## 3 Problem Formulation
### Secrecy Outage Probability  
Considering that an eavesdropper has access to the signal provided by the source and according to [24], the secrecy capacity associated with the two fading channels can be obtained as  
\[C=\begin{cases}\ln\left(1+\gamma_{D}\right)-\ln\left(1+\gamma_{E}\right)& \gamma_{D}>\gamma_{E}\\ 0&\gamma_{D}\leq\gamma_{E}\end{cases}, \tag{16}\]  
where \(\gamma_{E}\) is the SNR of the link between the source and the eavesdropper. Therefore, the SOP is defined as the probability that the instantaneous secrecy capacity, \(C\), be less than or equal to a given capacity threshold, \(\ln\left(1+\gamma_{th}\right)\), which is expressed as  
\[\text{SOP}=Pr[\ln\frac{(1+\gamma_{D})}{(1+\gamma_{E})}\leq\ln\left(1+\gamma_{ th}\right)]=\int_{0}^{\infty}\int_{0}^{(1+\gamma_{E})(1+\gamma_{D})-1}f_{ \gamma_{E}}(w)f_{\gamma_{D}}(u)dudw, \tag{17}\]  
where \(Pr[.]\) denotes the probability of a random event.  
Considering a Nakagami-\(m\) distributed eavesdropper channel, the SOP can be obtained as follows:  
\[\text{SOP}=\int_{0}^{\infty}\int_{0}^{(\gamma_{th}+1)(x+1)-1}\frac{\beta^{ \alpha}(2m^{m}x^{2m-1})y^{\alpha-1}\exp(-\beta y)\exp(-\frac{mx^{2}}{\Omega}) }{\Gamma(\alpha)(\Omega^{m}\Gamma(m))}dydx \tag{18}\]  
Solving the first integral, the remaining expression becomes  
\[\text{SOP}=\int\limits_{0}^{\infty}\frac{2e^{-\frac{m^{2}}{\Omega}m}m^{m}x^{2 m-1}\Omega^{-m}[\Gamma(\alpha)-\Gamma(\alpha,\beta(x+\gamma_{th}(1+x)))]}{ \Gamma(m)\Gamma(\alpha)}dx, \tag{19}\]  
where the term \(\Gamma(\alpha)-\Gamma(\alpha,\beta(x+\gamma_{th}(1+x)))\) can be rewritten as function of the lower incomplete gamma function  
\[\gamma(s,x)\triangleq\int\limits_{0}^{x}t^{s-1}e^{-t}dt, \tag{20}\]  
considering that \(\Gamma(s)=\gamma(s,x)+\Gamma(s,x)\).  
Representing the exponential in terms of power series, it follows that the incomplete Gamma function can be written as  
\[\int\limits_{0}^{x}t^{s-1}e^{-t}dt=\int\limits_{0}^{x}\sum\limits_{k=0}^{ \infty}(-1)^{k}\frac{t^{s+k-1}}{k!}dt=x^{s}\sum\limits_{k=0}^{\infty}\frac{(- x)^{k}}{k!(s+k)}. \tag{21}\]  
Applying the expansion (21) in (19), and using the result obtained by the reference [22] in its table of integrals for integrands of type \(x^{\alpha}e^{-px^{2}}\gamma(\
u,cx)\), the SOP can be rewritten as (22), where \({}_{p}\bar{F}_{q}\) is the regularized \({}_{p}F_{q}\) hypergeometric function and \(v=\frac{1+\gamma_{th}}{\gamma_{th}}\). Althoughthis expression is an infinite sum, it is possible to verify, in Section 4, that the error is small if only the first term is considered to compute the SOP:  
\[\begin{array}{l}\text{SOP}=\sum\limits_{k=0}^{\infty}\frac{(-1)^{k}p^{\alpha +k}z_{k}+k}{\Gamma(\alpha)\Gamma(k+1)}\\ \times(\frac{\pi m^{m}2^{-\alpha-k}-p^{-2m}\Gamma(m+\frac{1}{2})\csc(\pi(a+k+2 m))z_{2}(m,m+\frac{1}{2};\frac{1}{2}(k+2m+\alpha+1),\frac{1}{2}(k+2m+\alpha+2);-\frac{m}{ 2\Gamma})}{\Gamma(-k+\alpha+1)}\\ +\frac{\pi^{3/2}\Omega^{\frac{k+1}{2}}\frac{1}{2}(-k-\alpha)v^{\alpha+k}}{2 \Gamma(m)}\big{(}\frac{2\csc(\frac{1}{2}\pi(a+k+2m))z_{2}(\frac{1}{2}(-k- \alpha),\frac{1}{2}(-k-\alpha+1);\frac{1}{2}(-k-2m-\alpha+2);-\frac{m}{2\Gamma })}{a+k}\\ -\frac{\sqrt{m}\sec(\frac{1}{2}\pi(a+k+2m))z_{2}(\frac{1}{2}(-k-\alpha+1), \frac{1}{2}(-k-\alpha+2);\frac{1}{2}(-k-2m-\alpha+3);-\frac{m}{2\Gamma})}{v \sqrt{\Omega}})),\end{array} \tag{22}\]