## 2 Capture Probabilities Under Different Fading Channels
### Rayleigh Fading Channel  
The Rayleigh fading channel is frequently used to model multi-path fading with no direct line-of-sight (LOS) path. The instantaneous received power in Rayleigh channel is given by [29]  
\[f_{\gamma}(\gamma)=\dfrac{1}{\gamma}e^{-\frac{\gamma}{\gamma}},\gamma\geq 0. \tag{2}\]  
According to the theorem of Laplace transform, we can obtain  
\[f_{\gamma}(\gamma)\leftrightarrow F_{\gamma}(s)=\dfrac{1}{\gamma}\dfrac{1}{s+ \frac{1}{\gamma}}. \tag{3}\]  
To do the convolution of \(n-1\) independent and identically distributed (i.i.d.) signals, we apply the property of Laplace transform. Then, we have  
\[f_{\gamma}(\gamma)^{\otimes n-1}\leftrightarrow G_{\gamma_{n-1}}(s)=\dfrac{1} {\gamma^{n-1}}\dfrac{1}{(s+\frac{1}{\gamma})^{n-1}}. \tag{4}\]By applying the inverse Laplace transformation, we can obtain  
\[G_{\gamma_{n-1}}(s)\leftrightarrow g_{\gamma_{n-1}}(\gamma_{n-1})=\frac{\gamma_{n -1}^{n-2}}{\tilde{\gamma}^{n-1}(n-2)!}e^{-\frac{\gamma_{n-1}}{\tilde{\gamma}}}. \tag{5}\]  
After substituting (2) and (5) into (1), the capture probability in Rayleigh fading channel can be written as  
\[\begin{split} p_{cap}^{Ray}&=n\int_{0}^{\infty} \frac{1}{\tilde{\gamma}}e^{-\frac{\gamma_{t}}{\tilde{\gamma}}}\left[\int_{0}^{ \frac{\gamma_{t}}{\tilde{\gamma}}}\frac{\gamma_{n-1}^{n-2}}{\tilde{\gamma}^{n- 1}(n-2)!}e^{-\frac{\gamma_{n-1}}{\tilde{\gamma}}}d\gamma_{n-1}\right]d\gamma_{ t}\\ &=\frac{n}{\tilde{\gamma}^{n}(n-2)!}\int_{0}^{\infty}e^{-\frac{ \gamma_{t}}{\tilde{\gamma}}}\left[\int_{0}^{\frac{\gamma_{t}}{\tilde{\gamma}}} \gamma_{n-1}^{n-2}e^{-\frac{\gamma_{n-1}}{\tilde{\gamma}}}d\gamma_{n-1}\right] d\gamma_{t}.\end{split} \tag{6}\]  
According to [30] (p. 900), i.e.,  
\[\gamma(k,x)=(k-1)!\left[1-e^{-x}\sum_{j=0}^{k-1}\frac{x^{j}}{j!}\right],k=1,2,\ldots \tag{7}\]  
where \(\gamma(\cdot,\cdot)\) is the lower incomplete gamma function defined as \(\gamma(\alpha,x)=\int_{0}^{\infty}y^{\alpha-1}e^{-y}dy\), we have  
\[\begin{split}\int_{0}^{\frac{\gamma_{t}}{\tilde{\gamma}}}\gamma_{ n-1}^{n-2}e^{-\frac{\gamma_{n-1}}{\tilde{\gamma}}}d\gamma_{n-1}& =\tilde{\gamma}^{n-1}\int_{0}^{\frac{\gamma_{t}}{\tilde{\gamma}} }\left(\frac{\gamma_{n-1}}{\tilde{\gamma}}\right)^{n-2}e^{-\frac{\gamma_{n-1 }}{\tilde{\gamma}}}d\left(\frac{\gamma_{n-1}}{\tilde{\gamma}}\right)\\ &=\tilde{\gamma}^{n-1}\gamma\left(n-1,\frac{\gamma_{t}}{z\gamma} \right)\\ &=\tilde{\gamma}^{n-1}(n-2)!\left[1-e^{-\frac{\gamma_{t}}{z\gamma }}\sum_{k=0}^{n-2}\frac{\left(\frac{\gamma_{t}}{z\gamma}\right)^{k}}{k!} \right].\end{split} \tag{8}\]  
After substituting (8) into (6), we can obtain  
\[\begin{split} p_{cap}^{Ray}&=n\left[\frac{1}{\tilde{ \gamma}}\int_{0}^{\infty}e^{-\frac{\gamma_{t}}{\tilde{\gamma}}}d\gamma_{t}- \frac{1}{\tilde{\gamma}}\int_{0}^{\infty}e^{-\frac{\gamma_{t}}{\tilde{\gamma}} \frac{z+1}{z}}\sum_{k=0}^{n-2}\frac{\left(\frac{\gamma_{t}}{z\gamma}\right)^{k }}{k!}d\gamma_{t}\right]\\ &=n\left[1-\frac{1}{\tilde{\gamma}}\int_{0}^{\infty}e^{-(z+1) \frac{\gamma_{t}}{\tilde{\gamma}}}\sum_{k=0}^{n-2}\frac{\left(\frac{\gamma_{t }}{z\gamma}\right)^{k}}{k!}d\gamma_{t}\right].\end{split} \tag{9}\]  
Let \(\frac{\gamma_{t}}{z\gamma}=u\) and then  
\[\begin{split}\frac{1}{\tilde{\gamma}}\int_{0}^{\infty}e^{-(z+1) \frac{\gamma_{t}}{z\gamma}}\sum_{k=0}^{n-2}\frac{\left(\frac{\gamma_{t}}{z \gamma}\right)^{k}}{k!}d\gamma_{t}&=z\left[\int_{0}^{\infty}e^{-( z+1)u}du+\int_{0}^{\infty}e^{-(z+1)u}udu+\cdots\\ &\qquad\qquad\qquad\qquad+\int_{0}^{\infty}e^{-(z+1)u}\frac{u^{ n-2}}{(n-2)!}du\right]\\ &=z\left[\frac{1}{z+1}+\frac{1}{(z+1)^{2}}+\cdots+\frac{1}{(z+1)^{ n-1}}\right]\\ &=z\sum_{k=0}^{n-2}\frac{1}{(z+1)^{k+1}}.\end{split} \tag{10}\]Then, substituting (10) into (9), the capture probability can be written as  
\[p_{cap}^{Ray}=n\left(1-z\sum_{k=0}^{n-2}\frac{1}{(z+1)^{k+1}}\right). \tag{11}\]