# Gradient of an integral

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. – achichi Sep 15 '11 at 16:52
Yes. Sorry for not suggesting that earlier. I reedited the question. – achichi Sep 15 '11 at 17:10
Okay, in which case you may be interested in this answer (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 and a smooth f'n. – Willie Wong Sep 15 '11 at 18:49
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 $dxdydz$. – achichi Sep 16 '11 at 8:20

## 1 Answer

The operator $\nabla$ can be passed inside the integral if some suitable conditions on $g$ are fulfilled. There are appropriate theorems on differentiating integrals with respect to parameter. It can be done with potential $V\;$ if the function $\rho$ (say) is bounded in $\mathbb R^3$ and has bounded support, since in this case the integral $\int_{\mathbb{R}^3}\nabla{\!\!}_{x}\frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})d^3\textbf{y}$ will be converging absolutely and uniformly.

The Laplace operator cannot be put inside the integral because it would mean that $\Delta V(x)\equiv0$ and for smooth enough $\rho$ actually $\Delta V(x)=\rho(x)$. The theorem aplied above doesn't work because the integral $\int_{\mathbb{R}^3}\left|\Delta \frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})\right| d^3\textbf{y}$ may diverge. The expression $\left|\Delta \frac{1}{|\textbf{x}-\textbf{y}|}\right|$ has non-integrable singularity at $y=x$.

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