I need to evaluate the following expression:
$\int \mathrm{d}\boldsymbol{r} \left[\nabla_{\boldsymbol{R}_\alpha}\delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\right]v(\boldsymbol{r})$
and I want to make use of the fact, that the gradient can be transferred to the function v, I know that in the 1d case
$\int dx \frac{d\delta(x-a)}{dx}f(x) = -\int dx \delta(x-a)\frac{f(x)}{dx}$
But somehow it does not help me a lot in solving the above expression. I guess the solution is not difficult, but I want to be 100% sure about the sign also, since the delta-function has the $\boldsymbol{R}_\alpha$ with a negative sign ahead, or does that not matter and the solution is just
$\int \mathrm{d}\boldsymbol{r} \left[\nabla_{\boldsymbol{R}_\alpha}\delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\right]v(\boldsymbol{r}) = -\int \mathrm{d}\boldsymbol{r} \delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\left[\nabla_{\boldsymbol{R}_\alpha}v(\boldsymbol{r})\right]$???
Please help me in my confusion...
(v depends also on $\boldsymbol{R}_\alpha$ via $v(\boldsymbol{r}) = \sum_\beta v(|\boldsymbol{r}-\boldsymbol{R}_\beta|$)
So far the comments suggest:
\begin{align} \int \mathrm{d}\boldsymbol{r} \left[\nabla_{\boldsymbol{R}_\alpha}\delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\right]v(\boldsymbol{r}) &= -\int \mathrm{d}\boldsymbol{r} \left[\nabla_{\boldsymbol{r}}\delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\right]v(\boldsymbol{r}) \\&= \int \mathrm{d}\boldsymbol{r} \delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\left[\nabla_{\boldsymbol{r}}v(\boldsymbol{r})\right]= \int \mathrm{d}\boldsymbol{r} \delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\left[\sum_\beta \nabla_{\boldsymbol{r}} v_\beta(\boldsymbol{r}-\boldsymbol{R}_\beta)\right]\\ &=-\int \mathrm{d}\boldsymbol{r} \delta(\boldsymbol{r}-\boldsymbol{R}_\alpha)\left[\sum_\beta \nabla_{\boldsymbol{R}_\beta} v_\beta(\boldsymbol{r}-\boldsymbol{R}_\beta)\right] = -\sum_\beta \nabla_{\boldsymbol{R}_\beta} v_\beta(\boldsymbol{R}_\alpha-\boldsymbol{R}_\beta) \end{align}
Is that correct?