# Energy Harvesting over Rician Fading Channel:
## Appendix D Proof of Theorem 4  
The CDF of \(\gamma_{1}\) as high SNR regime can be written as  
\[F_{\gamma_{1}^{\infty}}(x)=\Pr(\gamma_{1}^{\infty}<x)=\Pr\left(\frac{X_{2}}{ aX_{2}+aX_{1}}<x\right)= \begin{cases}\Pr\left(X_{2}<\frac{axX_{1}}{1-ax}\right),& \text{if }x\leqslant\frac{1}{a},\\ 1,&\text{if }x>\frac{1}{a}.\end{cases}  \tag{20}\]  
Now, we can rewrite the formula of SER by substituting Equation (20) into Equation (24)\[SER_{1}^{\infty}=\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\int\limits_{0}^{\infty} \frac{e^{-\theta x}}{\sqrt{x}}F_{\gamma_{1}}^{\infty}(x)dx=\frac{\omega\sqrt{ \theta}}{2\sqrt{\pi}}\int\limits_{0}^{1/a}\frac{e^{-\theta x}}{\sqrt{x}}F_{ \gamma_{1}}^{\infty}(x)dx+\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\int\limits_{1 /a}^{\infty}\frac{e^{-\theta x}}{\sqrt{x}}dx. \tag{111}\]  
For \(x\leq\frac{1}{a}\), we can apply the result of Theorem 3 with \(\tilde{\gamma}_{th}=\frac{x}{1-ax}\) to get  
\[ \begin{split} F_{\gamma_{1}}^{\infty}(x)&=\Pr( \gamma_{1}^{\infty}<x)=1-\zeta_{1}e^{-2K\sum\limits_{l=0}^{\infty}\sum\limits_ {k=0}^{\infty}\sum\limits_{m=0}^{l}\frac{K^{l+k}\zeta_{1}^{k}\zeta_{2}^{m}}{ l!m!(k!)^{2}}\left(a\tilde{\gamma}_{th}(x)\right)^{m}\frac{\Gamma(m+k+1)}{[ \zeta_{2}a\tilde{\gamma}_{th}(x)+\zeta_{1}]^{m+k+1}}\\ &=1-\zeta_{1}e^{-2K\sum\limits_{l=0}^{\infty}\sum\limits_{k=0}^{ \infty}\sum\limits_{m=0}^{l}\frac{K^{l+k}\zeta_{1}^{k}\zeta_{2}^{m}a^{m}}{l!m!( k!)^{2}}\frac{x^{m}(1-ax)^{k+1}\Gamma(m+k+1)}{[(\zeta_{2}-\zeta_{1})ax+\zeta_{1}]^{m +k+1}}\\ &=1-\zeta_{1}e^{-2K\sum\limits_{l=0}^{\infty}\sum\limits_{k=0}^{ \infty}\sum\limits_{m=0}^{l}\sum\limits_{v=0}^{k+1}\left(\begin{array}{c}k+1 \\ v\end{array}\right)\frac{K^{l+k}\zeta_{1}^{k}\zeta_{2}^{m}a^{m+v}(-1)^{v}x^{m+v }\Gamma(m+k+1)}{l!m!(k!)^{2}[(\zeta_{2}-\zeta_{1})ax+\zeta_{1}]^{m+k+1}}\\ &=1-\zeta_{1}e^{-2K\sum\limits_{l=0}^{\infty}\sum\limits_{k=0}^{\infty}\sum \limits_{m=0}^{l}\sum\limits_{v=0}^{k+1}\left(\begin{array}{c}k+1\\ v\end{array}\right)\frac{K^{l+k}\zeta_{2}^{m}a^{m+v}(-1)^{v}x^{m+v}\Gamma(m+k +1)}{l!m!(k!)^{2}[(\zeta_{2}-\zeta_{1})^{m+k+1}]^{m+k+1}}.\end{split}  \tag{112}\]  
The aymptotic SER is now obtained by substituting Equation (112) into Equation (111):  
\[ \begin{split}& SER_{1}^{\infty}=\frac{\omega\sqrt{\theta}}{2 \sqrt{\pi}}\int\limits_{0}^{\infty}\frac{e^{-\theta x}}{\sqrt{x}}dx\\ &-\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\zeta_{1}e^{-2K\sum\limits _{l=0}^{\infty}\sum\limits_{k=0}^{\infty}\sum\limits_{m=0}^{l}\sum\limits_{v=0 }^{k+1}\left(\begin{array}{c}k+1\\ v\end{array}\right)\frac{K^{l+k}\zeta_{1}^{k}\zeta_{2}^{m}a^{m-k-1}(-1)^{v} \Gamma(m+k+1)}{l!m!(k!)^{2}(\zeta_{2}-\zeta_{1})^{m+k+1}}\int\limits_{0}^{ \infty}\frac{x^{m+v-\frac{1}{2}}e^{-\theta x}dx}{\left[x+\frac{\zeta_{1}}{( \zeta_{2}-\zeta_{1})ax}\right]^{m+k+1}}\\ &=\frac{\omega}{2}-\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\zeta_{1}e^{-2K \sum\limits_{l=0}^{\infty}\sum\limits_{k=0}^{\infty}\sum\limits_{m=0}^{l}\sum \limits_{v=0}^{k+1}\left(\begin{array}{c}k+1\\ v\end{array}\right)\frac{K^{l+k}\zeta_{1}^{k}\zeta_{2}^{m}a^{m-k-1}(-1)^{v} \Gamma(m+k+1)}{l!m!(k!)^{2}(\zeta_{2}-\zeta_{1})^{m+k+1}}\int\limits_{0}^{1/a} \frac{x^{m+v-\frac{1}{2}}e^{-\theta x}dx}{\left[x+\frac{\zeta_{1}}{(\zeta_{2} -\zeta_{1})ax}\right]^{m+k+1}}.\end{split}  \tag{113}\]  
We consider three cases:  
**Case 1:**\(\xi_{2}=\xi_{1}\). Equation (113) can be rewritten as  
\[