A straightforward way to obtain the value of

    1

I(m)= ───── ∬{-π}^{π} ∬{-π}^{π} 4π m cosx cosy – cosx – cosy ------------------------------------------------ dx dy (1) [ sin²x+sin²y+( m – cosx – cosy )² ]^{3/2}

is to look at it as an ordinary flux–integral and to keep track of the places where the denominator can vanish.
Everything can be done with nothing more than multivariable calculus, the divergence theorem and the elementary identity

    div  r / |r|³  =  4π δ³(r) .                                   (2)

No mention of Chern classes is necessary.

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1. Re-interpret the integrand as a flux density

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Put

    n(x,y,m)= ( sinx , siny , m–cosx–cosy ) ,          R(x,y,m)= n/|n|³.

Because

    R·(∂xn×∂yn)= n·(∂xn×∂yn)/|n|³ ,                     (3)

(1) can be rewritten

    I(m)= 1/(4π)  ∬_{T}  R·(∂xn×∂yn) dx dy              (4)

where T=[-π,π]×[-π,π] is our fundamental rectangle.
Equation (4) is the flux of the Coulomb field R through the two–dimensional surface Σm obtained by keeping m fixed and letting (x,y) run over T:

    Σm :  (x,y) ↦ n(x,y,m) ⊂ ℝ³.

So

    I(m) = (total Coulomb flux through Σm)/4π .         (5)

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2. How does I(m) change when m varies?

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Let m₁<m₂ and let

    D = T×[m₁,m₂] ⊂ ℝ³_{(x,y,m)} .

The boundary of D is the union of Σ_{m₂}, Σ_{m₁} and the four side faces x=±π, y=±π.
Because the integrand is 2π–periodic in x and y, the contributions of the four side faces cancel pairwise, hence

    I(m₂) – I(m₁) = (1/4π) ∬_{∂D}  R·dS
                  = (1/4π) ∭_{D}  div R  dV           (Gauss). (6)

With (2) we get

    I(m₂) – I(m₁) = ∭_{D} δ³(n)\det(∂(n)/∂(x,y,m)) dx dy dm. (7)

Thus I(m) can change only when n(x,y,m)=0 has a solution. Those solutions are easy to list:

    sinx = siny = 0,
    m  = cosx + cosy.                                   (8)

Because sinx = 0 ⇒ x ≡ 0,π (mod 2π), and the same for y, the four possibilities are

    (x,y,m) :  (0,0,2),   (π,π,-2),   (0,π,0),   (π,0,0).    (9)

Hence I(m) can jump only when m passes -2, 0 or 2.

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3. Size and sign of every jump

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The size of the jump produced by one zero of n is the sign of the Jacobian

    J = det[∂(n)/∂(x,y,m)] = cosx·cosy                      (10)

evaluated at that zero, because δ³(n) integrates to 1/|J| and (7) keeps the sign of J.

• (π,π,-2) : cosx cos y = (+)(+) = +1 ⇒ jump +1 when m crosses –2 • (0,0, 2) : (+)(+) +1 ⇒ jump +1 when m crosses 2 • (0,π, 0) : (+)(-) –1 ⇒ jump –1 • (π,0, 0) : (-)(+) –1 ⇒ jump –1

The two points with m=0 contribute together a jump –2.

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4. Assemble the result

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Start with very negative m.
For m<-2 the vector n points essentially in the negative z–direction for all (x,y); the surface Σm lies completely in the lower half–space and encloses no charge, so I(m)=0 for m<-2.

Now raise m and add the jumps just found:

• Crossing m=-2 : I jumps by +1 ⇒ I=+1 for –2<m<0 • Crossing m=0 : I jumps by –2 ⇒ I=-1 for 0<m<2 • Crossing m=2 : I jumps by +1 ⇒ I=0 for m>2

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5. Final value

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Therefore

    I(m)= {  1   for  –2 < m < 0
           –1   for   0 < m < 2
            0   otherwise. }                            (11)

which is exactly the piecewise constant function obtained from the abstract characteristic–class argument,

    I(m)= ( sign(m–2)+sign(m+2) )/2 – sign(m).          (12)

All we have used is the interpretation of the integrand as a flux, the divergence theorem, the elementary identity (2) for the Coulomb field, and a local Jacobian calculation at the finitely many points where the integrand can become singular.