How to find the expectation $\mathbb{E} \left[ \frac{|h|^4}{|h+w|^2} \right]$? Given the independent and complex Gaussian random variables $h$ and $w$, how does one can find the following expectation?
$$\mathbb{E} \left[    \frac{|h|^4}{|h+w|^2}    \right] = \int_{\mathbb{C}}\int_{\mathbb{C}}{\frac{|h|^4}{|h+w|^2}} f(h)g(w)d_{h}d_{w},$$ 
where $h \sim \mathcal{CN}\left(0,d\right)$ and $w \sim \mathcal{CN}\left(0,p\right)$. The pdf of $h$ and $w$ are defined as
$$f(h) = \frac{1}{\pi d} \text{e}^{-\frac{|h|^2}{d}},$$
$$g(w) = \frac{1}{\pi p} \text{e}^{-\frac{|w|^2}{p}}.$$
I've tried to change the variables like: $|h|^2 = r^2$ and $|w|^2 = n^2$, however, I could not apply this change of variables to the denominator: $|h + w|^2$. Note that $h$ and $w$ are complex random variables that can be written in rectangular form like: $h = a + i*b$ and $w = c + i*d$, where $i = \sqrt{-1}$.
$\textbf{UPDATE}$: After running some simulations, it seems as $p \to \infty$ the expectation above tends to $d$. It can be checked with the following matlab simulation script: https://pastebin.com/U48fcMZ9
 A: The expectation is infinite.
One way to see this is to condition on $H$.  Preliminary changes of variable (merely involving rescaling $H$ and $W$ and then shifting to a new origin) reduce the conditional expectation to a positive constant times a two-dimensional integral of the form
$$\mathcal{I}(\lambda)=\iint_{\mathbb{C}}\ \frac{1}{|z|^2} e^{-\lambda |z-1|^2}\ dz d\bar z$$
with $\lambda \gt 0.$
In polar coordinates $(r,\theta),$ $|z|^2 = r^2$ and $|z-1|^2 = r^2 - 2r\cos(\theta)+1,$ and the area element is $dzd\bar{z} = r dr d\theta,$ giving
$$\mathcal{I}(\lambda) = e^{-\lambda}\int_0^{2\pi}d\theta \int_0^\infty \frac{1}{r^2}e^{-\lambda(r^2 - 2r\cos\theta)}\ r\, drd.$$
For $0 \le r \le \sqrt{1 + 1/\lambda} -1 = u(\lambda)\gt 0,$ the expression in the exponent exceeds $-1,$ so we may underestimate this integral by replacing the exponential by $e^{-1}$ and limiting $r$ to this range:
$$\mathcal{I}(\lambda) \ge e^{-\lambda-1}\int_0^{2\pi}d\theta \int_0^{u(\lambda)}\frac{1}{r}dr = 2\pi e^{-\lambda-1} \lim_{\epsilon\to 0} \int_\epsilon^{u(\lambda)} \frac{dr}{r}\ \propto\ \lim_{\epsilon\to 0}\log(u(\lambda)) - \log(\epsilon),$$
which diverges to $+\infty.$
Since all conditional expectations are infinite, the expectation must be infinite.
A simulation bears this out.  For simplicity I chose $H$ and $W$ to have independent standard (Complex) Normal normal distributions, generated twenty million realizations $(h,w),$ and computed the running mean of $|h|^4/|h+w|^2.$  The periodic large jumps are characteristic of a divergent expectation: no matter how far out you run this simulation, these jumps will recur (whenever a tiny value of $|w+h|$ is generated compared to $|h|^2$) and its mean will never converge.

This plot shows the running mean "Mean" as a function of the number of simulated values "N" for $n=10^4$ through $n=2\times 10^7.$  Colors highlight the largest jumps.  Evidently one could be fooled by relying on a simulation to estimate the mean: notice how the purple segment from $N\approx 508,000$ to $N\approx 9,300,000$ seems to settle down--only to be followed by a large jump.  This indicates that the simulation-based estimate depends entirely on when you choose to end the simulation.
