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I'm interested in the behavior of Dirac deltafunctions within multivariate integrals. Here is a simple example to which I do not know the answer:

$\iint\limits_{[0,1]\times [0,1]} \delta\left(x - y\right)\, dA$

This integrates the constant function $1$ over the line $y=x$. Is there a prefactor involed though due to the delta? I can imagine this evaluating either to $1$ or to $\sqrt2$.

The information from the relevant Wikipedia page suggests the value $\sqrt2$ while WolframAlpha suggests the value $1$.

How is the behavior of a Dirac deltafunction within a multivariate integral defined? My goal is to create an algorithm to solve a broad class of these problems in general.

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    @Bill: If by that integral you mean what the Wikipedia article defines it to mean, then the factor $1/|\nabla g|=1/\sqrt2$ makes this come out as $1$, too. (Sorry about the multitude of comments I wrote and deleted -- I was trying to figure out what you meant by that integral.) Generally speaking, we should be able to treat an integral with a one-dimensional delta as a limiting case with a single parameter going to $0$, so there's no question of interchange of limits and the result shouldn't depend on technicalities like the order of integration.2011-12-22

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In this case, where the coordinates appear linearly in the argument of the delta function, no special treatment is required; you can just evaluate the inner integral as you'd normally evaluate an integral over a delta function, and then evaluate the outer integral:

$\int_{0}^{1}\int_{0}^{1} \delta\left(x - y\right)\, \mathrm dx\, \mathrm dy=\int_0^11\,\mathrm dy=1\;.$

In case you got the value $\sqrt2$ from the length of the diagonal, you didn't take into account the factor $1/|\nabla g|=1/\sqrt2$.

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    Ah, yes, in contrast to my (deleted after @joriki's correction) earlier comment, reducing to the 1-D case by rotating gives $\delta(\sqrt{2}x)$ along the diagonal of length $\sqrt{2}$, so the $\sqrt{2}$'s go away.2011-12-22