## 2 Capture Probabilities Under Different Fading Channels
### Rician Fading Channel  
The instantaneous received power in Rician fading channel is given by [31]  
\[f_{\gamma}(\gamma)=\frac{1+K}{\bar{\gamma}}e^{-K-\frac{\gamma(1+K)}{\bar{\gamma }}}I_{0}\left(2\sqrt{\frac{K(1+K)\gamma}{\bar{\gamma}}}\right),\gamma\geq 0, \tag{12}\]  
where the factor \(K\) corresponds to the ratio of the power of LOS component to the average power of scattered components and \(I_{0}(\cdot)\) is the modified Bessel function of the first kind, whose general form is given by  
\[I_{\alpha}(x)=\sum_{k=0}^{\infty}\frac{1}{k!\Gamma(k+\alpha+1)}\left(\frac{x} {2}\right)^{2k+\alpha}, \tag{13}\]  
where \(\Gamma(\cdot)\) is the gamma function defined as \(\Gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}dx\). According to [32] (p. 1026), i.e.,  
\[\left(\frac{x}{\beta}\right)^{\frac{n-1}{2}}I_{\mu-1}\left(2\sqrt{\beta x} \right)\leftrightarrow\frac{1}{s^{\mu}}e^{\frac{\beta}{s}}. \tag{14}\]  
Let \(\mu=0\), \(\beta=\frac{K(1+K)}{\bar{\gamma}}\) in (14) and apply the shifting property, we can obtain  
\[f_{\gamma}(\gamma)\leftrightarrow F_{\gamma}(s)=\frac{1+K}{\bar{\gamma}}e^{-K }\frac{1}{s+\frac{1+K}{\bar{\gamma}}}\exp\left[\frac{\frac{K(1+K)}{\bar{\gamma }}}{s+\frac{1+K}{\bar{\gamma}}}\right]. \tag{15}\]  
According to the convolution property of Laplace transform, the following expression can be easily obtained  
\[f_{\gamma}(\gamma)^{\otimes n-1}\leftrightarrow G_{\gamma_{n-1}}(s)=\left( \frac{1+K}{\bar{\gamma}}\right)^{n-1}e^{-(n-1)K}\left[\frac{1}{s+\frac{1+K}{ \bar{\gamma}}}\right]^{n-1}\exp\left[\frac{\frac{(n-1)K(1+K)}{\bar{\gamma}}}{ s+\frac{1+K}{\bar{\gamma}}}\right]. \tag{16}\]  
Taking advantage of the inverse Laplace transformation, (14) and [30] (p. 26), i.e., \(e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k! }\), we can obtain  
\[ \begin{split} G_{\gamma_{n-1}}(s)\leftrightarrow g_{\gamma_{n-1} }(\gamma_{n-1})=\left(\frac{1+K}{\bar{\gamma}}\right)^{\frac{n}{2}}\left[ \frac{\gamma_{n-1}}{(n-1)K}\right]^{\frac{n}{2}-1}e^{-(n-1)K-\frac{\gamma_{n- 1}(1+K)}{\bar{\gamma}}}\\ \cdot I_{n-2}\left(2\sqrt{\frac{K(1+K)(n-1)\gamma_{n-1}}{\bar{ \gamma}}}\right).\end{split}  \tag{17}\]  
Then, we substitute (12), (17) into (1) and utilize (13), (7) and [30] (p. 900), i.e.,  
\[\Gamma(k,x)=(k-1)!e^{-x}\sum_{j=0}^{k-1}\frac{x^{j}}{j!},k=1,2,\ldots \tag{18}\]Let \(\frac{1+K}{\gamma}=a\) for simplifying the writing, the capture probability in Rician fading channel can be written as  
\[ \begin{split} p_{cap}^{Ric}&=n\int_{0}^{\infty}ae^{-K-a \gamma_{t}}I_{0}\left(2\sqrt{aK\gamma_{t}}\right)\left\{\int_{0}^{\frac{\gamma_{ t}}{z}}a^{\frac{\gamma}{2}}\left[\frac{\gamma_{n-1}}{(n-1)K}\right]^{\frac{n}{2}-1} \right.\\ &\left.\cdot e^{-(n-1)K-a\gamma_{n-1}}I_{n-2}\left(2\sqrt{a(n-1)K \gamma_{n-1}}\right)d\gamma_{n-1}\right\}d\gamma_{t}\\ &=n\int_{0}^{\infty}ae^{-K-a\gamma_{t}}\sum_{j=0}^{\infty}\frac{ (aK\gamma_{t})^{j}}{(j!)^{2}}\left\{1-e^{-(n-1)K-\frac{\gamma_{t}}{z}}\sum_{i =0}^{\infty}\frac{[(n-1)K]^{i}}{i!}\sum_{k=0}^{n-2+i}\frac{(\frac{\sigma_{t} \gamma_{t}}{z})^{k}}{k!}\right\}d\gamma_{t}\\ &=n\left\{\int_{0}^{\infty}ae^{-K-a\gamma_{t}}\sum_{j=0}^{\infty} \frac{(aK\gamma_{t})^{j}}{(j!)^{2}}d\gamma_{t}\right.\\ &\left.-\int_{0}^{\infty}ae^{-nK-\frac{\sigma_{t}(z+1)}{z}}\sum_{ j=0}^{\infty}\frac{(aK\gamma_{t})^{j}}{(j!)^{2}}\sum_{l=0}^{\infty}\frac{[(n-1)K] ^{i}}{i!}\sum_{k=0}^{n-2+i}\frac{(\frac{\sigma_{t}\gamma_{t}}{z})^{k}}{k!}d \gamma_{t}\right\}\end{split}  \tag{19}\]  
Next, we calculate the two parts in the brace of (19), respectively. Let \(a\gamma_{t}=x\), the first part can be written as  
\[ \begin{split}\int_{0}^{\infty}ae^{-K-a\gamma_{t}}\sum_{j=0}^{ \infty}\frac{(aK\gamma_{t})^{j}}{(j!)^{2}}d\gamma_{t}&=e^{-K} \left[\int_{0}^{\infty}e^{-x}dx+K\int_{0}^{\infty}xe^{-x}dx+\frac{K^{2}}{(2!)^ {2}}\int_{0}^{\infty}x^{2}e^{-x}dx+\cdots\right]\\ &=e^{-K}\left[\Gamma(1)+K\Gamma(2)+\frac{K^{2}}{(2!)^{2}}\Gamma( 3)+\cdots\right]\\ &=e^{-K}\sum_{j=0}^{\infty}\frac{K^{j}}{j!}=e^{-K}e^{K}=1\end{split}  \tag{20}\]  
Let \(\frac{a\gamma_{t}(z+1)}{z}=y\), the second part in the brace can be written as  
\[