On Complex Improper Integrals and Rectifiable Paths I would like to know if we may generalize both definite and improper complex integral as follows:
I tried to write as Conway does in his book.
Let's define a path in $U \cup \partial{U}$, $\Gamma$, as follows
$$\Gamma: [0,1] \to U \cup \partial{U} \\ u \mapsto \Gamma(u) = p(z) + u \cdot [q(z) - p(z)]$$
Let $g: U \to \mathbb{C}$ and $p,q: D \to U \cup \partial{U}$ both analytic on their open and connected domains, and $G: U \cup \partial{U} \to \mathbb{C}$, primitive of $g$. Then $I$ converges for every $z \in D$.
$$ I: D \to \mathbb{C}  \\z \mapsto I(z) = \int_\Gamma g(s)\,ds = G(q(z)) - G(p(z)), \\ \Gamma: [0,1] \to U \cup \partial{U} \\ u \mapsto \Gamma(u) = p(z) + u \cdot [q(z) - p(z)]\\ \, \\$$
The main question: Is this how complex improper integrals are defined? I.e., we define a path which one of its end point is at the boundary of the integrand's domain? Is there anything missing? Most books emphazise solving complex improper integrals via residues, but that's not what I want now.
Does $\Gamma$ need to be rectifiable? I guess it is already
As far as I know, we have the following possibilities:
P1 $I(z)$ will be a definite integral, if  $z \in D$ such that $p(z), q(z) \in U$.
P2 On the other hand, $I(z)$ will be an improper integral, if $ z \in D$ such that $ p(z) \lor q(z) \in \partial{U}$. Would this be correct?
From Conway's Graduate book, we have Theorem 1.18 (which I'm using as an example):
Let $G$ be open in $\mathbb{C}$ and let $\gamma$ be a rectifiable path in $G$ with initial and end points $\alpha$ and $\beta$ resceptively. If $f: G \to \mathbb{C}$ is a continuous function with a primitive $F:G \to \mathbb{C}$, then
$$\int_\gamma f = F(\beta)-F(\alpha)$$
(Recall that $F$ is a primitive of $f$ when $F'=f$.)
I have read the whole chapter, but no detailed explanation was found. Thus, any other references would be appreciated.
Thanks
EDIT It's been a while since the bounty started and no answers. I'm starting to think that my question is "wrong" somehow. Please comment anything that may be corrected. Thanks again
EDIT 2 If $g$ is analytic on its open domain $U$, shouldn't it be also at $\partial{U}$? If so, we may use the Theorem 1.18. Right?
 A: My favorite reference for this is Rudin's book "Real and complex analysis." Instead of "rectifiable" Rudin says "bounded variation." Piecewise $C^1$ curve  implies absolutely continuous, which in turn implies bounded variation. But all these implications are not reversible. For a basic course in Complex Analysis, I strongly advise using piecewise-$C^1$-smooth curves and avoid all the complications coming from the more general notions.
Now, a curve $\gamma: I\to {\mathbb C}$ is piecewise $C^1$-smooth if it is continuous and its domain can be subdivided into intervals $I_k$ such that $\gamma$ is $C^1$ on each closed subinterval $I_k$ (but its left/right derivatives at the subdivision points may differ). Thus, each piecewise $C^1$-curve is continuously differentiable except for a discrete subset of its domain, an interval $I$ (bounded or unbounded). This allows you to get by the Riemann integral which is what likely you learned in  a calculus/real analysis class. Given a continuous function  $f$ on $\gamma(I)$, you can define the integral
$$
\int_\gamma f(z)dz
$$
as
$$
\int_a^b f(\gamma(t)) \gamma'(t)dt.
$$
You can also define this integral using an approximation procedure using line segments, as you probably saw in a calculus/ complex analysis class. The answer will be the same.
This makes sense since the integrand is defined and is continuous everywhere except for the subdivision points $t_k$ (where $\gamma'(t)$ is undefined). However, since the one-sided derivatives of $\gamma'_\pm(t_k)$ are assumed to exist and $\gamma'$ is continuous on each $[t_k, t_{k+1}]$, you get an integral (possibly improper) of the type you encountered in a calculus class. One still needs to take care of improper integration in case $I$ is unbounded or is not closed, but this becomes a "calculus issue."
More generally, things also work pretty much the same way when $\gamma$ is absolutely continuous, the formula
$$
\int_a^b f(\gamma(t)) \gamma'(t)dt
$$
still makes sense, but you have to be aware of the fact that $\gamma'(t)$ exists only "almost everywhere" and in order to define this integral, Riemann integration is not enough, you need Lebesgue integral.
A nice thing about absolutely continuous functions (whatever they are, see Rudin's book for the precise definition), is that you can assume that the ordinary rules of calculus apply; in particular, the length of $\gamma$ is still given by the formula
$$
\int_a^b |\gamma'(t)|dt. 
$$
Just remember that the integral is more complicated than the one you learned in a calculus class.
A general rectifiable function is a BV (bounded variation) function and you only have the inequality
$$
\int_a^b |\gamma'(t)|dt \le length(\gamma).  
$$
You can have a monotonic (hence, BV) continuous function $\gamma: [0,1]\to [0,1]$ satisfying $\gamma(0)=0, \gamma(1)=1$, but $\gamma'(t)=0$ almost everywhere, which means that the integral of the derivative vanishes, but the curve, of course, has unit length. Such issues one should avoid while learning basic complex analysis.
