# Multivariate integrals involving Dirac delta functions

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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I'm not totally sure this is absolutely correct, but I think the iterated integral you have listed above has value $1$. I believe the other answer, $\sqrt{2}$, maybe the value of $\iint\limits_{[0,1]\times[0,1]} \delta(x-y)\,dA$. There's no reason to believe Fubini's theorem should apply here. – Bill Cook Dec 22 '11 at 16:31
I'm particularly interested in the single integral over a region, not an iterated integral. I'll change the question. – MRocklin Dec 22 '11 at 16:51
@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. – joriki Dec 22 '11 at 17:28

$$\int_{0}^{1}\int_{0}^{1} \delta\left(x - y\right)\, \mathrm dx\, \mathrm dy=\int_0^11\,\mathrm dy=1\;.$$
In case you got the value $\sqrt2$ from the length of the diagonal, you didn't take into account the factor $1/|\nabla g|=1/\sqrt2$.
 Where does the $1/ \sqrt 2$ appear in your solution? What is the missing intermediate step? – MRocklin Dec 22 '11 at 16:48 @MRocklin: There's no need for that factor in my solution. That last sentence wasn't related to my solution; it was just a speculation on what might have gone wrong when you tried to follow the Wikipedia article and got $\sqrt2$. If it doesn't make sense, you can ignore it. – joriki Dec 22 '11 at 17:04 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. – paul garrett Dec 22 '11 at 17:20