Strange integral in 3 space (Maybe Divergence Theorem) I'm trying to find something called the density of states and the model that I am using specifies
$$E = \frac{h^2}{2 m} k^2$$
where $k = |\bf{k}|$.
The quantity I am trying to calculate is
$$D(E) = \int{(\nabla_k E)^{-1} \mathrm{d}E}$$.
I think how to simplify this is to substitute
$$\mathrm{d}E = \frac{h^2}{m} \mathrm{d}k$$
and 
$$\nabla_k E = \nabla_k [\frac{h^2}{2 m} (k_x^2 + k_y^2 + k_z^2)].$$
Therefore,
$$\nabla_k E = \frac{h^2}{m}(k_x, k_y, k_z).$$
However this leaves me with
$$D(E) = \int{\frac{1}{(k_x, k_y, k_z)} \mathrm{d}k},$$
which I am sure how to solve. Any idea how to solve this with a vector in the denominator?
Edit
I realized that $D(E) = \int{(\nabla_k E)^{-1} \mathrm{d}E}$ is actually
$$D(E) = \int{(\nabla_k E)^{-1} \mathrm{d}\bf{S}}.$$
 A: This is truly warped notation!
Let me explain this more straightforwardly:


*

*The wavevectors $\mathbf k \in \mathbb R^3$ are assumed to be spaced uniformly, like $(0,0,0),(0,0,\Delta),(0,\Delta,0),(\Delta,0,0),\cdots$ for instance. (This is usually justified by a particle in a box argument.)

*We want to calculate sums of functions $f(\mathbf k)$ over all states (with some uninteresting overall scaling by a partition function). We approximate the sum by an integral because it's easier. The sum should be approximated via $$\sum_{\text{integer vectors }\mathbf n} f(\Delta \mathbf n) \propto \int \mathrm d^3 \mathbf n \quad f(\Delta \mathbf n)$$for small $\Delta$.

*Then changing variables we get
$$\propto\int \mathrm d^3 \mathbf k \quad f(\mathbf k)$$
Assuming the functions we calculate depend only on $k=|\mathbf k|$, notice that spherical coordinates give
$$I = \int \mathrm d^3 \mathbf k \; f(k) = 4\pi \int_0^\infty \mathrm d k \; k^2 f(k)$$
Now we have a one-dimensional integral and everything is nice.

*We can change variables to $E(k)=\frac{\hbar^2}{2m}k^2$ in the usual way:
$$k=\sqrt{2mE/\hbar^2} \implies \mathrm dk=\sqrt{m/2\hbar^2E}\, \mathrm dE \implies I = 4\pi \int_0^\infty \mathrm d E \; \sqrt{\frac{m}{2\hbar^2E}} \frac{2mE}{\hbar^2} f(k(E))$$
or simplifying
$$I = \int_0^\infty \mathrm d E \; \hat f(E) \times \sqrt E\times 2\pi \left(\frac{2m}{\hbar^2}\right)^{3/2}$$

*The thing you are supposed to work out is what you get from setting $$\hat f(E)=\cases{1 & $E<E_0$ \\ 0 & $E\ge E_0$}$$ i.e. integrate from $E=0$ up to $E_0$.

*What the strange expression you had meant was "First, switch to $k=|\mathbf k|$, then use $\mathrm d k = k'(E) \mathrm d E = 1/E'(k) \mathrm d E$."

