What algebraic function will result in transcendental function by indefinite integral?

Question

Let $f(x)$ be a algebraic function over $\mathbb{C}$, under what condition will $\int f(x)dx$ be a transcendental function?

Answer 1

A function is algebraic if it is really a function on a Riemann surface $X$ covering the Riemann sphere $\Bbb C_\infty$. The surface $X$ will be compact, and the covering will be finite sheeted and ramified at finitely many points.

The differential $dz$ on $\Bbb C_\infty$ pulls back to a differential on $X$, and so we can regard $\omega=f(z)\,dz$ as a differential on $X$. This differential may have poles, but only finitely many, and so let $X'=X$ with these poles removed. Then topologically $X$ is a surface of finite genus with finitely many punctures, and so its first homology group is finitely generated Abelian.

If we pull $\omega$ back to the universal cover of $X'$ then it is the derivative of a holomorphic function $g$ there, but there may be topological obstructions to $g$ being a function on $X'$ itself.

If $\int_C\omega\ne0$ for some closed contour $C$ in $X'$ then any branch of $g$ on $X'$ can be analytically continued to infinitely many other branches of $g$. So $g$ cannot be defined on a finite-sheeted cover of $\Bbb C_\infty$. In this case $\int f(z)\,dz$ cannot be an algebraic function.

Otherwise if $\int_C\omega=0$ for all closed contours in $X'$ then $f$ has an indefinite integral on $X'$ and this means that $f$ has an algebraic indefinite integral on $\Bbb C_\infty$. As $H_1(X')$ is finitely generated, one only needs $\int_C\omega=0$ for a suitable finite collection of contours $C$.