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The integral $$ I(m)=\frac{1}{4\pi}\int_{-\pi}^{\pi}\mathrm{d}x\int_{-\pi}^\pi\mathrm{d}y \frac{m\cos(x)\cos(y)-\cos x-\cos y}{\left( \sin^2x+\sin^2y +(m-\cos x-\cos y)^2\right)^{3/2}} $$ 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 $\hat{\mathbf{n}}$, $$ I(m)=\frac{1}{4\pi}\int_{-\pi}^{\pi}\mathrm{d}x\int_{-\pi}^\pi\mathrm{d}y\, \hat{\mathbf{n}}\cdot\left(\partial_x \hat{\mathbf{n}}\times \partial_y \hat{\mathbf{n}}\right), $$ 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

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    $\begingroup$ It appears there are two cases missing: $I(\pm2)=\mp1/2$. $\endgroup$
    – TheSimpliFire
    Aug 8, 2023 at 16:35
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    $\begingroup$ The case $m=0$ is easy because we can rewrite as $-\pi I(0)=\int_0^\pi f(x)\,dx$ where $$f(x)=\int_0^\pi\frac{\cos x+\cos y}{(2+2\cos x\cos y)^{3/2}}\,dy$$ and we find that $f(x)=-f(\pi-x)$ for $x\in[0,\pi]$ after substituting $u=\pi-y$. So $I(0)=0$. $\endgroup$
    – TheSimpliFire
    Aug 9, 2023 at 13:03
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    $\begingroup$ @TheSimpliFire do you know$$\int_{0}^{\pi} \int_{0}^{\pi} \frac{\text{d}x\text{d}y}{\sqrt{\sin(x)^2+\sin(y)^2+\left ( 1-\cos(x)-\cos(y) \right )^2 } }=\frac{\Gamma\left ( \frac14 \right )^4 }{8\pi}?$$ $\endgroup$ Aug 18, 2023 at 8:40
  • $\begingroup$ @SetnessRamesory That case can probably be done with elliptic integrals but I still believe there must be an easier way to prove $I(m)=0\quad\forall|m|>2$, for instance. Due to the symmetry of the integrand and the exponent of $3/2$, an application of Green's theorem may be useful, similar to this approach. $\endgroup$
    – TheSimpliFire
    Aug 18, 2023 at 9:14
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    $\begingroup$ @xzd209 I would suggest you crosspost this question on MathOverflow, ensuring the post includes a link to this MathStackExchange post. $\endgroup$
    – TheSimpliFire
    Aug 23, 2023 at 21:29

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