# Vector analysis: \nabla applied on integral?

1. Dec 12, 2008

### Gerenuk

Is it possible to simply or rewrite
$$\nabla_{\vec{r}}\iiint\frac{f(\vec{r}\,')\mathrm{d}\vec{r}\,'}{4\pi|\vec{r}-\vec{r}\,'|}$$
$$\nabla_{\vec{r}}\times\iiint\frac{\vec{A}(\vec{r}\,')\mathrm{d}\vec{r}\,'}{4\pi|\vec{r}-\vec{r}\,'|}$$
?

What's a good reference (internet, book) to learn this sort of vector analysis?

2. Dec 13, 2008

### mathman

If the function is sufficiently well-behaved, you can put the nabla under the integral sign.

3. Dec 13, 2008

### jostpuur

Yes.

I would like to know too So far I've been collecting information from quite random sources.

Those equations look like they could be related to electromagnetism. I'm not sure if this is the case, but anyway, notably large portion of information sources that deal with concrete and useful formulas, are some kind of "methods for physicists" things. I don't know what kind of info you are after. The physicists' stuff isn't rigor enough for my taste at least.

Last month somebody asked about when the order of differentiation and integration can be changed, and I gave a rather detailed answer to it here: interchanging limits and differentiation/integration If you have not yet got introduced into real analysis, it could be that my post there is too heavy, but anyway... you should now that sometimes you can change the order of integration and differentiation, and sometimes you cannot.

hmhmhmhmhmhmmh........ yeeaaah.... well I think it is so, but that's a little bit dangerous. For example $\nabla^2$ would not commute with the integration!

This is a very useful trick, which should be remembered: Sometimes it happens that you have such functions $f,g$ that you cannot change the order of integration and differentiation in the expression

$$D_x \int dy\; f(y)g(x-y),$$

but it turns out that you can change the order, if you first perform a suitable change of variable in the integral, and write it like this

$$D_x \int du\; f(x-u)g(u) = \int du\; \big(D_x f(x-u)\big)g(u).$$

The first expression in the original post can be written like this

$$\int d^3r'\;\frac{1}{4\pi} \frac{f(r-r')}{\|r'\|}$$

and it could be useful when differentiating with respect to the $r$.

This is all related to a formula

$$\nabla_x\cdot\frac{x-x'}{\|x-x'\|^3} = 4\pi\delta^3(x-x').$$

A one possible more rigor formulation of this would be a formula

$$\int d^3x\; \big(\nabla f(x)\big)\cdot \frac{x-x'}{\|x-x'\|^3} = -4\pi f(x')$$

for suitable test functions $f$. Some time ago there was a discussion about spins, which slightly got distracted to the use of this delta function formula. I wrote some somewhat rigor stuff back then, so in case you are interested, check the post #37 from the thread What exactly is spin? There the delta function formula was calculated with some particular test function $f$, but the calculation would work out the same way, if a more arbitrary test function $f$ would have been used.

4. Dec 13, 2008

### jostpuur

I just realized I had not read your first sentence very carefully. What does it actually mean? I guess you wrote some correct sentence first, and then started editing it... and edited and posted it at the same time? It happens once in a while...

The question to which I gave the "yes" answer was "is it possible to simply write?".

5. Dec 13, 2008

### jostpuur

Here's the simplest (to my knowledge) example of the change of variable I talked about:

Consider some differentiable and integrable (over $\mathbb{R}$) test function $f:\mathbb{R}\to\mathbb{R}$, which satisfies a condition $$\lim_{x\to-\infty}f(x)=0$$, and the step function

$$\theta(x) =\left\{\begin{array}{ll} 1,\quad & x > 0\\ \frac{1}{2},\quad & x=0\\ 0,\quad & x < 0\\ \end{array}\right.$$

An incorrect calculation:

$$D_x\int\limits_{-\infty}^{\infty} dy\; f(y)\theta(x-y) = \int\limits_{-\infty}^{\infty} dy\; f(y)\big(\underbrace{D_x\theta(x-y)}_{=0,\;\textrm{a.e.}}\big) = \int\limits_{-\infty}^{\infty} dy\; 0 = 0$$

A correct calculation:

$$D_x\int\limits_{-\infty}^{\infty} dy\; f(y)\theta(x-y) = D_x\int\limits_{-\infty}^{\infty} du\; f(x-u)\theta(u) = \int\limits_{-\infty}^{\infty} du\; \big(D_x f(x-u)\big)\theta(u) = \int\limits_{0}^{\infty} du\; D_x f(x-u)$$
$$= \int\limits_{0}^{\infty} du\; D_u\big(-f(x-u)\big) = -f(-\infty)+f(x) = f(x)$$

Then there are some other variants known as "physicists' calculations", of course But it won't hurt to understand a correct one too.

edit: hmmh.. it could be that this is "too simple" in some way, because the expression could be calculated easily also without the change of variable in the beginning. Anyway, this is an example of how the technique can be used, and there are more complicated examples out there too (like the three dimensional one mentioned in the earlier posts), where the technique cannot be replaced easily with anything else.

Last edited: Dec 13, 2008