I am reading "Calculus of Several Variables 3rd Edition" by Serge Lang.
How to prove case 1 of Theorem 4.3 in "Calculus of Several Variables" by Serge Lang?
In Theorem 4.1, for any two points $P, Q$ in $U$, $\int_{P,C}^{Q} F$ was independent of the path $C$ in $U$ joining $P$ and $Q$.
But in Theorem 4.3, Lang specifies the path from $(1,0)$ to $X$ for $\int_{(1,0)}^{X} F$.
So I cannot mimic the proof of Theorem 4.1.
Theorem 4.3
Theorem 4.1
Even though the path is specified, you can still mimic the proof of Theorem 4.1.
Focus on case 1. We need to show that the gradient of $\phi$ is indeed $F=(f,g)$, i.e., $D_1\phi=f$ and $D_2\phi=g$. I’ll show how to mimic the proof that $D_1\phi=f$. (Showing $D_2\phi=g$ is very similar; I leave that to you.)
By definition, $D_1\phi(x)$ is
$$\lim_{h\to0}\frac{\phi(x+he_1)-\phi(x)}{h}$$
Denote by $c_1$ the specified path from $(1,0)$ to $x$, and denote by $c_3$ the specified path from $(1,0)$ to $x+he_1$. Then, by definition of $\phi$, the numerator in the difference quotient above can be written as a difference of line integrals:
$$\int_{c_3}F-\int_{c_1}F$$
In theorem 4.1 we simplify this expression by picking $c_3$ cleverly, but we don’t have that freedom here: $c_3$ is fixed for us. The trick is to notice that $c_3$ and $c_1$ can be combined with a third curve $c_2$ to give a closed path from $(1,0)$ back to itself, where $c_2$ is the line segment running from $x$ to $x+he_1$. In particular, the path $c_1+c_2-c_3$ is closed.
Now, for sufficiently small $h$, we can enclose the path $c_1+c_2-c_3$ in an open rectangle that does not contain the origin. This means that we can apply Theorem 2.1, which says that if $D_2f=D_1g$ throughput a rectangle, then $F$ has a potential in that rectangle. (Ignore how that potential may or may not relate to the $\phi$ we are studying.) You can see now why the assumption that $D_2f=D_1g$ is crucial: we need it to invoke Theorem 2.1.
Now that we know $F$ has a potential in the rectangle, we can apply Theorem 3.1 to conclude the integral of $F$ around the closed curve $c_1+c_2-c_3$ is zero. That means
$$\int_{c_2}F=-\int_{c_1-c_3}F=\int_{c_3-c_1}F$$
But the right-hand side is precisely the numerator of our difference quotient above!
The upshot is that the numerator works out to be exactly the same numerator as in Theorem 4.1, the integral of $F$ along the line segment from $x$ to $x+he_1$. The rest of the proof proceeds the same.
I’ll remark that we could also prove that $F$ vanishes over the closed curve by using Green’s theorem on the region enclosed by the closed curve, but Lang doesn’t prove that until later in the book.