[proofplan] The argument has two independent calculations. First, we differentiate $\alpha$ directly and verify that the mixed partials of the coefficient functions agree, so $d\alpha = 0$. Second, to show $\alpha$ is not exact, we integrate $\alpha$ around the unit circle $\gamma: [0, 2\pi] \to U$, $\gamma(t) = (\cos t, \sin t)$ and obtain $\oint_\gamma \alpha = 2\pi \neq 0$. By the theorem that exact $1$-forms have zero integral around any closed curve, $\alpha$ cannot be exact. The nontriviality of $H^1_{\mathrm{dR}}(U)$ follows immediately, since $[\alpha]$ is a nonzero class in $Z^1(U) / B^1(U)$. [/proofplan] [step:Verify that $\alpha$ is a well-defined smooth $1$-form on $U$] The coefficient functions \begin{align*} P: U &\to \mathbb{R}, & P(x, y) &= \frac{-y}{x^2 + y^2}, \\ Q: U &\to \mathbb{R}, & Q(x, y) &= \frac{x}{x^2 + y^2} \end{align*} are quotients of polynomials whose denominator $x^2 + y^2$ vanishes only at the origin. Since $0 \notin U$, both $P$ and $Q$ are smooth on $U$, and $\alpha = P\, dx + Q\, dy \in \Omega^1(U)$. [/step] [step:Compute $d\alpha$ and verify closedness] For a smooth $1$-form $\alpha = P\, dx + Q\, dy$ on an open subset of $\mathbb{R}^2$, \begin{align*} d\alpha = \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \wedge dy. \end{align*} We compute each partial derivative on $U$ via the quotient rule. For $Q(x, y) = x \cdot (x^2 + y^2)^{-1}$: \begin{align*} \frac{\partial Q}{\partial x}(x, y) = \frac{1 \cdot (x^2 + y^2) - x \cdot 2x}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}. \end{align*} For $P(x, y) = -y \cdot (x^2 + y^2)^{-1}$: \begin{align*} \frac{\partial P}{\partial y}(x, y) = \frac{-1 \cdot (x^2 + y^2) - (-y) \cdot 2y}{(x^2 + y^2)^2} = \frac{-(x^2 + y^2) + 2y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}. \end{align*} Subtracting, \begin{align*} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{y^2 - x^2}{(x^2 + y^2)^2} - \frac{y^2 - x^2}{(x^2 + y^2)^2} = 0. \end{align*} Therefore $d\alpha = 0$ on $U$, i.e., $\alpha$ is closed. [guided] The [exterior derivative](/theorems/1525) of a smooth $1$-form $\alpha = P\, dx + Q\, dy$ on an open subset of $\mathbb{R}^2$ is given in coordinates by \begin{align*} d\alpha = dP \wedge dx + dQ \wedge dy = \left(\frac{\partial P}{\partial x} dx + \frac{\partial P}{\partial y} dy\right) \wedge dx + \left(\frac{\partial Q}{\partial x} dx + \frac{\partial Q}{\partial y} dy\right) \wedge dy. \end{align*} Using $dx \wedge dx = dy \wedge dy = 0$ and $dy \wedge dx = -dx \wedge dy$, this simplifies to \begin{align*} d\alpha = \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \wedge dy. \end{align*} So showing $d\alpha = 0$ reduces to checking that the two mixed partial expressions $\partial_x Q$ and $\partial_y P$ agree. We compute via the quotient rule. For $Q(x, y) = x/(x^2 + y^2)$: \begin{align*} \frac{\partial Q}{\partial x}(x, y) = \frac{1 \cdot (x^2 + y^2) - x \cdot 2x}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}. \end{align*} For $P(x, y) = -y/(x^2 + y^2)$: \begin{align*} \frac{\partial P}{\partial y}(x, y) = \frac{-(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-(x^2 + y^2) + 2y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}. \end{align*} The two expressions agree pointwise on $U$, hence $\partial_x Q - \partial_y P \equiv 0$ on $U$, and consequently $d\alpha = 0$. It is worth recording the geometric meaning: away from the origin, $\alpha$ is the differential of the polar angle $\theta$. Locally on any simply-connected open subset $V \subseteq U$ one can choose a smooth branch of $\theta: V \to \mathbb{R}$ with $\alpha = d\theta$ on $V$, so $\alpha$ must be closed. The point of the next