Showing why a form is not exact Consider the differential form
$$\omega=\frac{y dx-x dy}{x^2+y^2}$$ on $\mathbb{R}^2-0$.
I have proved the following.
1-This $1$-form is closed on $\mathbb{R}^2-0$.
2-If $f_n:S^1 \to \mathbb{R}^2-0$ is the map $$f_n(\theta)=(\cos(n \theta), \sin(n \theta))$$
Then
$$\int_{S^1} f_n^* \omega= 2\pi n \neq 0$$
and this is where I am stuck.
3-Use the previous parts to show that the form $\omega$ is not exact.

My Thoughts-
If $\omega=d\tau$ is exact, then $d \omega =d^2 \tau=0$ and thus $f_n^* d \omega = d(f_n^* \omega)=0$ . But now I am not sure what to do next? I am thinking about using stokes theorem and looking for guidance in that direction.

 A: The comments give a clear explanation, which I'll re-present here.
Suppose for contradiction that $\omega$ is exact. Then $\omega = d\eta$ for some $0$-form $\eta$. Now
$$f_1^* \omega = f_1^*(d \eta) = d (f_1^* \eta),$$
so by Stokes' Theorem
$$2\pi = \int_{S^1} f_1^* \omega = \int_{S^1} d(f_1^* \eta) = \int_{\partial S^1} f_1^* \eta.$$
On the other hand, since $\partial S^1 = \varnothing$, this integral equals $0$. Contradiction!
A: For
$$\tag{1}
\omega=\frac{y\,dx-x\,dy}{x^2+y^2}=a\,dx+b\,dy
$$
we have
\begin{align}
d\omega&=(\partial_ya)\,dx\wedge dy+(\partial_xb)\,dy\wedge dx\\
&=\frac{(x^2-y^2)\,dx\wedge dy+(x^2-y^2)\,dy\wedge dx}{(x^2+y^2)^2}=0\tag{2}
\end{align}
because $dx\wedge dy$ is antisymmetric.

*

*$\omega$ is defined only on the punctured, and therefore non-simply connected, domain $\mathbb R^2\setminus\{(0,0)\}\,.$ The  Poincare lemma says that, in general, only on simply connected domains all closed $p$-forms, with $p\ge 1\,,$ are exact.


*Because the vector field of which $\omega$ is the dual, namely,
$$\tag{3}
\frac{1}{x^2+y^2}\begin{pmatrix}y\\-x\end{pmatrix}\,,
$$
is tangent to the circles around the origin it looks geometrically as if it should have rotation. But since $\omega$ is closed it doesn't! We have here an example of an irrotational vortex.


*Let's look for the function $\phi$ on any simply connected domain $D\subset\mathbb R^2\setminus\{(0,0)\}$ that satisfies
$$\tag{4}
\omega=d\phi\,.
$$
It turns out that this function is given in polar coordinates by
$$\tag{5}
\phi(r,\theta)=-\theta\,.
$$
The graph of this function is a helix.
In Cartesian coordinates this is
$$\tag{6}
\phi(x,y)=-{\rm sign}( y)\,\arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\,
$$
but we will not use that.
Since,
\begin{align}\tag{7}
\begin{pmatrix}\partial_x\\\partial_y \end{pmatrix}=\begin{pmatrix} \cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}=\begin{pmatrix}\partial_r\\\frac{1}{r}\partial_\theta\end{pmatrix}
\end{align}
it is easy to see from $\partial_r\phi=0$ and $\partial_\theta\phi=-1$ that
\begin{align}\tag{8}
\partial_x\phi=\frac{r\sin\theta}{r^2}=\frac{y}{x^2+y^2}\,,\quad\partial_y\phi=\frac{-r\cos\theta}{r^2}=\frac{-x}{x^2+y^2}
\end{align}
which shows
$$\tag{9}
d\phi=(\partial_x\phi)\,dx+(\partial_y\phi)\,dy=\omega\,.
$$


*The maximal domain on which this $\phi$ can be defined without becoming multivalued is $\mathbb R^2$ with a branch cut, say, $\mathbb R^2\setminus\{x\ge 0,y=0\}\,,$ similar to the complex logarithm.


*

*The example (1) is very popular in homework exercises. I am a bit surprised that the following closed form which is not exact is rarely seen: Instead of (4) and (5) one could take an arbitrary strictly decreasing function $f:[0,+\infty)\to\mathbb R$ and define
$$\tag{10}
\omega=d\phi=df(\theta)\,.
$$
Clearly, $\omega$ is closed, $d\omega=dd\phi=0$ but its integral
around the circle $\theta\in[0,2\pi)$ is $f(2\pi)-f(0)<0\,.$ For example
when $f(\theta)=-\frac{1}{2}\theta^2$ then $\partial_r\phi=0$ and $\partial_\theta\phi=-\theta$ so that here, instead of (8), and using (6),
\begin{align}\tag{11}
\partial_x\phi&=\theta\frac{r\sin\theta}{r^2}={\rm sign}( y)\,
\arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\frac{y}{\sqrt{x^2+y^2}}\,,\\[2mm]
\partial_y\phi&=-\theta\frac{r\cos\theta}{r^2}={\rm sign}( y)\,
\arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\frac{-x}{\sqrt{x^2+y^2}}\,.\tag{12}
\end{align}
This leads to
$$\tag{13}
\omega={\rm sign}( y)\,\arccos\Big(\textstyle\frac{x}{\sqrt{x^2+y^2}}\Big)\displaystyle\frac{y\,dx-x\,dy}{x^2+y^2}\,.
$$
