Let $X$ be a random variable that follows the "version 1" generalised normal distribution described here, with p.d.f.
$$f_X(x;\mu,\alpha,\beta)=k\exp\left\{-\left(\frac{|x-\mu|}{\alpha}\right)^\beta\right\}\enspace,$$
where
$$k=\frac{\beta}{2\alpha\Gamma(1/\beta)}\enspace,$$
and let $Y=e^X$. Is there a closed-form expression for $E[Y]$? If so, what is it? More generally, is there a closed-form expression for the moment-generating function $M_X(t)=E\left[e^{tX}\right]$?
I began with
\begin{align} E[Y]&= E\left[e^X\right] \\ &= \int_{-\infty}^\infty e^xf_X(x) dx \\ &= k\int_{-\infty}^\infty \exp\left\{x-\left(\frac{|x-\mu|}{\alpha}\right)^\beta\right\}dx \enspace, \end{align}
but I'm not sure how to find a closed-form solution, or if one exists. Presumably the integral diverges when $\beta< 1$, as $f_X$ decays subexponentially in this case.
I also tried to solve it in terms of the p.d.f. $f_Y$ of $Y$, which is:
$$f_Y(y)=\frac{f_X(\ln y)}{y}\enspace.$$
This leads to the expression:
$$E[Y]=\int_{0}^\infty f_X(\ln y) dy \enspace,$$
but again I'm not sure how to proceed from here.