By Jan-Hendrik Evertse
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Proof. Fix z ∈ U . Choose r > 0 such that D(z, r) ⊂ U , and let 0 < δ < 21 r. We show that for w ∈ D(z, δ), F (w) can be expanded into a Taylor series around z; then it follows that F is analytic on D(z, δ) and so in particular in z. Let M (x) : D → R be a measurable function such that |f (x, w)| M (x) for x ∈ D, w ∈ D(z, r) and M (x)dx < ∞. 5, 1 2πi f (x, w)dx = F (w) = D D γz,2δ f (x, ζ) · dζ ζ −w dx. By inserting f (x, ζ) f (x, ζ) f (x, ζ) = = ζ −w (ζ − z) − (w − z) ζ −z ∞ = n=0 w−z 1− ζ −z −1 f (x, ζ) · (w − z)n n+1 (ζ − z) we obtain 1 2πi F (w) = D ∞ γz,2δ ∞ 1 = 0 D n=0 n=0 f (x, ζ) (w − z)n (ζ − z)n+1 dζ dx f (x, z + 2δe2πit ) (w − z)n dt dx.
First let q = 1. Choose r > 0 such that γz1 ,r lies in the interior of γ. 4, 1 2πi f (z)dz = γ 1 2πi f (z)dz = res(z1 , f ). γz1 ,r Now let q > 1 and assume the Residue Theorem is true for fewer than q points. We cut γ into two pieces, the piece γ1 from a point w0 to w1 and the piece γ2 from w1 to w0 so that γ = γ1 + γ2 . Then we take a path γ3 from w1 to w0 inside the interior of γ without self-intersections; this gives two contours γ1 + γ3 and −γ3 + γ2 . We choose γ3 in such a way that it does not hit any of the points z1 , .
Given z ∈ U , we define F (z) by f (w)dw, F (z) := γz where γz is any path in U from z0 to z. This does not depend on the choice of γz . For let γ1 , γ2 be any two paths in U from z0 to z. Then γ1 − γ2 (the path consisting of first traversing γ1 and then γ2 in the opposite direction) is homotopic to z0 since U is simply connected, hence f (z)dz − γ1 f (z)dz = f (z)dz = 0. γ1 −γ2 γ2 F (z+h)−F (z) To prove that limh→0 = f (z), take a path γz from z0 to z and then the h line segment [z, z + h] from z to z + h.