In the introduction chapter of the book Navier-Stokes Equations and Turbulence, the following is stated with only a one-sentence proof:
Suppose $u$ is a smooth velocity field on ${\bf R}^d$ ($d=2,3$) decaying sufficiently rapidly at infinity and it is divergence free: $\nabla\cdot u=0$. Then integration by parts implies that $$ \sum_{i=1}^d\int_{{\bf R}^d}|\nabla u_i(x)|^2\ dx=\int_{{\bf R}^d}|\nabla\times u|^2\ dx.\tag{1} $$
None of the versions of integration by parts I know involves the curl operator.
Could anyone elaborate how (1) can be proved?
Well. For $n=2$, you have \begin{align} \int_{\mathbb{R}^2} \left|\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\right|^2\ dxdy =&\ \int_{\mathbb{R}^2}|\partial_x u_2|^2+|\partial_y u_1|^2-2(\partial_x u_2 \partial_y u_1)\ dxdy\\ =&\ \int_{\mathbb{R}^2}|\partial_x u_2|^2+|\partial_y u_1|^2+(\partial_x\partial_y u_2) u_1+u_2(\partial_y\partial_x u_1)\ dxdy. \end{align} Since \begin{align} \nabla\cdot u = \partial_x u_1 + \partial_y u_2 = 0 \ \ \implies \ \ \ \partial_y u_2 = -\partial_x u_1 \end{align} then we see that \begin{align} \int_{\mathbb{R}^2}(\partial_x\partial_y u_2) u_1+u_2(\partial_y\partial_x u_1)\ dxdy =&\ -\int_{\mathbb{R}^2}(\partial_x^2 u_1) u_1+u_2(\partial_y^2 u_2)\ dxdy\\ =&\ \int_{\mathbb{R}^2}|\partial_x u_1|^2+|\partial_y u_2|^2\ dxdy. \end{align} Hence it follows that \begin{align} \int_{\mathbb{R}^2} \left|\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\right|^2\ dxdy = \sum^2_{i=1}\int_{\mathbb{R}^2}|\nabla u_i|^2\ dxdy. \end{align}
For the $n=3$ case, we will need to use the following vector identities \begin{align} \nabla \cdot (\mathbf{F}\times \operatorname{curl}\mathbf{G}) =&\ ( \operatorname{curl}\mathbf{F})\cdot(\operatorname{curl} \mathbf{G})-\mathbf{F}\cdot(\operatorname{curl}(\operatorname{curl}\mathbf{G}))\\ =&\ (\operatorname{curl} \mathbf{F})\cdot(\operatorname{curl}\mathbf{G})- \mathbf{F}\cdot (\nabla(\nabla\cdot \mathbf{G})-\nabla^2\mathbf{G}) \end{align} where $\nabla^2$ is the vector Laplacian. Any how, set $\mathbf{F} = \mathbf{G} = u$, then we have \begin{align} \nabla\cdot (u\times \operatorname{curl} u) = |\operatorname{curl}u|^2+u\cdot \nabla^2 u \end{align} which means \begin{align} \int_{\mathbb{R}^3}|\nabla\times u|^2\ d^3x =&\ \int_{\mathbb{R}^3}\nabla\cdot(u\times [\nabla \times u])-u\cdot \nabla^2u\ d^3x\\ =&\ \sum^3_{i=1}\int_{\mathbb{R}^3} |\nabla u_i|^2\ d^3x. \end{align} Note that \begin{align} \nabla^2 u = [ \nabla^2 u_1, \nabla^2 u_2, \nabla^2 u_3]. \end{align}