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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:

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


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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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Thanks, did you mean $\delta=x^3$ instead of $\not=$, therefore I can ignore $x^3\int f(a,a,a)\textrm{ d}a$? I am reserved about this though hmm. I know that theoretically the space has zero volume. To tell the truth what I am trying to calculate has physical meaning, and so I wouldn't say it has exactly zero volume, i.e. small but not insignificant. It is akin to real world implementation of Kronecker delta function in signal processing, it is a spike of which the width is not exactly zero. – Mobius Pizza Mar 7 '12 at 18:25
Actually I am not convinced with my own answer, let's say we have $y_0=\int\int f(a,b)\textrm{d}a\textrm{d}b$. Within this Riemann sum, there are total contributions of 2$\int f(k,k)\textrm{d}k$ from the 'diagonal' plane of $f$. If we have a $\delta$ function here which only scales the diagonal contributions differently, then $y=\int\int f(a,b)\delta(a,b) \textrm{d}a\textrm{d}b$ will result in tangible difference s.t. $y$ becomes: $y=y_0-2\int f(k,k)\textrm{d}k + 2\delta(k,k)\int f(k,k)\textrm{d}k$? – Mobius Pizza Mar 7 '12 at 19:03
@Mobius: I was thinking that we can ignore the points with at least two identical coordinates. That would be ignoring the domain where $\delta \not= x$. The argument is based on that idea even though I incorrectly wrote $\delta \not= x^3$. – Patrick Mar 12 '12 at 22:32
@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. – Patrick Mar 12 '12 at 22:41

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