## Appendix A Proof of Theorem 1 in (16) and (17)  
From Equation (7) we have,  
\[\hat{\gamma}_{IoT_{R}}=\delta X\text{ where, }|h_{IoT_{R}}|^{2}=X\]  
Also, from Equation (9), we have,  
\[\hat{\gamma}_{s_{rec}}=\frac{\phi_{1}\hat{P}_{IoT_{R}}|h_{s_{rec}}|^{2}}{\sigma _{s_{rec}}^{2}}\triangleq\delta XYk\]  
where  
\[Y=|h_{s_{rec}}|^{2},\sigma_{s_{rec}}^{2}=1,k=\frac{2\alpha\eta P_{s}}{(1- \alpha)}\]  
From Equation (11), the outage probability of the source is:  
\[\begin{array}{l}\hat{P}_{OutS}=Pr(min(\hat{\gamma}_{IoT_{R}},\hat{\gamma}_{s _{rec}})<\hat{\psi})\\ =1-Pr(min(\hat{\gamma}_{IoT_{R}},\hat{\gamma}_{s_{rec}})\geq\hat{\psi})\\ =1-Pr(\hat{\delta}X\geq\hat{\psi},\hat{\delta}XY\geq\hat{\psi})\\ =1-Pr(X\geq\frac{\hat{\psi}}{\delta},Y\geq\frac{\hat{\psi}}{\delta kX})\end{array}\]Let  
\[\begin{array}{l}x_{0}=\frac{\dot{\Psi}}{\delta},\\ =1-Pr\big{(}X\geq x_{0},Y\geq\frac{x_{0}}{K\lambda}\big{)}\\ =1-\int_{x_{0}}^{\infty}\int_{X}(x)\left(\int_{x_{0}}^{\infty}f_{Y}(y)dy\right) dx\\ =1-\int_{x_{0}}^{\infty}\lambda_{h}e^{-\lambda_{h}x}\big{(}\int_{x_{0}}^{ \infty}\lambda_{g}e^{-\lambda_{g}y}dy\big{)}dx\\ =1-\int_{x_{0}}^{\infty}\lambda_{h}e^{-\lambda_{h}x}\big{(}e^{-\lambda_{g} \frac{x_{0}}{K\lambda}}\big{)}dx\\ =1-\int_{x_{0}}^{\infty}\lambda_{h}\big{(}e^{-\lambda_{h}x-\lambda_{g}\frac{x_ {0}}{K\lambda}}\big{)}dx\\ =1-\left(\underbrace{\lambda_{h}\int_{x=0}^{\infty}\big{(}e^{-\lambda_{h}x} \frac{x_{0}}{K\lambda}-\lambda_{h}x}_{I_{1}}\big{)}dx}_{I_{1}}-\underbrace{ \lambda_{h}\int_{x=0}^{x_{0}}\big{(}e^{-\lambda_{g}\frac{x_{0}}{K\lambda}- \lambda_{h}x}\big{)}dx}_{I_{2}}\right)\end{array}\]  
Let us first evaluate the integral \(I_{1}\) by using the formula (36), Equation 3.324.1)  
\[\begin{array}{l}\int_{0}^{\infty}e^{-\frac{\dot{\Psi}}{\delta}-\gamma x}dx= \sqrt{\frac{\dot{\Psi}}{\gamma}}K_{1}\sqrt{\dot{\Psi}\gamma}\\ I_{1}=\lambda_{h}\sqrt{\frac{4\lambda_{g}x_{0}}{K\lambda_{h}}}K_{1}\bigg{(} \sqrt{\frac{4\lambda_{g}x_{0}\lambda_{h}}{k}}\bigg{)}\\ I_{1}=2\sqrt{\frac{\lambda_{h}\lambda_{g}x_{0}}{k}}K_{1}\bigg{(}2\sqrt{\frac{ \lambda_{h}\lambda_{g}x_{0}}{k}}\bigg{)}\end{array}\]  
Now, let us evaluate the integral \(I_{2}\)  
\[I_{2}=\lambda_{h}\int_{x=0}^{x_{0}}\big{(}e^{-\lambda_{h}x-\frac{\lambda_{g}x _{0}}{K\lambda}}\big{)}dx\]  
Expanding the term \(e^{-\lambda_{h}x}\) in Taylor series  
\[=\lambda_{h}\sum_{n=0}^{\infty}\frac{(-1)}{n!}(\lambda_{h})^{n}\int_{x=0}^{x_ {0}}x^{n}e^{-\frac{\lambda_{g}x_{0}}{K\lambda}}dx\]  
Substituting \(y=\frac{1}{x}\to dx=-\frac{1}{y^{2}}dy\)  
\[=\sum_{n=0}^{\infty}\frac{(-1)}{n!}(\lambda_{h})^{n+1}\int_{y=\frac{1}{x_{0}}} ^{\infty}y^{-n-2}e^{-\frac{\lambda_{g}x_{0}y}{K\lambda}}dy\]  
Now, substituting further \(t=x_{0}y\to dt=x_{0}dy\)  
\[=\sum_{n=0}^{\infty}\frac{(-1)}{n!}(\lambda_{h}x_{0})^{n+1}\int_{t=1}^{\infty} t^{-n-2}e^{-\frac{\lambda_{g}x_{0}^{t}}{K}}dt\]  
Now, by definition of exponential integral of order \(n\), we have,  
\[\begin{array}{l}E_{n}(a)=\int_{y=1}^{\infty}y^{-n}e^{-ay}dy\\ I_{2}=\sum_{n=0}^{\infty}\frac{(-1)}{n!}(\lambda_{h}x_{0})^{n+1}E_{n+2}\Big{(} \frac{\lambda_{g}}{K}\Big{)}\end{array}\]  
Therefore,  
\[\hat{P}_{Out_{S}}=1-I_{1}+I_{2}\]  
Putting the value of \(\hat{P}_{Out_{S}}\) in Equation (13), we get,\[\hat{Thr}_{S}=\frac{R(1-\alpha)}{2}\left(2\sqrt{\frac{\lambda_{h}\lambda_{g}x_{0}}{k }}K_{1}\Big{(}2\sqrt{\frac{\lambda_{h}\lambda_{g}x_{0}}{k}}\Big{)}-\sum_{n=0}^{ \infty}\frac{(-1)^{n}}{n\,!}(\lambda_{h}x_{0})^{n+1}E_{n+2}\Big{(}\frac{\lambda _{g}}{k}\Big{)}\right)\]  
This ends the proof of Theorem 1.