with 0 < y < 1 and consider the discontinuity 6 Employing the Rodriguez formula for the Gegenbauer polynomials C 5 2 $_{n}$( z ) [46]: one can derive the following basic relation Now being based on the result (A4) we introduce the function whose discontinuity at z = x reads The asterisk in the sum in (A6) denotes that for ν = 0 the terms with l = 0 and l = 1 are actually absent. 29 disc$_{z}$$_{=}$$_{x}$ F (2 ν $^{)}$( z, y ) = disc$_{z}$$_{=}$$_{x}$ ( 1 + y ∂ ∂y + 1 6 y 2 ∂ 2 ∂y 2 ) ∫ 1 − 1 dsξ 2 ν ∞ ∑ l =0 ξ l $_{z}$− 2 ν − l +1 s (1 − s $^{2}$) P$_{l}$ ( 1 ξ ) = θ ( 1 − x 2 ξ 2 ) ( 1 − x 2 ξ 2 ) 2 ∞ ∑ l =0 ∗$_{y}$$_{2}$$_{ν}$$_{+}$$_{l}$ − $^{1}$C$^{5}$ 2 2 ν + l − 2 ( x ξ ) ξP$_{l}$ ( 1 ξ ) . (A6) F (2 ν $^{)}$( z, y ) = ( 1 + y ∂ ∂y + 1 6 y 2 ∂ 2 ∂y 2 ) ∫ 1 − 1 dsξ 2 $^{ν}$z 2 − 2 ν s 1 − s 2 √ z 2 s 2 z$_{s}$ + ξ 2 (A5) (1 − z $^{2}$)$^{2}$C$^{5}$ 2 n − $_{1}$( z ) = ( − 1) n − 1 2 $^{n}$Γ( n ) ( 1 + 5 n 6 + n 2 6 ) ( ∂ ∂z ) n − 1 (1 − z $^{2}$) n +1 (A3) disc$_{z}$$_{=}$$_{x}$ ∫ 1 − 1 ds 1 − s 2 z N s = ( − 1) N − 1 Γ( N ) ∫ 1 − 1 ds (1 − s $^{2}$) δ ( N − $^{1)}$( x$_{s}$ ) = ( − 1) N − $^{1}$θ ( 1 − x 2 ξ 2 ) y N 2 $^{N}$ξ$^{N}$Γ( N ) ( ∂ ∂s ) N − 1 (1 − s $^{2}$) N +1 ∣ s = x ξ (A2) disc$_{z}$$_{=}$$_{x}$ ( 1 + y ∂ ∂y + 1 6 y 2 ∂ 2 ∂y 2 ) ∫ 1 − 1 ds 1 − s 2 z N s = y N ξ N θ ( 1 − x 2 ξ 2 ) ( 1 − x 2 ξ 2 ) 2 C 5 2 N − 1 ( x ξ ) . (A4) 6 The discontinuity of the function f ( z ) is defined as disc z = x f ( z ) = 1 2 πi ( f ( x − i 0) − f ( x + i 0)) = 1 $_{π}$Im f ( x − i 0)