B. Solution in the positive region ( x > 0) Let us study the other region in which x > 0, from Eq. 7 using the variable 1 /z = 1 + e a ( x − L ) and the transformation φ = z − $^{ω}$(1 − z ) − $^{ρ}$g ( z ), we obtain: Following a line of thought similar to the one outlined in the previous subsection we obtain the mass function The final result for the parameters ω, ρ and δ (the coefficient introduced requiring that the term multiplying the factor 1 / [ z (1 − z )] be a costant) is found to be: The differential equation of the function g ( z ) in Eq. (20) becomes: with the general solution ( d$_{1}$ and d$_{2}$ constants): Let us briefly comment on the change of variables chosen in the negative region x ∈ [ −∞ , 0]: y = − e − a ( x + L ) and in the positive region x ∈ [0 , + ∞ ]: 1 /z = 1 + e a ( x − L $^{)}$. This implies clearly that y ∈ [ −∞ , − e − $^{aL}$] and z ∈ [(1 + e − $^{aL}$) − $^{1}$, 0] (which reduce to y ∈ [ −∞ , 0] and z ∈ [1 , 0] with the assumption aL ≫ 1). The reader might worry that the choice in the negative region is not appropriate because then the Hypergeometric series $_{2}$F$_{1}$ ( a, b, c ; y ) would not be convergent 8 g ( z ) = d$_{1 2}$F$_{1}$ ( − ρ − ν − λ, − ρ − ν + λ ; 1 − 2 ν ; z )+ d$_{2}$ z 2 ν $_{2}$F$_{1}$ ( − ρ + ν − λ, − ρ + ν + λ ; 1+2 ν ; z ) . (24) z (1 − z ) d $^{2}$g ( z ) dz 2 + [1 − 2 ω − z (1 − 2 ρ − 2 ω )] d g ( z ) dz − ( − ρ − ω − λ ) ( − ρ − ω + λ ) g ( z ) = 0 , (23) ρ = − i ( E − W ) a = µ (22a) ω = i$^{√}$ E 2 − m 2 0 a 2 = i k a = ν (22b) δ = λ 2 − µ 2 − ν 2 − 2 ν (22c) m ( z ) = m$_{0}$ (1 − z ) . (21) z (1 − z ) d $^{2}$g ( z ) dz 2 + [ 1 − 2 ω − z (2 − 2 ρ − 2 ω ) − ˙ m m z (1 − z )$^{]}$ d g ( z ) dz + { 2 ω (1 − ρ ) − ˙ m m z (1 − z )$^{[}$ − ω z + ρ 1 − z$^{]}$ + 1 z (1 − z ) [ ω ( ω + 1)(1 − z ) 2 + ρ (1 + ρ ) z 2 − ω (1 − z ) + zρ − (2 z $^{2}$) ρ + 1 a 2 [ ( E − W z ) 2 − m ( z ) 2 − iaW z (1 − z ) − ia ( E − W z ) ˙ m m z (1 − z )$^{]]}}$ g ( z ) = 0 (20)