Let's assume we have an integration like this:
$$I = \int d^2x_1 \ d^2x_2 \ \delta^2 (x_1 + x_2 -p) \ f(x_1) \ f(x_2) $$
For a simple 1D delta function we can use this simply:
$\int \delta (x-a)\phi (x)dx=\phi(a)$
………………………………………………………
An Example:
Let's use functions for $f(x_1)$ and$f(x_2)$ :
$$I = \int d^2x_1 \ d^2x_2 \ \delta^2 (x_1 + x_2 -p) \ \ e ^{-x_1^2 + x_2^2 } $$
How would I treat $\delta^2$??
According to the 1D properties of delta function $\int \delta (x-a)\phi (x)dx=\phi(a)$, the integration is applicable, when $p = x_1 + x_2$, but I'm not sure how to deal with the 2D delta function in the integration and the $d^2 x_1,$ or $d^2 x_2$,term.
Best Answer
First a remark about notations. To be sure not to make forbidden manipulations, it is advised to see the Dirac delta as a distribution or as a measure, and so, for example in dimension $1$, it is better to write $$ \int_{\mathbb{R}^d} \varphi(x)\,\delta_a(\mathrm{d}x) = \langle \delta_a,\varphi\rangle = \varphi(a) $$ (unless you really know what you are doing).
About $I$, what is meant with this kind of expression is the lineic measure on the set $$ \int_{\mathbb{R^{2d}}} \varphi(x_1,x_2)\,\delta_{\{x_1+x_2=p\}}(\mathrm{d}x_1\mathrm{d}x_2) := \int_{\mathbb{R}^d} \varphi(x,p-x)\,\mathrm{d}x $$ which in this particular case can be seen as a generalized Fubini theorem (since I am in some sense using first the one dimensional definition of the Dirac with variable $x_2$ and then integrating with respect to $x_1$), but one could exchange $\{x_1+x_2=p\}$ with more general set. In your case this yields $$ I(p) = \int_{\mathbb{R}^d} f(x)\,f(p-x)\,\mathrm{d}x = (f*f)(p) $$ which is just the convolution of $f$ with itself (this is sometimes called the autocorrelation).
Remarks:
Following my first remark about notations, notice that in particular, I wrote $\delta_{\{x_1+x_2=p\}}(\mathrm{d}x_1\mathrm{d}x_2)$ what you call $\delta^2(x_1+x_2-p)dx_1dx_2$
Other ways of writing the above formula for $I(p)$ that are perhaps more clear for the people not understanding the meaning of the first ones are $$ \begin{align*} I(p) &= \int_{\mathbb{R^{2d}}} f(x_1)f(x_2)\,\delta_{\{x_1+x_2=p\}}(\mathrm{d}x_1\mathrm{d}x_2) \\ &= \int_{\mathbb{R^{d}}} f(p-x_2)f(x_2)\,\mathrm{d}x_2 \\ &= \int_{\mathbb{R^{d}}} f(x_1)f(p-x_1)\,\mathrm{d}x_1 \end{align*} $$ (but of course the name of the variable does not matter) or also by defining the space $S_p := \{(x_1,x_2)\in\mathbb{R}^{2d}, x_1+x_2=p\} = \{(x_1,x_2)\in\mathbb{R}^{2d}, x_1+x_2-p=0\}$ and defining $\mathrm{d}s$ as the uniform measure on this space $$ \begin{align*} I(p) &= \int_{(x_1,x_2)\in S_p} f(x_1)f(x_2)\,\mathrm{d}s \end{align*} $$
In the particular case of Guassians $$ \begin{align*} I(p) &= \int_{\mathbb{R^{2d}}} e^{-|x_1|^2-|x_2|^2}\,\delta_{\{x_1+x_2=p\}}(\mathrm{d}x_1\mathrm{d}x_2) \\ &= \int_{\mathbb{R^{2d}}} e^{-|x_1|^2}\,e^{-|x_2|^2}\,\delta_{\{x_1+x_2=p\}}(\mathrm{d}x_1\mathrm{d}x_2) \\ &= \int_{\mathbb{R^{d}}} e^{-|x_1|^2}e^{-|p-x_1|^2}\,\mathrm{d}x_1 \\ &= (e^{-|x|^2} * e^{-|x|^2})(p) \\ &= \left(\frac{\pi}{2}\right)^{d/2} e^{-|p|^2/2} \end{align*} $$
In some sense, the Dirac Delta removes one integral, so in dimension one no integral are left, in dimension $2$ we pass from $2$ to $1$ integral.