula for density function $$ f(x) = \int_{\mathbb R}\mathrm e^{-ixs} \varphi(s)\operatorname{d}\!s $$ into definition of characteristic function and got \begin{align} \varphi_{|W_1|}(t) &= \int_{\mathbb R}\exp(it\mathrm e^x) \left(\int_{\mathbb R}\mathrm e^{-ixs} \varphi_{\ln|W_1|}(s)\operatorname{d}\!s\right) \operatorname{d}\!x \\ & \overset ? = \int_{\mathbb R} \varphi_{\ln|W_1|}(s) \left(\int_{\mathbb R}\exp(it\mathrm e^x-ixs)\operatorname{d}\!x\right) \operatorname{d}\!s. \end{align} I didn't bother checking Fubini's theorem assumptions in (?), because Wolfram said that inner integral cannot be expressed in terms of elementary functions, and when I tried to express it as some series I failed to produce anything meaningful.