## Appendix A  
\[ \begin{array}{l}I_{2}^{(3)}=-\frac{\eta^{2}R}{2}J_{10}-\\ \frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\big{\{}\eta^{2}R\cos(4\beta)-2 \big{[}R_{1}-\eta\big{(}z_{Q}-z_{1}\big{)}\big{]}\cos(2\beta)\big{\}}\text{ arsinh}\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt{D_{10}}}d\beta\end{array}  \tag{1}\]  
This integral will be solved by the partial integration. \[v=\int\limits_{0}^{\frac{\pi}{2}}\Big{\{}\eta^{2}R\cos(4\beta)-2\big{[}R_{1}- \eta\big{(}z_{Q}-z_{1}\big{)}\big{]}\cos(2\beta)\Big{\}}d\beta\]  
or  
\[v=\frac{\sin(2\beta)}{2}\Big{\{}\eta^{2}R\cos(2\beta)-2\big{[}R_{1}-\eta\big{(} z_{Q}-z_{1}\big{)}\big{]}\Big{\}}\]  
\[u=\text{arsinh}\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt{D_{10}}}\]  
\[du=\frac{R_{1}\sqrt{\eta^{2}+1}\ k_{1}\sin(2\beta)[\cos(2\beta)+a_{22}]d\beta }{\eta\sqrt{RR_{1}}\Delta[\cos^{2}(2\beta)-a_{11}\cos(2\beta)-a_{33}]}\]  
with  
\[a_{11}=\frac{2\big{[}R_{1}-\eta\big{(}z_{Q}-z_{1}\big{)}\big{]}}{\eta^{2}R}\]  
\[a_{22}=\frac{\eta R^{2}+\eta\big{(}z_{Q}-z_{1}\big{)}^{2}-R_{1}\big{(}z_{Q}-z _{1}\big{)}}{\eta RR_{1}}\]  
\[a_{33}=\frac{\big{[}R_{1}-\eta\big{(}z_{Q}-z_{1}\big{)}\big{]}^{2}+\big{(}\eta ^{2}+1\big{)}R^{2}}{\eta^{2}R^{2}}\]Now,  
\[ \begin{array}{c}I_{2}^{(3)}=-\frac{\eta^{2}R}{2}I_{10}-\frac{1}{2}\left\{uv\, \sqrt{\frac{\alpha}{2}}\,\right.\left.\begin{array}{c}-\frac{\pi}{2}\,\end{array} vdu\right\}=-\frac{\eta^{2}R}{2}I_{10}+\\ \int\limits_{0}^{\frac{\pi}{2}}\frac{\eta^{2}RR_{1}\sqrt{\eta^{2}+1}\,k_{1} \sin(2\beta)(\cos(2\beta)+a_{22})\sin(2\beta)[\cos(2\beta)-a_{11}]d\beta}{4\sqrt {RR_{1}}A[\cos^{2}(2\beta)-a_{11}\cos(2\beta)-a_{33}]}=\\ \frac{\eta^{2}R}{2}I_{10}-\frac{\eta\sqrt{\eta^{2}+1}\,k_{1}\sqrt{RR_{1}}}{4} \int\limits_{0}^{\frac{\pi}{2}}\frac{[\cos^{2}(2\beta)-1][\cos(2\beta)-a_{11} ][\cos(2\beta)+a_{22}]d\beta}{[\cos^{2}(2\beta)-a_{11}\cos(2\beta)-a_{33}]A_{1 }}=\\ \frac{\eta^{2}R}{2}I_{10}-\frac{k_{1}\sqrt{\eta^{2}+1}\sqrt{RR_{1}}}{4} \int\limits_{0}^{\frac{\pi}{2}}\{\cos^{2}(2\beta)+a_{22}\cos(2\beta)+a_{33}- 1\}\,\frac{d\beta}{A_{1}}-\\ \frac{k_{1}\eta\sqrt{\eta^{2}+1}\sqrt{RR_{1}}}{4}\int\limits_{0}^{\frac{\pi}{ 2}}\frac{(A\cos(2\beta)+B)d\beta}{\cos^{2}(2\beta)-a_{11}B_{0}(2\beta)-a_{33} ]A_{1}}d\beta\end{array}  \tag{10}\]  
with  
\[A=(a_{22}+a_{11})a_{33}-a_{2},\]  
\[B=a_{33}^{2}-a_{33}+a_{11}a_{22}\]  
The expression (10) can be written as follows. \[ \begin{array}{c}I_{2}^{(3)}=-\frac{\eta^{2}R}{2}I_{10}-\frac{k_{1}\eta\sqrt{ \eta^{2}+1}\sqrt{RR_{1}}}{4}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{ \frac{\pi}{2}}\{\cos^{2}(2\beta)+a_{2}\cos(2\beta)-1+a_{3}\}\frac{d\beta}{A}- \\ \frac{k_{1}\eta\sqrt{\eta^{2}+1}\sqrt{RR_{1}}}{4}\int\limits_{0}^{\frac{\pi} {2}}\frac{(A\cos(2\beta)+B)d\beta}{[\cos(2\beta)-p_{1})(B_{0}(2\beta)-p_{2}) ]\overline{A}_{1}}\end{array}  \tag{11}\]  
Let us find the solutions of the following equation. \[\cos^{2}(2\beta)-a_{11}\cos(2\beta)-a_{33}=(\cos(2\beta)-p_{1})(\cos(2\beta)- p_{2})=0\]  
\[p_{1,2}=\frac{a_{11}\pm\sqrt{a_{11}^{2}+4a_{33}}}{2}=\frac{a_{11}\pm\sqrt{D_{ 1}}}{2}=\frac{\left[R_{1}-\eta\left(z_{Q}-z_{1}\right)\right]}{\eta^{2}R}+ \frac{\sqrt{(\eta^{2}+1)}}{\eta^{2}R}\sqrt{T_{10}}\]  
\[D_{1}=a_{11}^{2}+4a_{33}=4\frac{\left[R_{1}-\eta\left(z_{Q}-z_{1}\right) \right]^{2}}{\eta^{4}R^{2}}+4\frac{\left[R_{1}-\eta\left(z_{Q}-z_{1}\right) \right]^{2}+(\eta^{2}+1)R^{2}}{\eta^{2}R^{2}}\]  
\[D_{1}=\frac{4\left(\eta^{2}+1\right)}{\eta^{4}R^{2}}\Big{\{}\left[R_{1}-\eta \left(z_{Q}-z_{1}\right)\right]^{2}+\eta^{2}R^{2}\Big{\}}=\frac{4\left(\eta^{ 2}+1\right)}{\eta^{4}R^{2}}T_{10}>0\]  
\[T_{10}=\Big{\{}\left[R_{1}-\eta\left(z_{Q}-z_{1}\right)\right]^{2}+\eta^{2}R^ {2}\Big{\}}\]  
The following fraction can be obtained in the following form. \[ \begin{array}{c}\frac{(A\cos(2\beta)+B)}{[\cos^{2}(2\beta)-a_{11}\cos(2 \beta)-a_{33}]\Delta}=\frac{A\cos(2\beta)+B}{(\cos(2\beta)-p_{1})(\cos(2\beta) -p_{2})}=\\ \frac{V_{1}}{\cos(2\beta)-p_{1}}+\frac{V_{2}}{\cos(2\beta)-p_{2}}\end{array}