## Appendix C Proof of Theorem 1 in (36) and (37)  
From Equation (27), we have,  
\[\gamma_{IoT_{R}}=(1-\varepsilon)\delta X\text{ where }|h_{IoT_{R}}|^{2}=X\]  
Also, from Equation (29), we have,  
\[\gamma_{s_{src}}=\frac{\phi_{1}P_{IoT_{R}}|h_{s_{src}}|^{2}}{\sigma_{s_{src}}^ {2}}\triangleq\delta XYa\]  
where  
\[Y=|h_{s_{src}}|^{2},\sigma_{s_{src}}^{2}=1,a=\eta\varepsilon\phi_{1}\]  
From Equation (31), the outage probability of the source is:  
\[\begin{array}{l}P_{Out_{S}}=Pr(min(\gamma_{IoT_{R}},\gamma_{s_{src}})<\psi) \\ =1-Pr(min(\gamma_{IoT_{R}},\gamma_{s_{src}})\geq\psi)\\ =1-Pr((1-\varepsilon)\delta X\geq\psi,\delta aXY\geq\psi)\\ =1-Pr\big{(}X\geq\frac{\psi}{(1-\varepsilon)\delta},Y\geq\frac{\psi}{\delta aX }\big{)}\end{array}\]  
Let  
\[\begin{array}{l}x_{0}=\frac{\psi}{(1-\varepsilon)\delta}\\ =1-Pr\big{(}X\geq x_{0},Y\geq\frac{(1-\varepsilon)x_{0}}{aX}\big{)}\\ =1-\int_{x_{0}}^{\infty}f_{X}(x)\big{(}\int_{(1-\varepsilon)x_{0}}^{\infty}f_ {Y}(y)dy\big{)}dx\\ =1-\int_{x_{0}}^{\infty}\lambda_{h}e^{-\lambda_{h}x}\big{(}\frac{\delta x_{0} }{\frac{\delta x_{0}}{\delta x_{0}}}\lambda_{g}e^{-\lambda_{g}y}dy\big{)}dx\\ =1-\int_{x_{0}}^{\infty}\lambda_{h}e^{-\lambda_{h}x}\big{(}e^{-\lambda_{h}^{ \frac{1-\varepsilon)x_{0}}{\delta x}}}\big{)}dx\\ =1-\int_{x_{0}}^{\infty}\lambda_{h}\big{(}e^{-\lambda_{h}x-\lambda_{g}}\frac{( 1-\varepsilon)x_{0}}{ax}\big{)}dx\\ =1-\left(\underbrace{\lambda_{h}\int_{x=0}^{\infty}\big{(}e^{-\lambda_{h}x}- \lambda_{h}x\big{)}dx}_{I_{1}}-\underbrace{\lambda_{h}\int_{x=0}^{x_{0}}\big{(} e^{-\lambda_{g}\frac{(1-\varepsilon)x_{0}}{ax}-\lambda_{h}x}\big{)}dx}_{I_{2}} \right)\end{array}\]  
Let us first evaluate the integral \(I_{1}\) by using the formula [36], Equation 3.324.1)  
\[\begin{array}{l}\int_{0}^{\infty}e^{-\frac{\beta}{4x}-\gamma x}dx=\sqrt{ \frac{\beta}{\gamma}}K_{1}\big{(}\sqrt{\beta\gamma}\big{)}\\ I_{1}=\lambda_{h}\sqrt{\frac{4\lambda_{g}(1-\varepsilon)x_{0}}{a\lambda_{h}}}K_ {1}\bigg{(}\sqrt{\frac{4\lambda_{g}(1-\varepsilon)x_{0}\lambda_{h}}{a}}\bigg{)} \\ I_{1}=2\sqrt{\frac{\lambda_{h}\lambda_{g}(1-\varepsilon)x_{0}}{a}}K_{1}\bigg{(} 2\sqrt{\frac{\lambda_{h}\lambda_{g}(1-\varepsilon)x_{0}}{a}}\bigg{)}\end{array}\]  
Now, let us evaluate the integral \(I_{2}\)  
\[I_{2}=\lambda_{h}\int_{x=0}^{x_{0}}\big{(}e^{-\lambda_{h}x-\frac{(1-\varepsilon )\delta x_{0}}{ax}}\big{)}dx\]  
Expanding the term \(e^{-\lambda_{h}x}\) in Taylor series  
\[=\lambda_{h}\sum_{n=0}^{\infty}\frac{(-1)}{n!}(\lambda_{h})^{n}\int_{x=0}^{x_{ 0}}x^{n}e^{-\frac{(1-\varepsilon)x_{0}x_{0}}{ax}}dx\]Substituting \(y=\frac{1}{\varepsilon}\), and further \(t=x_{0}y\), we get,  
\[=\sum_{n=0}^{\infty}\frac{(-1)}{n!}(\lambda_{h}x_{0})^{n+1}\int_{t=1}^{\infty}t^ {-n-2}e^{-\frac{(1-t)\lambda_{g}t}{a}}dt\]  
Now, by definition of exponential integral of order \(n\), we have  
\[\begin{array}{l}E_{n}(a)=\int_{y=1}^{\infty}y^{-n}e^{-ay}dy\\ I_{2}=\sum_{n=0}^{\infty}\frac{(-1)}{n!}(\lambda_{h}x_{0})^{n+1}E_{n+2}\Big{(} \frac{(1-\varepsilon)\lambda_{g}}{a}\Big{)}\end{array}\]  
Therefore,  
\[P_{Out_{S}}=1-2\sqrt{\frac{\lambda_{h}\lambda_{g}(1-\varepsilon)x_{0}}{a}}K_{ 1}\bigg{(}2\sqrt{\frac{\lambda_{h}\lambda_{g}(1-\varepsilon)x_{0}}{a}}\bigg{)} +\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}(\lambda_{h}x_{0})^{n+1}E_{n+2}\Big{(} \frac{(1-\varepsilon)\lambda_{g}}{a}\Big{)}\]  
Putting the value of \(P_{Out_{S}}\) in Equation (33), we get,  
\[Thr_{S}=\frac{R}{2}\bigg{(}2\sqrt{\frac{\lambda_{h}\lambda_{g}(1-\varepsilon) x_{0}}{a}}K_{1}\bigg{(}2\sqrt{\frac{\lambda_{h}\lambda_{g}(1-\varepsilon)x_{0}}{a}} \bigg{)}-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}(\lambda_{h}x_{0})^{n+1}E_{n+2} \Big{(}\frac{(1-\varepsilon)\lambda_{g}}{a}\Big{)}\bigg{)}\]  
This ends the proof of Theorem 3.