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If I am integrating a function such as:

$X=\iiint f(a,b,c)\delta(a,b,c)\textrm{ d}a\textrm{ d}b\textrm{ d}c$

where $f$ is a multidimensional (in this example, 3-dimensional) smooth function that is symmetric, i.e. the result is independent of the 3-tuple permutations of its arguments (e.g., f(a,b,c) = f(b,a,c) = f(b,c,a), etc.), $\delta$ is discontinuous and numerically equal to:

$x$ to the power of the number of arguments sharing a common value

Since the 'region' when a=b or a=c or b=c is infinitesimal, I was thinking if I can ignore $\delta$ - the answer I think is no. Is it correct if I simplify it to:

$X= \iiint f(a,b,c)\textrm{ d}a\textrm{ d}b\textrm{ d}c + 3x^2\iint f(a,a,b)\textrm{ d}a\textrm{ d}b + x^3\int f(a,a,a)\textrm{ d}a$ ?

Thanks!

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That's correct, you can ignore the domain where $\delta \not= x^3$. The triple integral is the limit of Riemann sums, each of which involves the volume $dV$ of a (small) box. The restriction of this box to the domain where no two coordinates are equal is a union of contiguous regions determined by the intersection of the three disallowed planes and the one disallowed line with the box. Since all of these disallowed sets have zero volume, the volume of the restricted box is the same as the volume of the original. Thus the Riemann sums aren't affected by the restriction.

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    @Mobius: The measure of a plane or line in $\mathbb{R}^3$ is zero. If you don't want to deal with measure theory, one could just consider a box (a "base" in the 3d Riemann sum), except for the points belonging to a plane (or line). One can imagine filling up the points in the deleted box with ever smaller and smaller boxes, so that in the limit, the sum of these volumes equals the volume of the original (undeleted) box. Hence the deleted regions have no effect on the limit.2012-03-12