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Integral 1/r grad 1/r

  1. Apr 29, 2013 #1

    Jano L.

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    Gold Member

    I would like to calculate the integral

    \int \frac{1}{|\mathbf y|} \nabla \frac{1}{|\mathbf y -\mathbf c|}~d^3\mathbf y,

    but I do not know how. I was wondering whether somebody has encountered such integral already. Do you know how to calculate it? Do you know some resource where similar integrals are calculated or tabulated ?
  2. jcsd
  3. Apr 30, 2013 #2
    There are basically two techniques for any integrals.
    1. brute force and 2. clever guessing.
    The integral you' posted can be solved with either method.

    Since i'm too lazy to type a lot, I'll give you a slick
    "physics proof" derivation. (which in retrospect still requires
    too much typing!) This solution requires that you believe
    the identity
    \nabla \frac{1}{r} = - 4 \pi \delta^3({\vec r}),
    familiar from electrostatics.

    Take your integral which is a function of the vector c, and rewrite it.
    \vec{f}({\vec c}) = \int d^3y \frac{1}{y} \nabla \frac{1}{|{\vec y} - {\vec c}|}
    = -\int d^3y \frac{1}{y} \nabla_c \frac{1}{|{\vec y} - {\vec c}|},
    where [itex]\nabla_c[/itex] has the meaning of differentiating as
    if c were variable.

    Now calculate the c-divergence of f. We find
    $$ \nabla_c \cdot {\vec f}
    = -\int d^3y \frac{1}{y} \nabla_c^2 \frac{1}{|{\vec y} - {\vec c}|}
    = -\int d^3y \frac{1}{y} (-4 \pi \delta^3({\vec y} - {\vec c}))
    = 4 \pi \frac{1}{c}.

    now from the symmetry of the problem we can argue that
    $${\vec f}({\vec c}) = f_c({\vec c}) {\widehat c}$$

    And thus we use the spherical form of the divergence (in c-space)
    to find fc.
    $$\frac{1}{c^2}\frac{\partial }{\partial c} (c^2 f_c) = 4 \pi \frac{1}{c}$$
    and the solution with [itex]f_c[/itex] well defined as [itex]c\rightarrow 0[/itex] is
    $$f_c = 2 \pi, $$
    and so the original integral is
    $$ \int d^3 y \frac{1}{y} \nabla \frac{1}{|{\vec y}- {\vec c}|}
    = 2 \pi {\widehat c}
    = \frac{2 \pi}{c} {\vec c}.$$
  4. May 1, 2013 #3

    Jano L.

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    Gold Member

    That is very nice way to calculate it. Thank you. I might add that for ##\mathbf c = \mathbf 0##, the integral is zero.
    The result is quite interesting, if we interpret it as scalar product of potential and electric field due to two point charges. It seems that this product is independent of the distance between the two.
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