I am trying to prove that path independence of integral of 1-form implies the form is exact.
Suppose we have 1-form $\omega=Pdx+Qdy$.
We need to show that there exist function $F: \mathbb{R}^2 \supset U \rightarrow \mathbb{R}$ such that $\omega=dF$.
I am trying to show that
$$F(x, y) = \int_{(0, 0)}^{(x, y)} Pdx + Qdy$$
satisfies $\omega=dF$.
(since the integral is path-independent, it doesn't matter how we get from $(0, 0)$ to $(x, y)$)
It obviously can be done by calculating $\partial F / \partial x$ and $\partial F / \partial x$.
Ok, now, I am doing the trick. For calculating $\frac{\partial F}{\partial y}$ let's move towards
$(x,y)$ like that:
(path-independence)
It's not hard to parametrize $\gamma_1(t)=\begin{bmatrix} 0 \\ yt \end{bmatrix}$, where $t \in [0, 1]$
Let's denote $\gamma = \gamma_1 \gamma_2$
Evaluating $F(x, y):$
$$
\int_\gamma Pdx + Qdy = \\
=\int_{\gamma_1} (Pdx + Qdy) + \int_{\gamma_2} (Pdx + Qdy) = \\
=\int_{\gamma_1} Qdy + \int_{\gamma_2} Pdx
$$
And it looks like if we calculate the derivatives ${\partial}/{\partial y}$ and ${\partial}/{\partial x}$ componentwise we get what we want. But how to show that
$$
\frac{\partial}{\partial y} \int_{\gamma_1} Qdy = Q \quad ?
$$
I've tried this way:
$$
\frac{\partial}{\partial y} \int_{\gamma_1} Qdy = \frac{\partial}{\partial y} \int_0^1 \langle \begin{bmatrix}
0 \\
Q(\gamma_1(t)) \\
\end{bmatrix} , \dot{\gamma_1}(t) \rangle dt = \\
=\frac{\partial}{\partial y} \int_0^1 Q(0, yt) \cdot y dt=\text{but it doesn't seem to help}
$$
Best Answer
We see that $\frac{\partial F}{\partial y}|_{(x_0, y_0)} = \frac{d}{dt} G(t)|_{t = 0}$, where $G(t) = F(x_0, y_0 + t)$.
Consider that $F(x_0, y_0 + t) = F(x_0, y_0) + \int\limits_{(x_0, y_0)}^{(x_0, y_0 + t)} \omega$. This follows directly from path independence, since we can get a path from $(0, 0)$ to $(x_0, y_0 + t)$ by a path from $(0, 0)$ to $(x_0, y_0)$ with a path from $(x_0, y_0)$ to $(x_0, y_0 + t)$.
In other words, $G(t) = G(0) + \int\limits_{(x_0, y_0)}^{(x_0, y_0 + t)} \omega$.
Now the path does not matter. So we take the path parametrized by $[0, t]$ which maps $h \mapsto (x_0, y_0 + h)$. Then we see that $\int\limits_{(x_0, y_0)}^{(x_0, y_0 + t)} \omega = \int\limits_0^t Q(x_0, y_0 + h) dh$.
Thus, we see that $G(t) = G(0) + \int\limits_0^t Q(x_0, y_0 + h) dh$.
Now by the fundamental theorem of calculus, we see that $G'(t) = Q(x_0, y_0 + t)$. Thus, $G'(0) = Q(x_0, y_0)$ in particular. So $\frac{\partial F}{\partial y}|_{(x_0, y_0)} = Q(x_0, y_0)$.
The same argument holds for $\frac{\partial F}{\partial x}|_{(x_0, y_0)}$. Thus, we see that $dF = \omega$.
It's not hard to see that this proof generalises fully to 1-forms in $n$ dimensions for all $n$. Any 1-form in $n$-dimensional space which has the path-invariance property is exact.
You were very close to getting the answer. But you parametrized paths on $[0, 1]$ instead of parametrizing the path on $[0, t]$, which was the necessary step to use the fundamental theorem.