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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?

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    Look at Morera's theorem and the [related problem](http://math.stackexchange.com/questions/203902/how-to-prove-error-function-mboxerf-is-entire-i-e-analytic-everywhere/203920#203920).2012-11-19

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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 $a$.

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    ....$F_2$ 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.2012-11-19