We can derive the asymptotic expression as x → −∞ ( y → −∞ ) by using the following formula for the asymptotic behaviour of the hypergeometric function [26] and obtain: where and A, B, C, D are given by: Similarly we can derive the asymptotic form of lower component χ ( x ) from Eq. (5a): Similarly for the solution in the positive region we have from Eq. (24): Now we recall that z → 0 when x → ∞ and imposing the boundary condition of the scattering problem that in the ( x > 0 region) we only have a wave travelling to the right (only the transmitted wave) we find: and χ$_{R}$ ( x ) is found again through Eq. (5a) in terms of φ$_{R}$ : 11 lim x → + $_{∞}$χ$_{R}$ ( x ) = d$_{1}$ ( E − k ) m$_{0}$ e ik ( x − L ) (34) lim x → + $_{∞}$φ$_{R}$ ( x ) = d$_{1}$ e ik ( x − L ) (33) φ$_{R}$ ( z ) = d$_{1}$ z − ν (1 − z ) − ρ $_{2}$F$_{1}$ ( − ρ − ν − λ, − ρ − ν + λ ; 1 − 2 ν ; z ) + d$_{2}$ z ν (1 − z ) − ρ $_{2}$F$_{1}$ ( − ρ + ν − λ, − ρ + ν + λ ; 1 + 2 ν ; z ) . (32) lim x $_{→−∞}$χ ( x )$_{L}$ = G ( E + k ) m$_{0}$ e − ik ( x + L ) + H ( E − k ) m$_{0}$ e ik ( x + L ) (31) A = Γ(2 µ )Γ(2 ν ) Γ( µ + ν − λ )Γ( µ + ν + λ ) (30a) B = Γ(2 µ )Γ( − 2 ν ) Γ( µ − ν − λ )Γ( µ − ν + λ ) (30b) C = Γ(2 − 2 µ )Γ(2 ν ) Γ(1 − µ + ν − λ )Γ(1 − µ + ν + λ ) (30c) D = Γ(2 − 2 µ )Γ( − 2 ν ) Γ(1 − µ − ν − λ )Γ(1 − µ − ν + λ ) (30d) H = D$_{1}$ B e iπµ − D$_{2}$ D e − iπµ (29b) G = D$_{1}$ A e iπµ − D$_{2}$ C e − iπµ (29a) φ$_{L}$ ( x ) ∼ G e − ik ( x + L ) + H e ik ( x + L ) (28) $_{2}$F$_{1}$ ( a, b, c ; y ) = Γ( c )Γ( b − a ) Γ( b )Γ( c − a ) ( − y ) − a + Γ( c )Γ( a − b ) Γ( a )Γ( c − b ) ( − y ) − b , (27)