# How do I calculate integral analytically for small $k$?

In a Heisenberg antiferromagnet, the dispersion relation is $$\omega_{\mathbf{k}} =JSz\sqrt{ 1-\gamma_{\mathbf{k}}^2}$$ where $\gamma_{\mathbf{k}}=\frac{1}{d}\sum_{i=1}^d \cos k_ia$, where $d$ is the dimensionality of the cubic lattice.

In staggered magnetisation calculation, we get an integral which is given by

$$\delta m =\int_{BZ} \frac{d^dk}{(2\pi)^d}\frac{1}{\sqrt{ 1-\gamma_{\mathbf{k}}^2}}.$$

This integral is dominated by small $k$, since we have Goldstone modes at $k=0,(\pi,\pi,\dots)$. Using this property, we can extend the limits of this integral from Brillouin zone to infinite $k$-space and we replace $\gamma_{\mathbf{k}}$ by its small $k$ limit:

\begin{eqnarray} \delta m=\frac{\sqrt{d}}{a}\int_0^\infty \frac{d^dk}{(2\pi)^d}\frac{1}{k} \end{eqnarray}

In 1D, its gives the logarithmic divergence which is expected. But, in 2D and 3D we get \begin{eqnarray} \delta m &\sim& [k]_0^\infty, \quad \text{in}\, 2D\\ &\sim&[k^2]_0^\infty, \quad \text{in}\, 3D \end{eqnarray} which still show divergence. This is not correct as if we solve the Brillouin zone integral we get some finite value. I do not understand what is wrong with the infinite k space integral? Could anybody please tell me how to solve such integral analytically for small $k$?

• Your logic is wrong. For instance, the integral $I=\int_0^{+\infty}dx e^{-x}=1$ is dominated by small $x$. So, with your logic, we can replace $e^{-x}$ by $1$, because $x$ is small, and we have the integral $\int_0^{+\infty}dx$... Unfortunately, the latter integral is infinite... – Trimok Aug 19 '13 at 18:50

I don't what you mean by "analytically" or for "small $k$", but if you are interested in the qualitative result of the integral then you almost did the correct thing. Just don't send the cutoff to infinity. (BTW you shouldn't do this in $d=1$ either, since the integral is logarithmically sensitive to the upper bound). The normal way to do such a calculation is:
$$\delta m =\int_{BZ} \frac{d^dk}{(2\pi)^d}\frac{1}{\sqrt{ 1-\gamma_{\mathbf{k}}^2}}\sim\int_0^{1/a}k^{d-1}dk\,\frac{1}{k}\sim a^{2-d}.$$