“…Therefore, the condition on D 0,0 = d can be removed to calculate q κ,m,0 as follows q κ,m,0 (p) = R 0 q κ,m,0,D 0,0 (p, d) f D 0,0 (d)dd (41) Therefore, the proof is completed. Following the same steps of Appendix E, q κ,a,0,D a,0 (p, d) and q C,a,D C,a (p, d) can be obtained as in (42) and (43), as shown at the bottom of the page, respectively, where s = Next, we remove the conditions on the distance to obtain q κ,a,0, (p) and q C,a (p) as follows q κ,a,0 (p) =…”