# Energy Harvesting over Rician Fading Channel:
## Appendix D Proof of Theorem 4
\begin{split}& SER_{1}^{\infty}=\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+v}(-1)^{v} \Gamma(m+k+1)}{l!m!(k!)^{2}\zeta_{1}^{m+1}}\int\limits_{0}^{1/a}x^{m+v-\frac{ 1}{2}}e^{-\theta x}dx\\ &=\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_{v=0}^{l}\sum \limits_{v=0}^{l+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} \Gamma(m+k+1)}{l!m!(k!)^{2}\zeta_{1}^{m+1}}\theta^{-m-v-\frac{1}{2}}\gamma \left(m+v+\frac{1}{2},\frac{\theta}{a}\right),\end{split}  \tag{114}\]  
where \(\gamma(a,z)\triangleq\int_{0}^{z}e^{-t}t^{a-1}dt\) is the incomplete gamma function as defined in Equation (8.350.1) in [32] and the last equality comes from the Formula (3.381.1) in [32]. **Case 2:**\(\xi_{2}<\xi_{1}\). For all \(x\in[0,\frac{1}{a}]\), we have \(x<\frac{1}{a}.\frac{\xi_{1}}{\xi_{1}-\xi_{2}}\). Hence, we can apply the Taylor's series expansion of \(\frac{1}{\left[1-\frac{(\xi_{1}-\xi_{2})ax}{\xi_{1}}\right]^{m+k+1}}\) as in [37]:  
\[\frac{1}{\left[1-\frac{(\xi_{1}-\xi_{2})ax}{\xi_{1}}\right]^{m+k+1}}=\sum \limits_{p=0}^{\infty}\frac{(m+k+1)_{p}}{p!}\left[\frac{(\xi_{1}-\xi_{2})ax}{ \xi_{1}}\right]^{p}, \tag{115}\]where \((\cdot)_{p}\) is the Pochhammer symbol, which is defined as  
\[(\alpha)_{p}\triangleq \begin{cases}1,\quad\text{if }p=0,\\ \alpha(\alpha+1)...(\alpha+p-1),\quad\text{if }p\
eq 0.\end{cases}  \tag{101}\]  
Now, we can obtain the following equation by substituting Equation (100) into Equation (100) and applying (3.381.1) in [32]:  
\[SER_{1}^{\infty}=\frac{\omega}{2}\] \[-\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\xi_{1}e^{-2K}\sum_{l=0}^ {\infty}\sum_{k=0}^{\infty}\sum_{m=0}^{l}\sum_{v=0}^{k+1}\left( \begin{array}[] {c}k+1\\ v\end{array} \right)\frac{K^{l+k}\xi_{1}^{k}\xi_{2}^{m}\alpha^{-k-1}(-1)^{\rm T }\Gamma(m+k+1)}{l!m!(k! )^{2}(\zeta_{2}-\zeta_{1})^{m+k+1}}\int\limits_{0}^{1/a }\frac{x^{m+v-1}\epsilon^{-\theta x}dx}{\left[x+\frac{\zeta_{1}}{(\zeta_{2}- \zeta_{1})a}\right]^{m+k+1}}\] \[=\frac{\omega}{2}-\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\xi_{1} e^{-2K}\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}\sum_{m=0}^{l}\sum_{v=0}^{k+1} \left( \begin{array}{c}k+1\\ v\end{array} \right)\frac{K^{l+k}\xi_{2}^{m}\alpha^{m+v}(-1)^{\rm T}\Gamma(m+k +1)}{l!m! (k)^{2}\xi_{1}^{m+1}}\] \[\times\int\limits_{0}^{1/a}x^{m+v-1}\xi_{1}e^{-\theta x}\sum_{p=0 }^{\infty}\frac{(m+k+1)_{p}}{p!}\left[\frac{(\xi_{1}-\xi_{2})a\pi}{\xi_{1}} \right]^{p}dx \tag{102}\] \[=\frac{\omega}{2}-\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\xi_{1} e^{-2K}\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}\sum_{m=0}^{l}\sum_{v=0}^{k+1} \sum_{v=0}^{\infty}\left(m+k+1\right)_{p}\left[(\xi_{1}-\xi_{2})\right]^{p} \left( \begin{array}{c}k+1\\ v\end{array} \right)\frac{K^{l+k}\xi_{2}^{m}\alpha^{m+v+p}(-1)^{\rm T}\Gamma(m+ k+1)}{l!m!(k! )^{2}\xi_{1}^{m+p+1}}\] \[\times\int\limits_{0}^{1/a}x^{m+v+p-1}\frac{1}{e}^{-\theta x}dx\] \[=\frac{\omega}{2}-\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\xi_{1} e^{-2K}\sum_{l=0}^{\infty}\sum_{v=0}^{\infty}\sum_{m=0}^{l}\sum_{v=0}^{k+1} \sum_{v=0}^{\infty}\left(m+k+1\right)_{p}\left[(\xi_{1}-\xi_{2})\right]^{p} \left( \begin{array}{c}k+1\\ v\end{array} \right)\frac{K^{l+k}\xi_{2}^{m}\alpha^{m+v+p}(-1)^{\rm T}\Gamma(m+ k+1)}{l!m!(k! )^{2}\pi_{1}^{l+p+1}\theta^{m+v+p+\frac{1}{2}}}\] \[\times\gamma\left(m+v+p+\frac{1}{2},\frac{\theta}{a}\right).\]  
**Case 3:**\(\xi_{2}>\xi_{1}\). Using the same approach as in Case 2, with the [PERSON]'s series expansion \(\frac{1}{\left[1+\frac{(\xi_{2}-\xi_{1})a\pi}{\xi_{1}}\right]^{m+k+1}}=\sum \limits_{p=0}^{\infty}(-1)^{p}\frac{(m+k+1)_{p}}{p!