In three dimensions, the Dirac delta function $\delta^3 (\textbf{r}) = \delta(x) \delta(y) \delta(z)$ is defined by the volume integral:
$$\int_{\text{all space}} \delta^3 (\textbf{r}) \, dV = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \delta(x) \delta(y) \delta(z) \, dx \, dy \, dz = 1$$
where
$$\delta(x) = 0 \text{ if } x \neq 0$$
and
$$\delta(x) = \infty \text{ if } x = 0$$
and similarly for $\delta(y)$ and $\delta(z)$.
Does this mean that $\delta^3 (\textbf{r})$ has dimensions of reciprocal volume?
As an example, a textbook that I am reading states:
For a collection of $N$ point charges we can define a charge density
$$\rho(\textbf{r}) = \sum_{i=1}^N q_i \delta(\textbf{r} - \textbf{r}_i)$$
where $\textbf{r}_i$ and $q_i$ are the position and charge of particle $i$, respectively.
Typically, I would think of charge density as having units of charge per volume in three dimensions: $(\text{volume})^{-1}$. For example, I would think that units of $\frac{\text{C}}{\text{m}^3}$ might be possible SI units of charge density. If my assumption is true, then $\delta^3 (\textbf{r})$ must have units of $(\text{volume})^{-1}$, like $\text{m}^{-3}$ for example. Is this correct?
