Reading derivation of Helmholtz decomposition in wikipedia, there are two points I do not understand. First one appears in the step:
$$
\begin{align}
\dots
&=-\frac{1}{4\pi}\left[\nabla\left(\nabla\cdot\int_V\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)-\nabla\times\left(\nabla\times\int_V\frac{\mathbf{F}(\mathbf{r}')}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right] \\
&= -\frac{1}{4\pi} \left[\nabla\left(\int_V\mathbf{F}(\mathbf{r}')\cdot\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)+\nabla\times\left(\int_V\mathbf{F}(\mathbf{r}')\times\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}'\right|}\mathrm{d}V'\right)\right] \\
&= \dots
\end{align}
$$
where sign of second term jumps from "-" to "+" at same time than curl is inserted in the integral, something I do not see why must change the sign.
The second doubts appears a few lines after, where I can read:
$$
\mathbf{a}\times\nabla\psi =\psi(\nabla\times\mathbf{a})-\nabla \times (\psi\mathbf{a})
$$
because, starting from product rule of curl over scalar and vector:
$$
\nabla \times (\psi\mathbf{a}) = \psi(\nabla\times\mathbf{a}) + \mathbf{a}\times\nabla\psi
$$
I expected:
$$
\mathbf{a}\times\nabla\psi = -\psi(\nabla\times\mathbf{a})+\nabla \times (\psi\mathbf{a})
$$
Could be two obvious errors from my side, but I can not find them. Any help is welcome, thanks.
Best Answer
In both cases: the sign CHANGES if you reverse order of cross product.
$${\bf a}\times {\bf b}=-{\bf b}\times{\bf a}$$
There is a reverse of order in cross product from $\nabla\times {\bf F}f({\bf r})$ to $ {\bf F}\times \nabla f({\bf r})$. Note that the derivative works on $\bf r$, not $\bf r'$ so ${\bf F}$ acts as a constant. So, the minus sign is correct, because the reverse of order happened.
In the second one, you are again reversing the order without paying attention. The product you wrote is wrong, it should be:
$$\nabla\times (\psi {\bf a})=\psi(\nabla\times {\bf a})+(\nabla\psi)\times{\bf a}$$
Notice how ${\bf a}$ which gives the vector direction to the right operand of $\times$, stays on the right, and $\nabla$ which acts as a vector on the left, stays on the left.