I'm having trouble understanding the following. Say I have a quantity of objects arranged spatially, with distribution $dn/dx$. For one reason or another*, I want to sum auto-pairs of spatial points, but only pairs at the same place. Thinking about this discretely, we have
$$ \sum_{i,j} (\bar{n}_i \Delta x) (\bar{n}_j \Delta x) \delta_{ij} = \sum_i (\bar{n}_i \Delta x)^2 $$
Clearly this is dimensionless on both sides. However, moving into the continuous limit, I expect to have
$$ \int\int \frac{dn}{dx_i}\frac{dn}{dx_j}\delta(x_i-x_j)dx_idx_j = \int \left(\frac{dn}{dx}\right)^2 dx. $$
Here both sides have the same dimensions (inverse length), but this is different from the sum, which is dimensionless. What am I doing wrong??
Thanks for any help!
* For those who are interested, this problem arises (or at least I think it does) when finding the variance of a sum of correlated variables. In particular, let the number of objects be a function of $x$ and $y$, and let them be spatially correlated in $y$, but not $x$. Then
$$ {\rm Var}(\sum N_i) = \sum_{x_i}\sum_{x_j}\sum_{y_i}\sum_{y_j} (\bar{n}_{x_i} \Delta x \Delta y) (\bar{n}_{x_j} \Delta x \Delta y)C_{y_i y_j} \delta_{x_i x_j} = \sum_{x_i}\sum_{y_i}\sum_{y_j} (\bar{n}_i \Delta x \Delta y)^2 C_{y_i y_j} .$$
The rest follows as above (note that I have that the spatial covariance in $y$ at a given $x$ is $(\bar{n}_x \Delta x \Delta y)^2 C_{y_i y_j}$).
I think that the discrepancy is that your discrete case is not the strict analogue of your continuous case.
If I were summing up just values of $n$, and using the Kronecker delta to pull out one particular value, say $k$, then we'd have $$ \sum_i a_i\delta_{ik} = a_k, $$ and its continuous analogue would be $$ \int a(x)\delta(x-k)\,dx = a(k). $$ In all cases, the units are the same (whatever units $a$ has). Notice that the discrete version lacks the extra $\Delta x$ that you included.
Working backwards from your continuous version, we have \begin{align} \int \frac{dn}{dx_i}\frac{dn}{dx_j}\delta(x_i-x_j)\,dx_i\,dx_j &= \int \left(\frac{dn}{dx}\right)^2\,dx \\ \text{(Going to discrete case)}&\Rightarrow \sum_i \bar{n}^2_i \Delta x \\ &= \left[\sum_{i,j}\bar{n}_i\bar{n}_j \delta_{ij}\right]\Delta x, \end{align} so both versions have units of inverse length.