step is that no such $\theta$ exists globally on $U$. [/guided] [/step] [step:Compute $\oint_\gamma \alpha$ over the unit circle] Define the smooth closed curve \begin{align*} \gamma: [0, 2\pi] &\to U, \\ t &\mapsto (\cos t, \sin t). \end{align*} This is well-defined since $|\gamma(t)| = 1 \neq 0$ for all $t$, so $\gamma(t) \in U$. Moreover $\gamma(0) = \gamma(2\pi) = (1, 0)$. The pullback $\gamma^* \alpha \in \Omega^1([0, 2\pi])$ is computed by substituting $x(t) = \cos t$, $y(t) = \sin t$, $dx = -\sin t\, dt$, $dy = \cos t\, dt$, and noting that $x(t)^2 + y(t)^2 = 1$: \begin{align*} \gamma^* \alpha = \frac{-\sin t \cdot (-\sin t) + \cos t \cdot \cos t}{1}\, dt = (\sin^2 t + \cos^2 t)\, dt = dt. \end{align*} Therefore \begin{align*} \oint_\gamma \alpha = \int_0^{2\pi} \gamma^* \alpha = \int_0^{2\pi} 1\, d\mathcal{L}^1(t) = 2\pi. \end{align*} [/step] [step:Conclude $\alpha$ is not exact via the closed-curve criterion] By [Exact 1-Forms Have Zero Integral Around Closed Curves](/theorems/3584), if $\alpha = df$ for some $f \in C^\infty(U)$, then $\oint_\gamma \alpha = 0$ for every smooth closed curve $\gamma$ in $U$. The hypotheses of theorem 3584 are satisfied here: $U$ is a smooth manifold, $\gamma: [0, 2\pi] \to U$ is a smooth map with $\gamma(0) = \gamma(2\pi)$, and we would be assuming $\alpha$ is exact. But the previous step gave $\oint_\gamma \alpha = 2\pi \neq 0$. This contradicts exactness, so no such $f$ exists. Hence $\alpha$ is closed but not exact. [guided] The strategy is: the only obstruction to a closed $1$-form being exact is a non-vanishing period — a nonzero integral around some closed loop. We exhibit one such loop. Suppose, for contradiction, that $\alpha = df$ for some $f \in C^\infty(U)$. By [Exact 1-Forms Have Zero Integral Around Closed Curves](/theorems/3584), the integral of any exact $1$-form along any smooth closed curve in $U$ vanishes. Let us verify the hypotheses of theorem 3584 explicitly. The theorem requires (i) a smooth manifold — here $U = \mathbb{R}^2 \setminus \{0\}$ is open in $\mathbb{R}^2$, hence a smooth $2$-manifold; (ii) a smooth closed curve $\gamma: [a, b] \to U$ with $\gamma(a) = \gamma(b)$ — here $\gamma(t) = (\cos t, \sin t)$ is smooth on $[0, 2\pi]$ with $\gamma(0) = \gamma(2\pi) = (1, 0)$; (iii) an exact $1$-form $\alpha = df$ — this is the assumption we are contradicting. All hypotheses hold under the supposition. Applying the theorem would give $\oint_\gamma \alpha = 0$. But Step 3 computed $\oint_\gamma \alpha = 2\pi \neq 0$. This contradiction shows that no antiderivative $f$ exists on the whole of $U$, so $\alpha$ is not exact. A direct way to see what fails: on the slit plane $V := U \setminus \{(x, 0) : x \le 0\}$, $\alpha$ does admit an antiderivative — namely the principal branch of the argument function $\theta: V \to (-\pi, \pi)$. But $\theta$ cannot be extended continuously across the negative real axis: walking once around the origin increases $\theta$ by $2\pi$. The integral $\oint_\gamma \alpha = 2\pi$ is precisely this monodromy, and the global obstruction it measures is the topological fact that $U$ is not simply connected. [/guided] [/step] [step:Deduce $H^1_{\mathrm{dR}}(U) \neq 0$] By definition, the first de Rham cohomology of $U$ is the quotient \begin{align*} H^1_{\mathrm{dR}}(U) := \frac{Z^1(U)}{B^1(U)} = \frac{\{\omega \in \Omega^1(U) : d\omega = 0\}}{\{df : f \in C^\infty(U)\}}. \end{align*} Step 2 shows $\alpha \in Z^1(U)$, and Step 4 shows $\alpha \notin B^1(U)$. Hence the cohomology class \begin{align*} [\alpha] \in H^1_{\mathrm{dR}}(U) \end{align*} is nonzero, and therefore $H^1_{\mathrm{dR}}(U) \neq 0$. This completes the proof. [/step]