Evaluating the dirac delta expression $\int_{\mathbb{R}^3} \sum_{n \in \mathbb{Z}^3}\delta(x - n) \bigg(x_1 + x_3 n_2||n + x||\bigg) dx$?

Question

Ok so I know from my previous question that it is straightforward to evaluate a dirac delta $\delta(x - n)$ expression when the function $f$ is also a function of $n$, that is $f = f(x, n)$.

What I really want to evaluate is a more complicated expression involving vectors in $\mathbb{R}^3$.

$$\int_{\mathbb{R}^3} \sum_{n \in \mathbb{Z}^3}\delta(x - n) \bigg(x_1 + x_3 n_2||n + x||\bigg) dx.$$

So this is essentially:

$$\int_{\mathbb{R}^3} \sum_{n \in \mathbb{Z}^3}\delta(x - n) f(x, n) dx$$

with

$$ f(x,n) = x_1 + x_3 n_2||n + x||.$$

So can we do the following? $$\int_{\mathbb{R}^3} \sum_{n \in \mathbb{Z}^3}\delta(x - n) \bigg(x_1 + x_3 n_2||n + x||\bigg) dx = \sum_{n \in \mathbb{Z}^3} n_1 + n_3 n_2||2n||.$$

Basically do we get to drop the integral and replace the components of $x \in \mathbb{R}^3$ with $n \in \mathbb{R}^3$?

Answer 1

The answer to the question at the end of the OP is "yes."

To show this, note that we can write the Dirac Delta as

$$\delta (\vec x-\vec n)=\delta(x_1-n_1)\delta(x_2-n_2)\delta(x_3-n_3)$$

Hence, we have

$$\begin{align} \iiint_{\mathbb{R^3}} f(\vec x,\vec n)\delta (\vec x-\vec n)\,dx_1\,dx_2\,dx_3&=\int_{\mathbb{R}}\delta(x_3-n_3)\int_{\mathbb{R}}\delta(x_2-n_2)\int_{\mathbb{R}}f(\vec x,\vec n)\delta(x_1-n_1)\,dx_1\,dx_2\,dx_3\\\\ &=\int_{\mathbb{R}}\delta(x_3-n_3)\int_{\mathbb{R}}\delta(x_2-n_2)f(n_1,x_2,x_3,\vec n)\,dx_2\,dx_3\\\\ &\int_{\mathbb{R}}\delta(x_3-n_3)f(n_1,n_2,x_3,\vec n)\,dx_3\\\\ &=f(\vec n,\vec n) \end{align}$$

For $f(\vec x,\vec n)=x_1+n_2x_3|\vec n+\vec x|$, we have

$$\iiint_{\mathbb{R^3}} f(\vec x,\vec n)\delta (\vec x-\vec n)\,dx_1\,dx_2\,dx_3=n_1+2n_2n_3|\vec n|$$

whereupon summing over $n_1$, $n_2$, and $n_3$ yields the expected result.