Given the following p.d.f., which is the p.d.f. of the real and imaginary parts of a random variable that is the ratio between a complex Gaussian and a Chi-squared RVs:


\begin{equation\*}
 f\_U(u)=\exp\Big\{{-\frac{1}{4 u^2}}\Big\} \,\frac{\left(8 n u^2-1\right) I\_n\left(\frac{1}{4
 u^2}\right)+I\_{n+1}\left(\frac{1}{4 u^2}\right)}{4 |u|^3}, 
\end{equation\*}
where $n$ is a constant integer and $I\_n(z)$ is the modified Bessel function of the first kind.


I'd like to find its Moment-generating function or at least a closed form expression for its mean and variance.


**UPDATE 17/02/2019**
Further information on this p.d.f. and how it was obtained can be found at: [Distribution of ratio between complex Gaussian and Chi-square R.V.s](https://mathoverflow.net/questions/290092/distribution-of-ratio-between-complex-gaussian-and-chi-square-r-v-s)



MGF of a RV that is the ratio between a complex Gaussian and a Chi-squared RVs
As discussed in the [MO question](https://mathoverflow.net/questions/290092/distribution-of-ratio-between-complex-gaussian-and-chi-square-r-v-s) linked to in the OP, the variable $u$ is defined as $u=R^{1/2}\cos\phi$, where $\phi$ is uniformly distributed in $(0,2\pi)$ and $R=\xi\_{2}/(\xi\_2+\xi\_{2n-2})^2$ is composed from independent chi-squared distributions with $2$, respectively, $2n-2$ degrees of freedom. So all odd moments of $u$ vanish, while the even moments are given by
$$\mu\_{2p}\equiv\mathbb{E}(u^{2p})= \frac{(n-p-1)!}{(n+p-1)!}\frac{\Gamma(p+1/2)}{2^p\sqrt\pi}.
$$
We can then reconstruct the moment generating function, which is a hypergeometric function:
$$F(x)=\sum\_{p=0}^\infty x^{2p}\mu\_{2p}={}\_2\!F\_2\left[\left\{\tfrac{1}{2},1\right\},\{1-n,n\},-\tfrac{1}{2}x^2\right].$$


