Problem:
For my thesis I am trying to prove that
$$ \int \nabla_\eta g(\eta)p(x-\eta)d\eta= \nabla_x \int g(\eta) p(x-\eta)d\eta $$
where $x,\eta \in \mathbb{R}^n$
(Equation (12) in https://arxiv.org/pdf/1709.03749.pdf)
My approach:
Since the $\nabla $-operator within the integral is not with respect to $x$, I cannot use Leibnitz rule directly.
Hence I tried to use rules that involve convolution:
$$
\begin{align}
\int \nabla_\eta g(\eta)p(x-\eta)d\eta &= \int g'(\eta)p(x-\eta)d\eta \\
&= (g'*p)(x) \\
&= (g*p)'(x) \\
&=\nabla_x \int g(\eta) p(x-\eta)d\eta
\end{align}
$$
I'm very much aware that my proof isn't accurate (e.g. I ignored that things are multidimensional) but that was the best try I could come up with and it is also the way the authors suggested it on the related GIT-hub site.(https://github.com/siavashBigdeli/DMSP/issues/1)
I would very much appreciate any help with this !
$\require{cancel}$ First off, note that
$$ \nabla_{\eta}p(x - \eta) = -\nabla_x p(x - \eta) \tag{1} $$
and now use the product rule
\begin{eqnarray} \int_V \nabla_\eta g(\eta) p(x - \eta)~{\rm d}\eta &=& \int_V\left\{\nabla_\eta[g(\eta)p(x - \eta)] - g(\eta) \nabla_\eta p(x - \eta)\right\}{\rm d}\eta \\ &=& \int_V \nabla_\eta[g(\eta)p(x - \eta)]~{\rm d}\eta - \int_V g(\eta) \nabla_\eta p(x - \eta){\rm d}\eta \\ &=& \cancelto{0}{\left. g(\eta) p(x - \eta)\right|_{S}} - \int_V g(\eta)\nabla_\eta p(x - \eta)~{\rm d}\eta \\ &\stackrel{(1)}{=}& + \int_V g(\eta)\nabla_xp(x - \eta)~{\rm d}\eta \\ &=& \nabla_x\int_V g(\eta)p(x - \eta)~{\rm d}\eta \tag{2} \end{eqnarray}
where I have assumed the product $g(\eta) p(x - \eta)$ goes to zero on the boundary $S$ of the integration volume $V$