focus on $m=1$, i.e. the Rayleigh fading, and present a closed-form approximation of the success probability. In section \[simsec\], we will evaluate tightness of this expression.

\[t1\] For $m=1$, success probability in packet transmission can be approximated as: $$\begin{aligned}
&\text{p}_{\text{s}}(i,{\bf z})\approx \text{P}_{{\text{\tiny N} }} \big[\exp\big(-\sum\limits_{j\in\{1,2\}} \sum\limits_{k\in\mathcal K}\lambda_k\hat \upsilon_{k,j} \text{H}({\bf z},1, \frac{Q_j P_k\gamma_{\text{th}}}{\Omega  P_i})\big)\big]\nonumber\\
&\hspace{1cm}\times\exp\big(-\sum\nolimits_{j\in\{1,2\}}{\hat\upsilon_{i,j}}  \text{H}({\bf z},\text{f}^*({\bf x}), \frac{Q_j\gamma_{\text{th}}}{\Omega })\big),\label{ps}\end{aligned}$$ where $\text{f}^*(\cdot)=\text{conv}\big(\text{f}(\cdot),\text{f}(\cdot)\big)$, $$\begin{aligned}
  \text{H}\big({\bf z},\text{f}^*({\bf x}), \xi)&=\int\nolimits_{x\in\mathbb R^2}\frac{\text{g}({\bf x})}{\text{g}({\bf  x})+\text{g}({\bf z})/\xi}\text{f}^*({\bf x})d {\bf x}\label{hf},\\
   \text{P}_{{\text{\tiny N} }}&=\exp\big(-\mathcal N\gamma_{\text{th}}/[\Omega  P_i \text{g}({\bf z})] \big)\label{pn},
  \end{aligned}$$ and $\mathcal N$ is the noise power.

The proof is given in theorem 3.2 of the extended version [@opd].

$\text{H}({\bf z},\text{f}^*({\bf x}),\xi)$ and $\text{H}({\bf z},1,\xi)$ could be derived in closed-form for most well-known pathloss and distribution functions, as follows.

For $\text{g}({\bf x})=\alpha||{\bf x}||^{-\delta}$, $$\label{hfd} \text{H}({\bf z},1,\xi)=  ||{\bf z}||^2 \xi^{\frac{2}{\delta}} 2\pi^{2} \text{csc}({2\pi/\delta})/\delta.$$

By change of coordinates, ${\bf x}\to (r,\theta)$, we have: $$\begin{aligned}
\text{H}\big({\bf z},1, \xi)&=\int\nolimits_{x\in\mathbb R^2}\frac{\alpha{||\bf x||}^{-\delta}}{\alpha{||\bf x||}^{-\delta}+\alpha{||\bf z||}^{-\delta}/\xi} d {\bf x},\nonumber\\
&=2\pi\int\nolimits_{0}^{\infty}\frac{1}{1+(r/||{\bf z}||)^\delta/\xi}  { r  d r}.\nonumber\end{aligned}$$ Solving this integral by using [@seri Eq. 3.352], is derived. [ &lt;1.5em - 1.5em plus0em minus0.5em height0.75em width0.5em depth0.25em]{}

\[cne\] For $\text{g}({\bf x})=\alpha||{\bf x}||^{-4}$, and $\text{f}({\bf x})$ given in , $$\begin{aligned}
 \text{H}({\bf z},\text{f}^*({\bf x}),\xi)=&  \frac{||{\bf z}||^2  }{4 \sigma^2\sqrt\xi    }\bigg[\text{ci}(\frac{ ||{\bf z}||^2 