What algebraic function will result in transcendental function by indefinite integral? Let $f(x)$ be a algebraic function over $\mathbb{C}$, under what condition will $\int f(x)dx$ be a transcendental function?
 A: 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 $C$. 
