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Let $f : \mathbb{C} \rightarrow \mathbb{C}$ and let $\gamma_a$ be a continuous family of paths in the complex plane going from $0$ to $a$.

Which restrictions have to be imposed on $f$ to make $F(a)=\int_{\gamma_a}f(x)\mathrm{d}x$ holomorphic on some open set U?

Friedrich
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1 Answers1

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"Holomorphic at $a$" is sometimes (always?) taken to mean "complex-differentiable at every point in some open neighborhood of $a$". If $F$ is holomorphic at $a$, then some standard results say that for some $c_n$, $n=0,1,2,\ldots\,{}$, $$ F(z) = \sum_{n=0}^\infty c_n (z-a)^n $$ for $|z-a|<\text{some positive number}$. The series converges, and what it converges to is the right thing, $F(z)$. Another standard theorem says that in the interior of the disk of convergence, power series can be differentiated term-by-term, and the derivative has at least as large a radius of convergence. Consequently $F$ has derivatives of all orders in some open neighborhood of $a$. And yet another standard result says $F'=f$ in that neighborhood. So $f$ must itself by expressible as a convergent power series in that neighborhood. Bottom line: $f$ must itself be holomorphic at $a$, in order that $F$ be holomorphic at $a$.

Michael Hardy
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  • The definition at the outset of my answer is intended to distinguish holomorphic functions from something like $z\mapsto|z|^2$, which is differentiable at $0$ (and only at $0$) but not holomorphic at $0$, in the sense defined. – Michael Hardy Nov 18 '12 at 23:37
  • If I am not mistaken, this is a necessary condition. Could you think of a sufficient one? I have updated my question to make it clearer that I am looking for a sufficient condition. – Friedrich Nov 19 '12 at 12:04
  • You speak of integrating along a "continuous family of paths" rather than along a path. That is at best unclear. Before I would attempt to be precise about sufficient conditions, I wouldn't mind seeing something precise about how $F$ is defined. Does the integral from $0$ to $a$ depend on which path is chosen? If so, then you may have a "multiple-valued" $F$ with different branches. On different branches $F$ would differ by a constant, so the derivative would still be $f$. But it would be holomorphic on each branch separately. – Michael Hardy Nov 19 '12 at 14:23
  • What I mean by "family of paths" is that there should be one path for each endpoint $a$. So the family could for example be the straight lines going from $0$ to $a$. If it helps, I don't mind restricting the analyis to this family. – Friedrich Nov 19 '12 at 19:59
  • You say $f:\mathbb C\to\mathbb C$. That makes things a lot simpler than if you had $f:\text{some subset of }\mathbb{C}\to\mathbb{C}$. It implies that $F$ will be the same regardless of which path one takes from $0$ to $a$. That being the case, the necessary condition is sufficient. Difficulties come when the domain has holes. If, for example, $f$ is undefined at $2$, One could take a path from $0$ to $3$ that goes "below" $2$ and another than goes "above" $2$, and get different results. In that case, you might have $F_1$ defined near $2$ and at every point on the first path, and..... – Michael Hardy Nov 19 '12 at 22:40
  • ....$F_2$ defined near $2$ and at every point on the second path, and $F_1'=F_2'=f$, so $F_1-F_2$ is constant but not zero. Then one would have defined an antiderivative of $f$ whose domain is a surface that covers the domain of $f$ and has branches. But it would still be complex-differentiable. – Michael Hardy Nov 19 '12 at 22:42