Existence of antiderivative without Cauchy-Goursat's theoremI wonder if anybody has tried the following kind of direct proof for the existence of an antiderivative of an analytic function on a star-shaped domain.Theorem: Let f : D → C be an analytic function on a star-shaped domain D. Then f has an antiderivative F on D."proof": For simplicity, assume that every point in D is connected to 0 ∈ C by a line segment. Define F ( z ) = ∫ 0 1 z f ( z t ) d t .Then lim h → 0 ( F ( z + h ) − F ( z ) h ) = lim h → 0 ∫ 0 1 ( ( z + h ) f ( z t + h t ) − z f ( t ) h ) d t .It can be checked that as h → 0, the integrand converges to d d t t f ( z t ) ..Therefore, F ′ ( z ) = f ( z ), provided that the limit and integral are interchangeable. qed.Of course, a limit and integral cannot always be interchanged. But I wonder if anybody seriously considered the above line of proof.Thanks. As far as I can search from several textbooks in complex variables, the above theorem is proved by using Cauchy-Goursat's theorem. More concretely, they use the equality ∫ z 0 z + h f ( z ) d z − ∫ z 0 z f ( z ) d z = ∫ z z + h f ( z ) d z ,, which can be justified by Cauchy-Goursat's theorem. The point of my question is: Is it possible to directly invoke to the computation as above, without using Cauchy-Goursat's theorem? If this is possible, we get another proof of Cauchy-Goursat's theorem, at least for star-convex domains.

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Existence of antiderivative without Cauchy-Goursat&#039;s theoremI wonder if anybody has tried the following kind of direct proof for the existence of an antiderivative of an analytic function on a star-shaped domain.Theorem: Let   f  :  D  →      C   be an analytic function on a star-shaped domain D. Then f has an antiderivative F on D.&quot;proof&quot;: For simplicity, assume that every point in D is connected to   0  ∈      C   by a line segment. Define  F  (  z  )  =      ∫    0    1    z  f  (  z  t  )  d  t  .Then       lim          h      →      0            (                  F        (        z        +        h        )        −        F        (        z        )            h        )    =      lim          h      →      0            ∫    0    1        (                  (        z        +        h        )        f        (        z        t        +        h        t        )        −        z        f        (        t        )            h        )    d  t  .It can be checked that as   h  →  0, the integrand converges to       d          d      t        t  f  (  z  t  )  ..Therefore,       F    ′    (  z  )  =  f  (  z  ), provided that the limit and integral are interchangeable. qed.Of course, a limit and integral cannot always be interchanged. But I wonder if anybody seriously considered the above line of proof.Thanks. As far as I can search from several textbooks in complex variables, the above theorem is proved by using Cauchy-Goursat&#039;s theorem. More concretely, they use the equality       ∫                  z        0                    z      +      h        f  (  z  )  d  z     −         ∫                  z        0              z    f  (  z  )  d  z     =         ∫    z          z      +      h        f  (  z  )  d  z  ,, which can be justified by Cauchy-Goursat&#039;s theorem. The point of my question is: Is it possible to directly invoke to the computation as above, without using Cauchy-Goursat&#039;s theorem? If this is possible, we get another proof of Cauchy-Goursat&#039;s theorem, at least for star-convex domains.

Xzavier Shelton · Accepted Answer

Explanation:I&#039;m not sure I understand exactly what you want to ask, but perhaps this is helpful:A continuous function   f  :  Ω  →      C   has an anti-derivative on   Ω if and only if       ∫    γ    f  (  z  )    d  z  =  0 for every simple closed curve   γ in   Ω.In other words, it seems difficult to completely circumvent some variant of Cauchy-Goursat.

Existence of antiderivative without Cauchy-Goursat's theorem I wonder if anybody has tried the foll

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New Questions in Calculus 1