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Im − f (0, z) − For x > 0 we have Im − f (x, z) f (x, z) = f f¯ − f f¯ , 2i|f |2 so that Im(−(f (x, z)/f (x, z))) will be strictly positive for x > 0 if d f f¯ − f f¯ dx 2i > 0. It can easily be verified that d f f¯ − f f¯ dx 2i = Im z|f |2 > 0 for f ≡ 0. So −(f (x, z)/f (x, z)) has strictly positive imaginary part for all x > 0. We can write − u (x, z) + m(1) v (x, z) f (x, z) , =− f (x, z) u(x, z) + m(1) v(x, z) where m(1) is given by −m(1) = − Thus Im m(1) f (0, z) . f (0, z) 0, and we can reformulate Lemma 2 as follows.

K + 1) ∂β |β|=k+1 x β f (x) β! µ|x| dt 1 (1 − t)k t . (12) Formula (12) is simpler than it looks: because of our previous remark on (10), all the µ-polynomial terms in the previous expression for Tµ f must cancel. Let us then denote, for k 1, k Hk := l=1 k 1 k (−)l+1 k 1 = − + · · · − (−)k . l l 1 2 2 k At least for µ|x| (13) 1, the expression for Tµ f becomes Tµ f (x) = (−)k (k + 1) ∂β |β|=k+1 x β f (x) (log µ|x| + Hk ) . β! By performing the derivative with respect to log µ directly on this formula, one obtains in the bargain interesting formulae for distribution theory.

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