Evaluating integrals in R^m Let $|\cdot|_m$ denote the Euclidean norm in $\mathbb{R}^m$.  Then
I wish to prove that
$\displaystyle\int\limits_{\mathbb{R}^m}|x|_me^{-|x|_m}dx<\infty$
It's kinda embarrassing to say this, but I don't really know where to start.  Probably spherical coordinates would help, but I don't recall how to handle spherical coordinates in m-dimension.
 A: To avoid a more general measure-theoretic approach, we could always use a "shell-method" of sorts.
In particular, we start by noting that the $n$-dimensional surface area of the $n$-sphere is given by $V(r) = \alpha_n r^n$ (where the coefficient $\alpha_n$ is generally going to be some function of $\pi$).
We can then write this as
$$
\int_{\mathbb{R}^m}f(r(x))dx =\\
\int_{r=0}^\infty f(r(x))V_{n-1}(r(x))\,dx =\\
\alpha_{n-1} \int_{r=0}^\infty r^{n-1}f(r(x))\,dx 
$$
which will be enough to get you your result.
A: Essentially all you need to know is that the "surface area" of the unit sphere in $\mathbb{R}^m$ is finite (Call it $S_m$). (what this actually means is doing the angular integrals, which are all of bounded functions on bounded intervals, so the result is bounded.)  Then the "surface area" of a sphere of radius $r$ is $S_m r^{m-1}$, by scaling considerations, and the volume is $S_m r^m/m$, by integrating. It follows that the volume of the space between two spheres is about
$$ \frac{1}{m}S_m (r+h)^{m}-\frac{1}{m}S_m r^{n} \sim S_m r^{n} h $$
as $h \to 0$. Therefore, since $r=|x|_m$, the integral becomes
$$ S_m \int_0^{\infty} r e^{-r} r^{m-1} \, dr, $$
which you can calculate using integration by parts to get $m! S_m$.
