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I would like to better understand how to use and manipulate the Dirac Delta function.

It seems to me that whenever the delta function appears in an integral, it reduces the dimension of the domain of integration by $1$, i.e. $$\int \limits _{\Omega\ \in\ \mathcal{R}^2}f(x_1,x_2)\ \delta(x_1-y)\ \mathrm{d}\Omega \rightarrow \int \limits _{\Omega \ \{x_1\rightarrow y\}\ \in \ \mathcal{R}}f(y,x_2)\ \mathrm{d}x_2$$ For very easy functions and domains I was able to manually check this.

I would like to exploit this property in more complicated situations, for example: $$\int \limits _{\Omega\ \in\ \mathcal{R}^4}f(x_1,x_2,x_3,x_4)\ \delta(x_1-x_2+1)\ \delta(x_2)\ \mathrm{d}\Omega \rightarrow \int \limits _{\Omega \ \{x_2\rightarrow 0,x_1\rightarrow-1\}\ \in \ \mathcal{R}^2}f(-1,0,x_3,x_4)\ \mathrm{d}\Omega$$ but it appears that this is not always possible (I have seen how certain examples work, but others do not$^{(1)}$). Can anyone explain to me why this is wrong? I was hoping to use this "trick" to numerically integrate functions that have a Dirac Delta inside.

$^{(1)}$ I have tried the following case using Mathematica v$11.0$:

1)The function is: $$f(x,y,z,\zeta)=\frac{z^4}{2\ (y^2+z^2)^{5/2}}\cdot \delta\ (x+y-8) $$ 2)The domain of integration is:$$[0\leq x\leq25]\cup[-5\leq y\leq 5]\cup[-0.5\leq z\leq0.5]\cup[0\leq\zeta\leq1]\cup[0\leq x+y\leq25]\cup[0\leq z+\zeta\leq1] $$ I obtain different results when doing the domain reduction for $x\rightarrow8-y$ or $y\rightarrow8-x$

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As stated in another answer I gave, integrals with the "delta-function" in them should (or at least may) be handled by considering the dirac measure $$ \delta_b(A)=\begin{cases}1&b\in A\\0&b\notin A\end{cases} $$ for some $b\in\mathbb R$ as integrating over this measure $\delta_b(dx)$ then corresponds to the "function" / "integrand" $\delta(x-b)\,dx$ as it is used in the physics literature.

This measure is what is called $\sigma$-finite (it even is finite), as is the usual Lebesgue-Borel measure on $\mathbb R$. Therefore we can apply Tonelli's theorem which for states that if $f:X\times Y\to [0,\infty)$ with $X,Y\subseteq\mathbb R$ is measurable then $$ \int_X\Big(\int_Y f(x,y)\,dy\Big)\,\delta_b(dx)=\int_Y\Big(\int_X f(x,y)\,\delta_b(dx)\Big)\,dy\tag{1} $$ so we can interchange the integrals. This interchanging also works if we use the dirac measure twice, i.e. for the $x$ and the y integral. Also I chose this version of Fubini-Tonelli because the mapping from your example is obviously non-negative.

Note that here, if $f$ is continuous on $X\times Y$ then it for sure is measurable and the right-hand side of (1) evaluates to $$ \int_Y\Big(\int_X f(x,y)\,\delta_b(dx)\Big)\,dy=\begin{cases} \int_Y f(b,y)\,dy&\text{if }b\in X\\0&\text{if }b\notin X \end{cases}\,. $$


However, if one of the delta "displacements" (the $b$) depends on the other integration variable, things start breaking down. Related to the second equation in your original post: for any continuous $f:\mathbb R^2\to\mathbb R$ one gets $$ \int_{\mathbb R}\int_{\mathbb R} f(x,y)\delta(x-y)\delta(y)\,dx\,dy=\int_{\mathbb R}\Big(\int_{\mathbb R} f(x,y)\,\delta_y(dx)\Big) \,\delta_0(dy)=\int_{\mathbb R} f(y,y)\,\delta_0(dy)=f(0,0) $$ but $$ \int_{\mathbb R}\Big(\int_{\mathbb R} f(x,y)\delta(x-y)\delta(y)\,dy\Big)\,dx $$ involves the product of two "delta functions" in a single integral which is problematic for multiple reasons.

All of this of course is a rather general answer / comment but just goes to show that once a dirac measure in some sense depends on two integration variables, (a) one should be quite careful on what to do precisely and (b) interchanging integrals is not a given anymore (although it might still be possible, varies case by case).

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  • $\begingroup$ There are consistent definitions both for $\delta(f(x, y))$ and for $\delta(f(x, y)) \delta(g(x, y))$. The necessary restrictions are essentially the same as what is required for excluding things like $\delta(x^2)$. $\endgroup$ – Maxim Jul 2 at 21:32
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There are two different questions here. First is what to do when the domain is not all of $\mathbb R^n$. Let $$P(x, y) = 0 < x < 25 \land -5 < y < 5 \land 0 < x + y < 25.$$ If we integrate wrt $y$ first, then $$I_1(x) = \int_{\mathbb R} \delta(x + y - 8) \, \phi(x, y) \, [P(x, y)] \, dy = \\ \phi(x, 8 - x) \, [0 < x < 25 \land -5 < 8 - x < 5].$$ If we integrate wrt $x$ first, then $$I_2(y) = \int_{\mathbb R} \delta(x + y - 8) \, \phi(x, y) \, [P(x, y)] \, dx = \\ \phi(8 - y, y) \, [0 < 8 - y < 25 \land -5 < y < 5].$$ Now $$\int_{\mathbb R} I_1(x) \, dx = \int_3^{13} \phi(x, 8 - x) \, dx, \\ \int_{\mathbb R} I_2(y) \, dy = \int_{-5}^5 \phi(8 - y, y) \, dy,$$ which are identical.

Secondly, the integral of $\delta(f) \delta(g)$ is also well-defined as long as the curves $f = 0$ and $g = 0$ are not tangent to one another. For two straight lines, $$\iint_{\mathbb R^2} \delta(y - y_0 - k_1 (x - x_0)) \, \delta(y - y_0 - k_2 (x - x_0)) \, \phi(x, y) \, dx dy = \frac {\phi(x_0, y_0)} {|k_2 - k_1|}.$$ You can verify that this is a special case of the general formula from here.

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