Substituting $_{0}$F ( α ) 0 ( A , Z ) and r ∏ j =1 ∏ p s $_{=1}$(1 − b$_{s}$z$_{j}$ ) − 1 /α in the above formula by their expansions (6) and (36) in the series of Jack polynomials, and interchang- ing the order of integration and summation, which is possible because the series converge uniformly over the unit torus, we obtain But ∏ r j $_{=1}$( z$_{j}$ ) ( p − r +1) /α − $^{1}$C ( α ) τ ( Z ) = C ( α ) ˜ τ ( Z ) , where ˜ τ denotes partition [ τ$_{1}$ + p − r +1 − α α , ..., τ$_{r}$ + p − r +1 − α α ]. Note that ˜ τ is well defined for α = 2 /β, where β is an even integer. If β is an odd integer, we need to assume that p − r + 1 is even. Therefore, using the orthogonality of the Jack polynomials with respect to the torus scalar product, we have Using Lemma A1, (37), and equality r ∏ j =1 a − ( p − r +1) /α +1 j C ( α ) ˜ τ ( A ) = C ( α ) τ ( A ) = C ( α ) τ ( A ) , we get after some cancellations where In the above expression for ˜ γ ( α $^{)}$, substitute C ( α ) τ ( I$_{p}$ ) and C ( α ) ˜ τ ( I$_{r}$ ) by their 25 ˜ γ ( α ) = α 2 | ˜ τ $^{|}$| ˜ τ | ! w (˜ τ, α ) w ( τ, α ) α 2 $^{t}$t ! C ( α ) τ ( I$_{p}$ ) C ( α ) ˜ τ ( I$_{r}$ ) r ∏ j =1 Γ (( p +1 − j ) /α ) Γ (1+( r j ) /α ) ∏ s ∈ ˜ τ r + a $^{′}$( s ) α − l $^{′}$( s ) r +( a $^{′}$( s )+1) α l $^{′}$( s ) − 1 . RHS = ∞ ∑ t =0 ∑ τ ⊢ t ˜ γ ( α $_{)}$1 t ! C ( α ) τ ( A ) C ( α ) τ ( B ) C ( α ) τ ( I$_{p}$ ) , (38) RHS = γ ( α ) r ∏ j =1 a − ( p − r +1) /α +1 j ∞ ∑ t =0 ∑ τ ⊢ t w ( τ, α ) | ˜ τ | ! ( α $^{t}$t !) 2 C ( α ) ˜ τ ( A ) C ( α ) τ ( B ) C ( α ) ˜ τ ( I$_{r}$ ) 〈 C ( α ) ˜ τ ( Z ) ,C ( α ) ˜ τ ( Z ) 〉 $_{α}$. RHS = γ ( α ) ∞ ∑ k =0 ∑ κ ⊢ k ∞ ∑ t =0 ∑ τ ⊢ t w ( τ, α ) k ! ( α $^{t}$t !) 2 C ( α ) κ ( A ) C ( α ) τ ( B ) C ( α ) κ ( I$_{r}$ ) × 〈 C ( α ) κ ( Z ) , r ∏ j =1 ( z$_{j}$/a$_{j}$ ) ( p − r +1) /α − $^{1}$C ( α ) τ ( Z ) 〉$_{α}$ .