## 2 Basic Formulas
\[I_{2}=\int\limits_{0}^{\pi/2}Vcos(2\beta)d\beta=I_{2}^{(1)}+I_{2}^{(2)}+I_{2}^ {(3)}+I_{2}^{(4)} \tag{11}\]  
\[I_{2}^{(1)}=\sqrt{\eta^{2}+1}\int\limits_{0}^{\pi/2}\sqrt{cR_{1}^{2}+bR_{1}+ acos(2\beta)d\beta}=\sqrt{\eta^{2}+1}\int\limits_{0}^{\pi/2}\sqrt{T_{1}}\cos(2 \beta)d\beta \tag{12}\]  
\[T_{1}=c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}=\eta^{2}\Big{[}\big{(}R+R_{1}\big{)}^{2} +\big{(}z_{Q}-z_{1}\big{)}^{2}-4 RR_{1}\ \sin^{2}(\beta)\Big{]}\]  
\[I_{2}^{(1)}=\sqrt{\eta^{2}+1}\sqrt{\eta^{2}\Big{[}\big{(}R+R_{1}\big{)}^{2}+ \big{(}z_{Q}-z_{1}\big{)}^{2}\Big{]}}\int\limits_{0}^{\pi/2}\Delta_{1}(1-2 \sin^{2}(\beta))d\beta\]  
\[\Delta_{1}=\sqrt{1-k_{1}^{2}\sin^{2}(\beta)}\,\ \ k_{1}^{2}=\frac{4 RR_{1}}{ \big{(}R+R_{1}\big{)}^{2}+\big{(}z_{Q}-z_{1}\big{)}^{2}} \tag{13}\]  
Using [25] one obtains the following. \[I_{2}^{(1)}=2\eta\sqrt{\eta^{2}+1}\frac{\sqrt{RR_{1}}}{k_{1}} \int\limits_{0}^{\pi}\Delta_{1}(1-2\sin^{2}(\beta))d\beta=\] \[V=\sqrt{\eta^{2}+1}\left[\sqrt{c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}}-\sqrt{c_{1}r_{1}^{2 }+b_{1}r_{1}+a_{1}}\right]-\]0  
or  
\[V=\sqrt{\eta^{2}+1}\left[\sqrt{c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}}-\sqrt{c_{1}r_{1}^{2 }+b_{1}r_{1}+a_{1}}\right]-\]1  
This solution is obtained in the close form where \(K(k_{1})\) and \(E(k_{1})\) are the complete elliptic integrals of the first and second kind [25; 26]. Similarly, the solution of the next integral,  
\[V=\sqrt{\eta^{2}+1}\left[\sqrt{c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}}-\sqrt{c_{1}r_{1}^{2 }+b_{1}r_{1}+a_{1}}\right]-\]2  
is also obtained in the close form as follows,  
\[V=\sqrt{\eta^{2}+1}\left[\sqrt{c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}}-\sqrt{c_{1}r_{1}^{2 }+b_{1}r_{1}+a_{1}}\right]-\]3  
with  
\[V=\sqrt{\eta^{2}+1}\left[\sqrt{c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}}-\sqrt{c_{1}r_{1}^{2 }+b_{1}r_{1}+a_{1}}\right]-\]4The next integral is  
\[ \begin{array}{l}I_{2}^{(3)}=-\int\limits_{0}^{\frac{\pi}{2}}\left[\eta^{2}R\cos( 2\beta)-R_{1}+\eta\big{(}z_{Q}-z_{1}\big{)}\right]\times\\ \cos(2\beta)\text{arsinh}\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt {D_{10}}}d\beta\end{array}  \tag{17}\]  
or  
\[ \begin{array}{l}I_{2}^{(3)}=-\frac{\eta^{2}R}{2}\int\limits_{0}^{\frac{\pi}{2 }}\text{arsinh}\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt{D_{10}}} d\beta-\\ \frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\left\{\eta^{2}R\cos(4\beta)-2\big{[} R_{1}-\eta\big{(}z_{Q}-z_{1}\big{)}\big{]}\cos(2\beta\big{)}\right\}\text{arsinh}\frac{ \eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt{D_{10}}}d\beta=\\ -\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\left\{\eta^{2}R\cos(4\beta)-2 \big{[}R_{1}-\eta\big{(}z_{Q}-z_{1}\big{)}\big{]}\cos(2\beta\big{)}\right\} \text{arsinh}\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt{D_{10}}}d \beta=\\ -\frac{\eta^{2}R}{2}J_{10}\end{array}  \tag{18}\]  
where  
\[J_{10}=\int\limits_{0}^{\frac{\pi}{2}}arsinh\frac{\eta R\cos(2\beta)+\eta R_{ 1}+z_{Q}-z_{1}}{\sqrt{D_{10}}}d\beta\]  
This integral \(J_{10}\) does not have the analytical solution, so it must be solved numerically. The kernel function of this integral is the continuous function of the interval of integration. Let us solve this integral by partial integration (Appendix A). Thus, the solution is  
\[ \begin{array}{l}I_{2}^{(3)}=-\frac{\eta^{2}R}{2}J_{10}+\frac{\eta\sqrt{\eta^ {2}+1}\sqrt{RR_{1}}}{12k_{1}^{3}}\big{\{}\left[3(a_{22}+a_{33})k_{1}^{4}-(6a_ {22}+8)k_{1}^{2}+8\right]K(k_{1})+\\ \left[(6a_{22}+4)k_{2}^{2}-8\right]E(k_{1})\big{\}}+\frac{k_{1}\eta\sqrt{\eta^ {2}+1}\sqrt{RR_{1}}}{4}\left[\frac{V_{1}}{1-p_{1}}\varPi(h_{1},k_{1})+\frac{V _{2}}{1-p_{2}}\varPi(h_{2},k_{1})\right]\end{array}  \tag{19}\]  
Similarly,  
\[ \begin{array}{l}I_{2}^{(4)}=\int\limits_{0}^{\frac{\pi}{2}}\left[\eta^{2}R \cos(2\beta)-R_{1}+\eta\big{(}z_{Q}-z_{1}\big{)}\right]\cos(2\beta)\text{arsinh }\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{2}}{\sqrt{D_{10}}}d\beta\end{array}  \tag{20}\]  
can be obtained as follows (Appendix A),  
\[