# Integral question: zeroes of the primitive.

Let $z$ be a complex number. Let $f(z)$ be an elementary function but not a polynomial. Let its integral $F(z)$ be impossible to express in elementary functions. If we define $F(z)$ as $\int$ from $A$ to $z$ (in a straight path from $A$ to $z$), how do we know if there is a finite complex $A$? What if there are multiple values where $F(z)$ = $0$; What $A$ to choose? How to compute $A$? What $A$ makes the computation of the integral the easiest? Is it the $A$ closest to $z$?

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Could you please take a little time to expand your question? Some details are unclear. For example, what is $A$, is it a complex number? When you integrate from $A$ to $z$, which path do you integrate along? Does $f$ have poles and if so, what kind? Finally, if you want to define $F$ as an integral then it's custom to use a dummy variable: $$\ln x := \int_1^x \frac{dt}{t} \, .$$ – Fly by Night Sep 9 '12 at 17:56
It usually makes more sense to start with some convenient $A$ and take the antiderivative (in some neighbourhood of $A$) that is $0$ at $A$. – Robert Israel Sep 9 '12 at 18:20
@Robert Israel : I think what you meant is we work with the integration constant and/or integrate f(z) + C_2 such that we can find an antiderivative at A that is 0 whereever we want (and then pick a convenient A and/or place). – mick Sep 9 '12 at 18:29
@Fly by Night : i edited it slightly. Im aware of the dummy variable. – mick Sep 9 '12 at 18:30

If $f$ is an entire function that is not a polynomial and $F$ is an antiderivative of it, the little Picard theorem says $F$ takes on every complex value with at most one exception. That exception could be $0$, in which case $F = \exp(G)$ for some entire function $G$. Offhand I couldn't think of an example of this where $f$ is elementary but $F$ is not.