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I have a very basic question on complex integration.

How is the definite integral

$$ \int_{z_1}^{z_2}{f(z)dz} $$

$z \in \Bbb{C}$ to be interpreted in the absence of a specific path over which integration is to take place?

I find many spacial functions which have integral representations of the above form, but which do not specify the path over which the line integral is to be performed. Also, what is the correct method for evaluating the real and imaginary parts of such integrals?

Any help will be appreciated. I suspect that this is a fundamental question, so any pointers to educational resources explaining this concept would also be parreciated.

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    $\begingroup$ This is something that would need to be specified, unless the functions were analytic (and thus path independent). $\endgroup$ – davidlowryduda Jun 4 '14 at 21:23
  • $\begingroup$ @mixedmath you would also need the domain to be simply connected. $\endgroup$ – Steven Gubkin Jun 4 '14 at 21:26
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    $\begingroup$ I would say, if that notation is used by a reasonable individual then there is data to yield the independence of path. $\endgroup$ – James S. Cook Jun 4 '14 at 21:28
  • $\begingroup$ Right, I see! I suspected that there must exist some basic result that would establish path independence for holomorphic functions. Thanks! $\endgroup$ – JMK Jun 4 '14 at 21:45
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    $\begingroup$ I suggest Markushevich. It's a classic text with lots of nice proofs and pictures. Also, for a measure-theoretic approach, Papa Rudin is nice as well. $\endgroup$ – Alex Schiff Jun 6 '14 at 15:36
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In the generality you have posed the question, there is no way to answer. After all:

$$\int_1^1 \frac{1}{z}dz$$

could be $0$ or $2\pi i$ depending on which path you take!

On the other hand, if the domain of the function you are studying is simply connected, and the integrand is holomorphic, then the Cauchy integral theorem says that any path will give the same answer, so that the expression you have written is well defined.

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It might be helpful to work our way backwards. This will not be formal, but hopefully it will be illuminating. Let $f:G\to\mathbb{C}$ be a function, where $G\subset\mathbb{C}$. Let $z_1,z_2\in G$. What restrictions do we need to put on $f$ and $G$ in order for the expression $\int_{z_1}^{z_2}f(z)\,dz$ to make sense?

Well first of all, it would be helpful if there was a way to get from $z_1$ to $z_2$ while staying within $G$, preferably in a nice (smooth) manner. More precisely, there should exist a function $\gamma:[0,1]\to G$ such that $\gamma(0)=z_1$, $\gamma(1)=z_2$, and $\gamma$ is smooth. Thus, there is a path from $z_1$ to $z_2$. Since we are trying to make $G$ sufficiently "nice", we may as well put on the restriction that $G$ is path connected.

What if there are two such paths $\gamma_1,\gamma_2:[0,1]\to G$ such that $\gamma_1(0)=\gamma_2(0)=z_1$ and $\gamma_1(1)=\gamma_2(1)=z_2$? Our expression $\int_{z_1}^{z_2}f(z)\,dz$ is somewhat ambiguous; it does not specify how we get from $z_1$ to $z_2$. Therefore, it is natural to place another restriction on $G$: $G$ must be simply connected. One should be able to deform $\gamma_1$ to $\gamma_2$ and vice-versa in a nice way. We don't really want to worry about a path $\gamma$ intersecting a boundary point of $G$ or any other weirdness, so let's assume that $G$ is open.

Finally, we want our function $f$ to behave well on $G$. Let's say that $f$ is continuous. If $\gamma:[0,1]\to G$ is a smooth function such that $\gamma(0)=z_1$ and $\gamma(1)=z_2$, we are justified in making the $u$-substitution, $z=\gamma(t)$. So when $z=z_1$, $t=0$; when $z=z_2$, $t=1$; and $dz=\gamma'(t)\,dt$. Thus: $$ \int_{z_1}^{z_2}f(z)\,dz:=\int_0^1f(\gamma(t))\cdot\gamma'(t)\,dt. $$ By the restrictions we placed on $f$ and $G$, this expression is well defined, as it is independent of the choice of $\gamma$.

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