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Homework Help: When can we move the del operator under an integral sign?

  1. Feb 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi, it's me again. I'm new to vector calculus so this might sound like a stupid question, but in relation to a specific problem, I was wondering when we could move the del operator under the integration sign - in relation to a specific problem, which is:

    A(r) = integral over some volume v r'f(|r-r'|)d3r'

    Where r' represents the gradient with respect to r'

    Now, we're supposed to show that x A = 0 by finding some function of which A is the gradient.

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure how to tackle the problem other than by saying that

    A(r) = ∇r' ∫ f(|r-r'|) d3r

    I.e taking the del operator "outside" of the integral. We then have that A is the gradient of some kind of scalar function (it's given that f is a "well-behaved function of a single variable" and the volume is fixed, so I'm guessing this will be just a single function.) I've seen this done, but never been sure about the mathematics behind it.

    I've looked online to see about "differentiation under the integral sign" but how it relates to the del operator, I haven't found anywhere.

    If, as I suspect, this approach is flawed, please advise on where to go next. Would using Divergence or Stokes' Theorem (or corollaries) be helpful?

  2. jcsd
  3. Feb 21, 2014 #2


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    This obviously doesn't make sense, because you integrate over [itex]\vec{r}'[/itex]. Then the resulting function doesn't depend on [itex]\vec{r}'[/itex] anymore, and your gradient is trivially 0. However, you can take the gradient with respect to [itex]\vec{r}[/itex] in front of the integral. Then you can also take it inside the integral (supposed the integrand is sufficiently well-behaved). Now think further, what you can say about this gradient wrt. [itex]\vec{r}[/itex] under the integral and the gradient wrt. [itex]\vec{r}'[/itex].
  4. Feb 21, 2014 #3
    Thank you for your reply. I'm still a little confused, but your answer has given me something to go on. Let me see how this sounds:

    We're considering the function

    ψ = ∫f(|r-r'|)d3r' which is a scalar function in terms of r (after the integration is performed.)

    As the function is well-behaved:




    noting that the function depends only on |r-r'| which is the same as |r'-r|, this implies that the vector field:

    A(r) = ∫f(|r'-r|)d3r'

    is the gradient of the function ψ as described above, and so its curl is zero. Then we can interchange what we call r and what we call r' to get the desired result.

    Is this along the right lines? I can see that there's a clear symmetry when we take the gradient with respect to r inside the integral, and so it seems like taking the gradient inside the integral for one variable implies we could have done things the other way around and showed that A(r') was the gradient of something, but I'm unsure
  5. Feb 21, 2014 #4


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    There are two things going on that are worth noting. I'll use a single partial derivative as it's easier to see what's happening: first

    [tex]ψ(x, y, z) = \int f(|r - r'|)d^3r'[/tex]
    Is a function of (x, y, z). And, if f is a well-behaved function, then:

    [tex]\frac{\partial ψ}{\partial x} \ = \ \frac{\partial}{\partial x}\int f(|r - r'|)d^3r' \ = \ \int \frac{\partial}{\partial x}f(|r - r'|)d^3r' [/tex]

    And, as the same applies to partial derivatives in y and x, you can take ∇ inside the integral.

    Next, we can use the chain rule to see that:

    [tex]\frac{\partial}{\partial x}f(|r - r'|) = f'(|r-r'|)\frac{x-x'}{|r-r'|}[/tex]
    But, note that:

    [tex]\frac{\partial}{\partial x'}f(|r - r'|) = f'(|r-r'|)\frac{x'-x}{|r-r'|}= -f'(|r-r'|)\frac{x-x'}{|r-r'|}[/tex]

    So, these two derivatives are numerically equal and opposite at every pair of points r and r'. So, we have:
    [tex]\int \frac{\partial}{\partial x}f(|r - r'|)d^3r' = - \int \frac{\partial}{\partial x'}f(|r - r'|)d^3r'[/tex]

    Again, since the same is true for partial derivatives in y and z, we can change the ∇ to -∇'. Note that this crucially depends on the integrand being a function of |r-r'|. It won't be true in general.
    Last edited: Feb 21, 2014
  6. Feb 21, 2014 #5
    Thank you so much! I think I get it now. I had a vague idea that it was to do with the symmetry between r and r' in this expression but for some reason didn't think to use chain rule. As usual, reducing the problem to a single component helps. Thank you.
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