Cross post from Maths stack exchange
The integral
$$
I(m)=\frac{1}{4\pi}\int_{-\pi}^{\pi}\mathrm{d}x\int_{-\pi}^\pi\mathrm{d}y\phantom{,} \frac{m\cos(x)\cos(y)-\cos x-\cos y}{\left( \sin^2x+\sin^2y +(m-\cos x-\cos y)^2\right)^{3/2}}
$$
gives the Chern number of a certain vector bundle 1 over a torus. It can be shown using the theory of characteristic classes that
$$
I(m) = \frac{\mathrm{sign}(m-2)+\mathrm{sign}(m+2)}{2}-\mathrm{sign}(m) = \begin{cases}1 & -2< m < 0 \\ -1 & 0 < m < 2 \\0 & \text{otherwise}\end{cases}.
$$
Is there any way to evaluate this integral directly (i.e. without making use of methods from differential geometry) to obtain the above result?
I should mention that the above integral can be written as ($1/4\pi$ times) the solid angle subtended from the origin of the unit vector $\hat{\mathbf{n}}$,
$$
I(m)=\frac{1}{4\pi}\int_{-\pi}^{\pi}\mathrm{d}x\int_{-\pi}^\pi\mathrm{d}y\phantom{,} \hat{\mathbf{n}}\cdot\left(\partial_x \hat{\mathbf{n}}\times \partial_y \hat{\mathbf{n}}\right),
$$
where $\mathbf{n}(m)=(\sin x, \sin y, m- \cos x-\cos y)$. While this form makes it very straightforward to evaluate $I(m)$, I am interested in whether there is a way to compute this integral using more standard techniques.
1 B. Bernevig Topological Insulators and Topological Superconductors Chapter 8
Best Answer
If Stokes' theorem counts as a standard technique, then here's an answer:
Introduce a "vector potential" \begin{equation} A_i = \frac{1-\hat n_z}{\hat n_x^2 + \hat n_y^2}(\hat n_x \partial_i \hat n_y - \hat n_y \partial_i \hat n_x)\,,\qquad i=x,y\,, \end{equation} which is defined everywhere, except where $\hat n_x^2 + \hat n_y^2=0$, which is $(x,y)\in\{(0,0),(0,\pi),(\pi,0),(\pi,\pi)\}$ (assuming $-\pi<x,y\leq \pi$). One can check that \begin{equation} \Omega(x,y) \equiv \hat{\mathbf{n}}\cdot(\partial_x \hat{\mathbf{n}} \times \partial_y \hat{\mathbf{n}}) = \partial_xA_y -\partial_y A_x = \nabla\times \mathbf{A}\,. \end{equation} Now, going back to the integral, since the integrand is periodic, we can move the integration limit, such that the singular points of $\mathbf{A}$ are not at the boundary of the integration region, for instance \begin{equation} I(m) = \frac{1}{4\pi}\int_{-\pi+1}^{\pi+1}\mathrm{d} x \int_{-\pi+1}^{\pi+1}\mathrm{d} y\, \Omega(x,y)\,. \end{equation} We can now exclude small disks centered at the singular points from the intragral and treat them separately: $$R = [-\pi+1,\pi+1]\times[-\pi+1,\pi+1] \\ R' = R \setminus D(0,0,\epsilon)\setminus D(0,\pi,\epsilon)\setminus D(\pi,0,\epsilon)\setminus D(\pi,\pi,\epsilon)$$ where $D(x,y,r)$ is a disk centered at $(x,y)$ with radius $r$, and $0<\epsilon<1$. $R'$ is now a square with 4 holes, and the integral is written as \begin{align} I(m) &= \frac{1}{4\pi}\int_{R'} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \frac{1}{4\pi}\int_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ &= \frac{1}{4\pi}\int_{R'} \mathrm{d} x \mathrm{d} y\, \nabla\times \mathbf{A} + \frac{1}{4\pi}\int_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ &= \frac{1}{4\pi}\oint_{\partial R'} \mathrm{d} \mathbf{l}\cdot \mathbf{A} + \frac{1}{4\pi}\int_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ &= \left(\frac{1}{4\pi}\oint_{\partial R} \mathrm{d} \mathbf{l}\cdot \mathbf{A} -\frac{1}{4\pi}\oint_{\partial D(0,0,\epsilon)} \mathrm{d} \mathbf{l}\cdot \mathbf{A} -\dots\right) + \frac{1}{4\pi}\int_{D(0,0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y) + \dots\\ \end{align} In the last two steps we used Stokes' theorem and that $\partial R' = \partial R \cup \partial D(0,0,\epsilon) \cup \dots$, and $\pm$ signs account for the orientation of the boundary. Since $\Omega(x,y)$ is finite everywhere, when $\epsilon\rightarrow 0$ \begin{equation} \left|\int_{D(x_0,y_0,\epsilon)} \mathrm{d} x \mathrm{d} y\, \Omega(x,y)\right| \leq \pi\epsilon^2 \max_{x,y} |\Omega(x,y)| \xrightarrow{\epsilon\rightarrow 0} 0 \end{equation} Furthermore, $\oint_{\partial R} \mathrm{d} \mathbf{l}\cdot \mathbf{A}=0$, because $\mathbf{A}$ is periodic, so the integrals over opposite sides of the square cancel out. We're left with \begin{equation} I(m) = -\frac{1}{4\pi}\lim_{\epsilon\rightarrow 0}\oint_{\partial D(0,0,\epsilon)} \mathrm{d} \mathbf{l}\cdot \mathbf{A} -\dots \end{equation} We have \begin{align} \frac{1}{4\pi}\lim_{\epsilon\rightarrow 0}\oint_{\partial D(x_0,y_0,\epsilon)} \mathrm{d}\mathbf{l}\cdot\mathbf{A} = \frac{1}{4\pi}\int_0^{2\pi} \mathrm{d} \phi \lim_{\epsilon\rightarrow 0} \epsilon (-\sin\phi,\cos\phi)\cdot\mathbf{A}(x_0 + \epsilon\cos\phi, y_0 + \epsilon\sin\phi)\,. \end{align} Now one just has to evaluate this for all 4 disks. The results are \begin{align} (0,0)&: \frac{-m+\sqrt{(m-2)^2}+2}{2\sqrt{(m-2)^2}} = \theta(-m+2)\\ (0,\pi)&: \frac{m-\sqrt{m^2}}{2\sqrt{m^2}} = -\theta(-m)\\ (\pi,0)&: \frac{m-\sqrt{m^2}}{2\sqrt{m^2}} = -\theta(-m)\\ (\pi,\pi)&: \frac{-m+\sqrt{(m+2)^2}-2}{2\sqrt{(m+2)^2}} = \theta(-m-2)\,, \end{align} where $\theta$ is Heaviside step function. The total integral then is \begin{equation} I(m) = 2\theta(-m) - \theta(-m+2) - \theta(-m-2) = \begin{cases} -1 & 0<m<2 \\ 1 & -2<m<0 \\ 0 & |m|>2 \\ \end{cases}\,. \end{equation}