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Let's suppose that we have a three dimensional function $f(\vec{x})$ which is the integral of some another function $g(\vec{x},\vec{y})$, i.e

$f(\vec{x})=\int_{\mathbb{R}^3}g(\vec{x},\vec{y})d^3 \vec{y}$

What is the gradient of the $f(\vec{x})$? Can the operator pass inside the integral?

$\nabla f(\vec{x})=\nabla\int_{\mathbb{R}^3}g(\vec{x},\vec{y})d^3 \vec{y}=\int_{\mathbb{R}^3}\left[\nabla g(\vec{x},\vec{y})\right]d^3 \vec{y}$

The quantity $\nabla g(\vec{x},\vec{y})$ is a vector and it doesn't make sense to me integrating a vector.

In the case of the Laplacian operator $\nabla^2$ can it pass inside the integral?

Edit: The question was inspired from a physics problem where we have a potential $V(\textbf{x})=-\int_{\mathbb{R}^3}\frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})d^3\textbf{y}$ and we take a gradient to find the accelaration: $g(\textbf{x})=-\nabla V(\textbf{x})=\nabla\int_{\mathbb{R}^3}\frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})d^3\textbf{y}$.

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    I made some corrections to the question, check it again please. The links you gave are not what I'm asking.2011-09-15
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    Yes. Sorry for not suggesting that earlier. I reedited the question.2011-09-15
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    Okay, in which case you may be interested in [this answer](http://math.stackexchange.com/questions/12909/will-moving-differentiation-from-inside-to-outside-an-integral-change-the-resul/12911#12911) (pay also attention to the comments). For the specific case you are considering, the rules given in that link is violated: the corresponding $g(\vec{x},\vec{y})$ is not differentiable in $\vec{x}$ for all $\vec{y}$. In this case the proper way to make sense of the operation is treating it as convolution between a [distribution](http://en.wikipedia.org/wiki/Generalized_functions) and a smooth f'n.2011-09-15
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    Thnx, but the gradient is different from the partial derivative of a function because it's a "vector operator". Also $d^3\vec{r}$ is not a vector, it's a infinitesimal volume $dxdydz$.2011-09-16

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