On evaluating an integral related to Chern numbers

Question

I am trying to evaluate the following integral in Mathematica

$$ ∫_{-\infty}^{\infty}\mathrm dk_1∫_{-\infty}^{\infty}\mathrm dk_2 \frac{mπ}{(k_1^2+k_2^2+m^2)^{3/2}}\frac{Δ(k_1^2-k_2^2)}{\sqrt{[Δ(k_1^2-k_2^2)]^2}} $$

Integrate[( m π)/(k1^2 + k2^2 + m^2)^(
  3/2) ((k1^2 - k2^2) Δ)/ 
  Sqrt[(((k1^2 - k2^2)  Δ)^2)], {k1, -Infinity, 
  Infinity}, {k2, -Infinity, Infinity}]

where the second piece is just $\text{sign}(Δ(k_1^2-k_2^2))$. The integrand will flip sign under $k_1\leftrightarrow k_2$, except along the lines $k_1=\pm k_2$ where the integrand is not determined, thus the integral is nontrivial. Indeed, if I naively plug this one into Mathematica, then I obtain $-2π^2\text{sign}(mΔ)$. Now if I perform the following coordinate transformation

$ k_1=k\cos\phi , k_2=k\sin\phi $

which gives

$$ ∫_{0}^{\infty}\mathrm dkk∫_{0}^{2π}\mathrm d\phi \frac{mπ}{(k^2+m^2)^{3/2}}\frac{Δ k^2\cos 2\phi}{\sqrt{(Δ k^2\cos 2\phi)^2}} $$

Integrate[(
  k^2 m π Δ Cos[2 ϕ])/((k^2 + m^2)^(3/2) Sqrt[
   k^4 Δ^2 Cos[2 ϕ]^2]) k, {k, 0, 
  Infinity}, {ϕ, 0, 2 Pi}]

which (not surprisingly) gives 0. In fact the following closely related integral is also evaluated to be zero by Mathematica.

$$ ∫_{-\infty}^{\infty}\mathrm dk_1∫_{-\infty}^{\infty}\mathrm dk_2 \frac{mπ}{(k_1^2+k_2^2+m^2)^{3/2}}\frac{Δ k_1 k_2}{\sqrt{(Δ k_1k_2)^2}} $$

Integrate[( m π)/(k1^2 + k2^2 + m^2)^(
  3/2) ((k1 k2) Δ)/ 
  Sqrt[(((k1 k2)  Δ)^2)], {k1, -Infinity, 
  Infinity}, {k2, -Infinity, Infinity}]

The null-result of the second and third integral is probably because Mathematica will first check the symmetry of the numerator before evaluating it.

Now my question is why Mathematica gives a nonzero result for the very first integral? Is it a bug (which I don't think it is likely)? How can I know how Mathematica evaluate the integrals (The command Trace gives something quite not clear.)?

Thank you!

Answer 1

If we can assume all variables are real, we can simplify one factor:

FullSimplify[((k1^2 - k2^2) Δ)/Sqrt[(((k1^2 - k2^2) Δ)^2)], {k1, k2, Δ} ∈ Reals]
(*  Sign[(k1 - k2) (k1 + k2) Δ]  *)

Now it's rather obviously zero by symmetry, unless m == 0:

AbsoluteTiming@ Integrate[(m π)/(k1^2 + k2^2 + m^2)^(3/2) Sign[(k1 - k2) (k1 + k2) Δ],
 {k1, -Infinity, Infinity}, {k2, -Infinity, Infinity}, Assumptions -> m > 0]
(*  {29.1478, 0}  *)

Of course, it's easier if we rotate the symmetry to align with the coordinate axes:

AbsoluteTiming@ Integrate[(m π)/(k1^2 + k2^2 + m^2)^(3/2) Sign[(k1 - k2) (k1 + k2) Δ] /.
    {k1 -> (k1 - k2)/Sqrt[2], k2 -> (k1 + k2)/Sqrt[2]},
  {k1, -Infinity, Infinity}, {k2, -Infinity, Infinity}, Assumptions -> m > 0]
(*  {0.458337, 0}  *)

It does seem the result without the assumption that m is a nonzero real number is unreliable. Note that if m is pure imaginary, the integral is divergent. It's a good idea to add the appropriate assumptions.

Answer 2

Another way to verify is to evaluate the integral in pieces without limits and manually apply the limits. It's extra work, but if you doubt the result, it may be worth checking. The inner k2 integral.

k2int[k2_] = Integrate[((m*Pi)*((k1^2 - k2^2)*Δ))/((k1^2 + k2^2 + m^2)^(3/2)*
      Sqrt[((k1^2 - k2^2)*Δ)^2]), k2]//Simplify

Differentiate to check the result.

D[k2int[k2], k2] // Simplify

And we get back to the integrand. Now apply the limits.

intk2 = Limit[k2int[k2], k2 -> ∞] - Limit[k2int[k2], k2 -> -∞] // Simplify
(*-((2 π Δ m)/(Sqrt[Δ^2] (k1^2 + m^2)))*)

Now integrate over k1.

intk1[k1_] = Integrate[intk2, k1]

check

D[intk1[k1], k1]//Simplify

checks ok. Apply limits to get the final integral.

int = Limit[intk1[k1], k1 -> ∞] - Limit[intk1[k1], k1 -> -∞]
(*-((2 π^2 Δ Sqrt[1/m^2] m)/Sqrt[Δ^2])*)

which could be further simplified with assumptions for Δ and m