Hello, I was going through Ahlfors' Complex Analysis where I found some difficulty to underestand the proof of the followning theorem. This is the way he explains it in the book:
Update : $\Omega$ is a region. i.e. an open connected subset of plane.
The following theorem gives a necessary and sufficient condition under which a line integral depends only on the end points.
Theorem I. The line integral $\int_{\gamma} {p dx + q dy} $, defined in $ \Omega $, depends only on the end points of $\gamma$ if und only if there exists a function $U(x,y)$ in $\Omega$ with the partial derivatives $ \dfrac{\partial U}{\partial x} = p $, $\dfrac{\partial U}{\partial y} = q $.
The proof goes like this :
The sufficiency follows at once, for if the condition is fulfilled we can write, with the usual notations, $ \int_{\gamma}{pdx+qdy} = \int_{a}^{b}{ (\dfrac{\partial U}{\partial x} x'(t) + \dfrac{\partial U}{\partial y} y'(t)) dt} = \int_{a}^{b}{\dfrac{d}{dt} U(x(t),y(t))dt} = U(x(b),y(b))-U(x(a),y(a)) $
and the value of this difference depends only on the end points.
To prove the necessity we choose a fixed point $(x_0,y_0) \in \Omega $ join it to $(x,y)$ by a polygon $\gamma$, contained in $\Omega$, whose sides are parallel to the coordinate axes and define a function by $ U(x,y) = \int_{\gamma}{p dx + q dy} $.
Since the integral depends only on the end points, the function is well defined. Moreover, if we choose the last segment of $\gamma$ horizontal, we can keep $y$ constant and let $x$ vary without changing the other segments.
On the last segment we can choose $x$ for parameter and obtain $ U (x,y) = \int_{}^{x} p(x,y) dx + \text{some constant} (*)$, the lower limit of the integral being irrelevant.
What I Realized from this proof : suppose the last horizontal segment mentioned above conects $(a,b)$ to $(x,y)$.
Now since $\Omega$ is an open set for every $(x',y)$ is near enough to $(x,y)$ ( $x'$ belongs to a $\delta$-neighbourhood of $x$ and $x'-\delta>a$ ) we can find a polygon path connecting $(x_0,y_0)$ to $(x',y)$ which its last horizontal segment begins connects $(a,b)$ to $(x,y)$ and the other sides of this path is as same as $\gamma$.
In fact we can right $ \int_{\gamma'}{pdx+qdy} + \int_{\gamma"}{pdx+qdy} $ where $\gamma'$ is always the first part of $\gamma$ connecting $(x_0,y_0) to (a,b) $ and $\gamma"$ depends on $x'$.
This explains the equation $(*)$ except I can't realize where did $q$ go from the integrals?!
Note That the book doesn't work with differential forms. from the text we have these two definitions:
$ \int{f(z)dx} = \dfrac{1}{2}(\int_{\gamma}{f(z)dz} + \overline{\int_{\gamma}{f(z)dz}})$
$ \int{f(z)dy} = \dfrac{1}{2i}(\int_{\gamma}{f(z)dz} - \overline{\int_{\gamma}{f(z)dz}})$
I appreciate your help.
Update : I'm thinking that now I doubt how the first equality of the first part of the proof came from in the first place as well.
Update 2: those definitions I wrote was wrong! These are correct:
$ \int{f(z)dx} = \dfrac{1}{2}(\int_{\gamma}{f(z)dz} + \overline{\int_{\gamma}{\overline{f(z)}dz}})$
$ \int{f(z)dy} = \dfrac{1}{2i}(\int_{\gamma}{f(z)dz} - \overline{\int_{\gamma}{\overline{f(z)}dz}})$