Say we have a distribution for the density of particles in a region of space $$ n(x) = \frac{N}{\pi^{1/2} R} \textrm{e}^{-x^{2}/R^{2}} $$ where $R = \sqrt{\frac{2 k_{B} T}{m \omega^{2}}}$. Multiplication by certain functions of random variables, followed by integration, yield moments of this distribution, which correspond to useful quantities such as the total number, the mean, the variance, etc. For example, the total number, mean, and variance are $$ N = \int_{\infty}^{\infty}dx \, n(x), \qquad \mu = \int_{\infty}^{\infty} dx \, x \, n(x) = 0, \qquad \sigma^{2} = \int_{\infty}^{\infty} dx \; x^{2} \, n(x) = \frac{N R^{2}}{2}. $$
But what if we want the "mean density", or average value of the density distribution function? (As opposed to $\mu$, which tells us that the mean position of particles in this system is 0.)
In a bounded system, we would take N/V, with V being the volume. But the above density distribution extends to infinity, and therefore V is infinite and the average of $n(x)$ is zero except for the case of $N \rightarrow \infty$. The question is then, "What is the most appropriate way to bound $n(x)$ when calculating the average?"
It is clear for functions that are zero outside of some region, but in situations where that is not the case, such as the normal distribution, I'm unsure what to do.
(Plot of $\int e^{x^{2}} dx$ compared with a bounded function.)
Additionally, how would we extend this to three-dimensions. In Cartesian, the density is $n(r)$, with position $r = (x,y,z)^{T}$ and $x^{2} \rightarrow r^{2} = x^2 + y^2 + z^2$; but similarly, how does one get the mean density?
I don't know if the so-called "mean density" has a solid meaning in physics or not. Mathematically you are asking for the average defined as $$ \bar{f} = \mathop{\lim}_{R\to\infty}\frac{1}{2R}\int_{-R}^Rf(t)\,dt $$
In $n-$dimensional this becomes $$ \bar{f} = \mathop{\lim}_{R\to\infty}\frac{1}{vol(B_0(R))}\int_{B_0(R)}f(x)\,dx $$ where $B_0(R)$ is an $n-$dimensional ball with radius $R$ centered at $0$.