Residues and existence of complex antiderivativeI know that there already is a similar question out there but I am not really satisfied with that answer. I think it is too complicated. Assume that G ⊆ C is open and simply connected, P ⊆ C has finite cardinality and f : G ∖ P → C is holomorphic. Thenf has an antiderivative on its domain ⟺ R e s p ( f ) = 0 ∀ p ∈ P.Can I argue the following way? The crucial part is the second implication. ⟹ : Since f has an antiderivative, all integrals over closed rectifiable curves vanish. If I let p ∈ P be arbitrary and ε &gt; 0 so small that ∂ B ε ( p ) ⊆ G ∖ P, then according to the residue theorem: 0 = ∫ ∂ B ε ( p ) f ( z ) d z = 2 π i R e s p ( f ) ⟸ : For any closed rectifiable curve γ we have: ∫ γ f ( z ) d z = ∑ p ∈ P I p ( γ ) R e s p ( f ) = 0Therefore the existence of a primitive.

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Residues and existence of complex antiderivativeI know that there already is a similar question out there but I am not really satisfied with that answer. I think it is too complicated. Assume that   G  ⊆      C   is open and simply connected,   P  ⊆      C   has finite cardinality and   f  :  G  ∖  P  →      C   is holomorphic. Thenf has an antiderivative on its domain     ⟺               R      e      s        p    (  f  )  =  0     ∀  p  ∈  P.Can I argue the following way? The crucial part is the second implication.    ⟹  : Since f has an antiderivative, all integrals over closed rectifiable curves vanish. If I let   p  ∈  P be arbitrary and   ε  &amp;gt;  0 so small that   ∂      B    ε    (  p  )  ⊆  G  ∖  P, then according to the residue theorem:  0  =      ∫          ∂              B        ε            (      p      )        f  (  z  )         d    z  =  2  π  i            R      e      s        p    (  f  )    ⟸  : For any closed rectifiable curve   γ we have:      ∫    γ    f  (  z  )         d    z  =      ∑          p      ∈      P            I    p    (  γ  )            R      e      s        p    (  f  )  =  0Therefore the existence of a primitive.

Alexia Hart · Accepted Answer

Step 1Under the given conditions on G, P, and f, the following three statements are equivalent:(a) f has an antiderivative in G.(b)       Res    p    ⁡  (  f  )  =  0 for all   p  ∈  P.(c)       ∫    γ    f  (  z  )    d  z  =  0 for all closed rectifiable curves   γ in G.Step 2What you have demonstrated is that   (  a  )    ⟹    (  b  ) and that   (  b  )    ⟹    (  c  ).It remains to show that   (  c  )    ⟹    (  a  ), and that can be done by showing that   F  (  z  )  =      ∫                  z        0              z    f  (  w  )    d  w (which is well-defined under the condition (c)) satisfies       F    ′    =  f

Residues and existence of complex antiderivative I know that there already is a similar question ou

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