Integral of Gaussian ring/shell I would like to know the integral of the function
$$f(\mathbf{x}) = {1 \over \sqrt{2\pi \sigma^2}} \exp\left\{- {(|\mathbf{x}| - \mu)^2 \over 2\cdot \sigma^2}\right\} $$
over an $n$-dimensional vector space. It is a ring of radius $\mu$ around the origin, where the deviation from the radius yields a fall-off according to a Gaussian with standard deviation $\sigma$.
I had no success with WolframAlpha/Maxima/SymPy, probably due to the norm $|\mathbf{x}|$.
Is the right approach to switch to spherical coordinates, and then integrate along the ring first, and finally over the radius?
The radial function would be:
$$g(R, \phi) = {1 \over \sqrt{2\pi \sigma^2}} \exp\left\{- {(R - \mu)^2 \over 2\cdot \sigma^2}\right\} $$
so that
$$\int_0^\infty dR \int_0^{2\pi} d\phi ~~ g(R, \phi) = 2\pi\cdot 1 = 6.28 $$
I do not think this result is correct, for 2d it should be 0.087 according to a paper I have (for 5d 0.0017; using $r=2$, $w=0.1$).
Update
I missed out to use the volume element dV. In the general integration over the n-sphere (from here):
$$ d^nV = r^{n-1}\sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\cdots \sin(\phi_{n-2})\, dr\,d\phi_1 \, d\phi_2\cdots d\phi_{n-1} $$
Now the integral 
$$ \int g(R) R^{n-1} dR $$
is the (n-1)th moment of the gaussian, $M_{n-1} = \operatorname{E}\left[X^{n-1}\right]$ (see here, and scipy.stats.norm.moment(n-1, loc=r, scale=sigma))
So then we are only missing the rest of the volume elements. This is the "surface area" of a unit n-sphere (from here and here).
$$ S_{n}={(n+1) V_{n+1}}$$
with 
$$V_n = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}$$ 
So I should be able to write the complete integral as
$$ I = \int_{\cal{R}^n} g(R, \mathbf{\theta}) dV = S_n \cdot M_{n-1} $$
For n=2: 
The first moment with $\mu=2$, $\sigma=0.1$ is $M_1=2$. The surface $S_2=4\pi$. This gives $I_2 = 25.1$, which is still not what it ought to be.
 A: Let $\varphi$ and $\Phi$ denote the standard normal PDF and CDF, that is,
$$
\varphi(t)=\frac1{\sqrt{2\pi}}\mathrm e^{-t^2/2},\qquad\Phi(t)=\int_{-\infty}^t\varphi=\int_{-t}^\infty\varphi.
$$
Using spherical coordinates as you explained one is led to compute $\displaystyle\int g(R)R^{n-1}\mathrm dR$ the integral in your post, which is
$$
A_n(\mu,\sigma^2)=\int_0^\infty r^{n-1}\varphi\left(\frac{r-\mu}\sigma\right)\mathrm dr/\sigma.
$$
The change of variable $r=\sigma t$ yields
$$
A_n(\mu,\sigma^2)=\sigma^{n-1}B_{n-1}(\mu/\sigma),\qquad B_k(u)=\int_0^\infty t^k\varphi(t-u)\mathrm dt.
$$
Now, for $k=0$,
$$
B_0(u)=\int_{-u}^\infty \varphi=\Phi(u).
$$
Using the fact that $\varphi'(t)=-t\varphi(t)$, one sees that
$$
B_k'(u)=\int_0^\infty t^k(t-u)\varphi(t-u)\mathrm dt=B_{k+1}(u)-uB_k(u).
$$
One deduces from this that there exists some sequences $(P_k)$ and $(Q_k)$ of polynomials such that, for every $k\geqslant0$,
$$
B_k=P_k\varphi+Q_k\Phi.
$$
These are uniquely determined by the initial condition $(P_0,Q_0)=(0,1)$ and by the recursion
$$
P_{k+1}(u)=P'_k(u)+Q_k(u),\qquad Q_{k+1}(u)=Q'_k(u)+uQ_k(u).
$$
For example,
$$
(P_1(u),Q_1(u))=(1,u),\quad (P_2(u),Q_2(u))=(u,1+u^2),\quad (P_3(u),Q_3(u))=(2+u^2,3u+u^3).
$$
One recognizes (or a direct computation shows that), for every $k$,
$$
Q_k(u)=\mathrm i^{-k}H_k(\mathrm iu),
$$
where $H_k$ is the $k$th Hermite polynomial.
