## Appendix  
Define:  
\[x=\sum_{j\in\mathsf{DV}_{All}}\frac{1}{d_{ij}^{2}}-\sum_{j\in\mathsf{DV}_{i}} \frac{1}{d_{ij}^{2}} \tag{1}\]Equation (A4) is a quadratic equation of \(s\), which can be solved by the quadratic formula as:  
\[s=\sqrt{R_{0}^{2}-D_{0i}^{2}\sin^{2}\theta}-D_{0i}\cos\theta\] (A6)  
Equation (A6) provides the relationship between \(s\) and \(\theta\). Note that the other solution is negative, which cannot be used to represent the distance.  
Based on the above analysis, \(x\) can be rewritten as:  
\[\begin{split} x&=\frac{N_{0}}{\pi R_{0}^{2}}\int_{0 }^{2\pi}\int_{r_{i}}^{\sqrt{R_{0}^{2}-D_{0i}^{2}\sin^{2}\theta}-D_{0i}\cos \theta}\frac{1}{s^{2}}sdsd\theta\\ &=\frac{2N_{0}}{\pi R_{0}^{2}}\Big{\{}\int_{0}^{\pi}[\ln(\sqrt{R _{0}^{2}-D_{0i}^{2}\sin^{2}\theta}-D_{0i}\cos\theta)]d\theta-\pi\ln r_{i}\Big{\}} \end{split}\] (A7)To solve the above integral, let:  
\[\begin{split} z&=\int_{0}^{\pi}\ln(\sqrt{R_{0}^{2}-D_{0i }^{2}\sin^{2}\theta}-D_{0i}\cos\theta)d\theta\\ &=\int_{0}^{\pi}\ln\Big{[}\frac{(\sqrt{R_{0}^{2}-D_{0i}^{2}\sin^{ 2}\theta}-D_{0i}\cos\theta)}{(\sqrt{R_{0}^{2}-D_{0i}^{2}\sin^{2}\theta}+D_{0i} \cos\theta)}\cdot(\sqrt{R_{0}^{2}-D_{0i}^{2}\sin^{2}\theta}+D_{0i}\cos\theta) \Big{]}d\theta\\ &=\int_{0}^{\pi}\ln\Big{[}\frac{R_{0}^{2}-D_{0i}^{2}}{\sqrt{R_{0} ^{2}-D_{0i}^{2}\sin^{2}\theta}+D_{0i}\cos\theta}\Big{]}d\theta\\ &=\pi\ln(R_{0}^{2}-D_{0i}^{2})-\int_{0}^{\pi}\ln\big{[}\sqrt{R_{ 0}^{2}-D_{0i}^{2}\sin^{2}\theta}+D_{0i}\cos\theta\big{]}d\theta\end{split} \tag{10}\]  
Let \(\Phi=\pi-\theta\), we find:  
\[\begin{split} z&=\pi\ln(R_{0}^{2}-D_{0i}^{2})-\int _{\pi}^{0}\ln\big{[}\sqrt{R_{0}^{2}-D_{0i}^{2}\sin^{2}(\pi-\Phi)}+D_{0i}\cos( \pi-\Phi)\big{]}d(\pi-\Phi)\\ &=\pi\ln(R_{0}^{2}-D_{0i}^{2})-\int_{0}^{\pi}\ln\big{[}\sqrt{R_{ 0}^{2}-D_{0i}^{2}\sin^{2}\Phi}-D_{0i}\cos\phi\big{]}d\Phi\\ &=\pi\ln(R_{0}^{2}-D_{0i}^{2})-z\end{split} \tag{11}\]  
We obtain:  
\[z=\frac{\pi}{2}\ln(R_{0}^{2}-D_{0i}^{2}) \tag{12}\]  
Taking Equation (12) into (10), we obtain:  
\[\begin{split} x&=\frac{2N_{0}}{\pi R_{0}^{2}}\big{[} \frac{\pi}{2}\ln(R_{0}^{2}-D_{0i}^{2})-\pi\ln r_{i}\big{]}\\ &=\frac{N_{0}}{R_{0}^{2}}\big{[}\ln(R_{0}^{2}-D_{0i}^{2})-\ln \frac{R_{0}^{2}}{N_{0}}-\ln n_{i}\big{]}\end{split} \tag{13}\]  
To calculate \(y\), we establish the polar coordinate system as shown in Figure 11b, then:  
\[y=\frac{N_{0}}{\pi R_{0}^{2}}\int_{0}^{r_{\alpha}}\int_{0}^{2\pi}\frac{s}{s^{ 2}+d_{\alpha j}^{2}-2 sd_{\alpha j}\cos\theta}d\theta ds \tag{14}\]  
Let \(a=d_{\alpha j}^{2}+s^{2}\) and \(b=-2d_{\alpha j}s\), we have \(a^{2}>b^{2}\), so the above integral becomes:  
\[\begin{split} y&=\frac{N_{0}}{\pi R_{0}^{2}}\int_{0 }^{r_{\alpha}}s\int_{0}^{2\pi}\frac{1}{a+b\cos\theta}d\theta ds\\ &=\frac{N_{0}}{\pi R_{0}^{2}}\int_{0}^{r_{\alpha}}\frac{2s}{ \sqrt{a^{2}-b^{2}}}\Big{[}\arctan\big{(}\sqrt{\frac{a-b}{a+b}}\cdot\tan\frac{ \theta}{2}\big{]}_{0}^{\pi}\big{)}+\arctan\big{(}\sqrt{\frac{a-b}{a+b}}\cdot \tan\frac{\theta}{2}\big{]}_{\pi}^{2\pi}\big{)}\Big{]}\\ &=\frac{N_{0}}{R_{0}^{2}}\int_{0}^{r_{\alpha}}\frac{2s}{\sqrt{a^{2 }-b^{2}}}ds\end{split} \tag{15}\]  
Since we assume that the picocells do not overlap with one another, we have \(D_{\alpha i}\geq s\), so:  
\[\begin{split} y&=\frac{N_{0}}{R_{0}^{2}}\int_{0}^{r_{ \alpha}}\frac{2s}{d_{\alpha j}^{2}-s^{2}}ds\\ &=\frac{N_{0}}{R_{0}^{2}}[\ln D_{\alpha i}^{2}-\ln(D_{\alpha i}^{ 2}-r_{\alpha}^{2})]\end{split} \tag{16}\]Finally, taking Equations (11) and (14) into Equation (13), \(I_{0i}\) is obtained as a function of \(n_{i}\), written as:  
\[I_{0i}(n_{i})=\frac{N_{0}GP_{0}}{(N_{0}-\sum_{\alpha}n_{\alpha})R_{0}^{2}}\Big{\{} \big{[}\ln(R_{0}^{2}-D_{0i}^{2})-\ln\frac{R_{0}^{2}}{N_{0}}-\ln n_{i}\big{]}- \sum_{\alpha\
eq i}\big{[}\ln D_{\alpha i}^{2}-\ln(D_{\alpha i}^{2}-\frac{R_{0 }^{2}n_{i}}{N_{0}})\big{]}\Big{\}} \tag{15}\]