I am trying to find the mean of the function $\exp(-2r)$ in 2 dimensions. This should then be the equivalent of doing
$$\frac{\int\!\int\!\sqrt{x^2+y^2}e^{-2\sqrt{x^2+y^2}}\,\mathrm{d}x\,\mathrm{d}y}{\int\!\int\!e^{-2\sqrt{x^2+y^2}}\,\mathrm{d}x\,\mathrm{d}y}$$
Unfortunately, do not know how to do either integral. I tried u substitution with the top one, but I keep ending up with an extra $2x$.
Any ideas?
For one dimension, the average $<f>$ of a function $F$ on the interval $(a,b)$ is
$$<f>=\frac{\int_a^{b} f(x)dx}{\int_a^{b} 1dx}=\frac{1}{b-a}\int_a^{b} f(x)dx$$
For $n$-dimensions, the average $<f>$ of a function $f$ defined on a finite domain $\mathscr{D}$ is given
$$\frac{\int_\mathscr{D} f d^nx}{\int_\mathscr{D} 1 d^nx}$$
In the $2$-D case presented, $f=e^{-2r}$, where $r=\sqrt{x^2+y^2}$ and the domain is $\mathscr{R}^2$. Thus,
$$\begin{align} <e^{-2r}>&=\lim_{R \to \infty} \frac{\int_0^{2\pi} \int_0^R e^{-2r} rdrd\phi}{\int_0^{2\pi} \int_0^R 1 rdrd\phi}\\\\ &=\lim_{R \to \infty} \frac{2\pi\left(\frac14-\frac14(2R+1)e^{-2R}\right)}{\pi R^2}\\\\ &=0 \end{align}$$
NOTE:
If one is finding the average of $r$ under the measure $e^{-2r}$, then
$$<r>=\lim_{R \to \infty} \frac{\int_0^{2\pi} \int_0^R e^{-2r} rdrd\phi}{\int_0^{2\pi} \int_0^R e^{-2r} drd\phi}=\lim_{R \to \infty} \frac{2\pi\left(\frac14-\frac14(2R+1)e^{-2R}\right)}{\pi (1-e^{-2R})}=\frac12$$