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By Leonard Lovering Barrett

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Then we have m 1 β = 0 if and only if α = γ = (L − 1) = . 26) This implies J2 = 2 · m! Re ! m2 ! 2 We have Iαγβ = 0 1 for 1 ≤ α, γ ≤ (L − 1) 2 and 1 ≤ β = m − α − γ. 28) Proof Let us first show that the product of the last complex components of γ ˆ β ˆ [Dtα Z(0)] w and [Dt Z(0)] w and of wDt φ(0) have a zero integral. In fact, this product has the form const(wRm wm−2α · Rm wm−2γ + · · ·)(w−2β−1 + · · ·) 2 1+2m−2(α+β+γ )−1 2 w + · · · = const ·Rm + ··· = const Rm since α + β + γ = L − 1 = m. The same holds true for the scalar product of the first two complex components, β multiplied by wDt φ(0).

With a suitable choice of c ∈ C we can arrange for E (3) (0) < 0 since Rm = 0 and (m − n)2 ≥ 1. In case (ii) we write w3 Xˆ ww · Xˆ ww as 2 2m+1 w + f (w), w 3 Xˆ ww (w) · Xˆ ww (w) = (m − n)2 Rm where ∞ f (w) := w 2m+2 aj ∈ C. aj w j , j =0 From τ3 = 3 3 τ0 + 3 2 τ02 τ1 + 3 τ0 τ12 + τ13 it follows that g(w) := w 3 Xˆ ww (w) · Xˆ ww (w)τ 3 (w) is meromorphic in B, continuous in {w : ρ < |w| ≤ 1} for some ρ ∈ (0, 1), and its Laurent expansion at w = 0 has the residue 2 Resw=0 (g) = 3 2 c3 (m − n)2 Rm + 3 3 c an−k , 1 < k ≤ n.

Now we proceed as follows: β Step 1. We choose τ = φ(0) and Dt φ(0) for β ≥ 1 in such a way that f and gL−M−1 are holomorphic. Then the integrands of the three integrals in J1 are holo- 40 3 Very Special Case; The Theorem for n + 1 Even and m + 1 Odd j ˆ j ˆ morphic because all w-derivatives [Dt Z(0)] w of the harmonic functions Dt Z(t) are holomorphic. Then it follows that J1 = 0, and thus we have E (L) (0) = J2 . 6) Step 2. Then it will be shown that E (L) (0) reduces to the single term E (L) (0) = 2 · m!

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