I working with the integral

Integrate[(4*(x^2 + y^2))/(1 + x^2 + y^2)^2 +
            Derivative[0, 1][w][x, y]^2 + Derivative[1, 0][w][x, y]^2 + 
            (4*x*Derivative[0, 1][w][x, y] -  4*y*Derivative[1, 0][w][x, y])/(1 + x^2 + y^2),
          {x, -Infinity,  Infinity},
          {y, -Infinity, Infinity}]

Mathematica is not able to reduce this any further even with FullSimplify.

I believe this integral diverges to positive infinity, since if we distribute the integrals over the addition, the integral of the first term clearly is positive and diverges and the second term is non-negative. The third term can be negative and arbitrarily large, but it is asymptotically smaller than the second term. If that is correct and provided w and its partial partial derivatves are defined throughout the domain, the answer should be Infinity, in the same way Mathematica returns Infinity for things like -Log[0], despite the usual mathematical issues with this.

How can I guide Mathematica to do this?


The above integral can also be rewritten as

Integrate[((2*x)/(1 + x^2 + y^2) + 
              Derivative[0, 1][w][x, y])^2 +
          (-((2*y)/(1 + x^2 + y^2)) + 
              Derivative[1, 0][w][x, y])^2, 
   {x, -Infinity, Infinity}, 
   {y, -Infinity, Infinity}]

but Mathematica cannot do anything with that either.

  • $\begingroup$ could you rewrite this without using the letter l as variable? Makes it hard to read as you have 1 and l next to each others. $\endgroup$ – Nasser Sep 4 '14 at 5:58
  • 1
    $\begingroup$ @Nasser Fixed now. $\endgroup$ – Daniel Mahler Sep 4 '14 at 6:36
  • $\begingroup$ First, I think you are wrong that the first term diverges. It is reduced to 2 Pi Integrate[2 u/(1 + u^2)^2, {u, 0, Infinity}] which is not infinite, and NIntegrate[2 u/(1 + u^2)^2, {u, 0, Infinity}]=1. Second, it seems that Mma cannot give such an answer in a general form, unless the function w is explicitly fixed. Third, if I fix w=y , its derivative w^(0,1)=1 and the integral of its square is infinite. Thus, there are functions with which the integral is infinite, if this is what you intend to prove. $\endgroup$ – Alexei Boulbitch Sep 4 '14 at 7:19
  • $\begingroup$ @AlexeiBoulbitch The first term does diverge. First if we set u=x^2 + y^2 then the first term is 4 u (1+u)^-2, but we are doing a surface integral. If we want to turn it into a line integral along the radius the you must account for the fact that the perimiter grows with the radius giving Integrate[8 Pi u^(3/2) (1 + u)^-2, {u, 0, Infinity}]. $\endgroup$ – Daniel Mahler Sep 4 '14 at 8:08
  • $\begingroup$ @Alexei: Daniel Mahler is right with the first integral: changing to cylindrical coordinates and integrating from r=0 to r=R gives $4 \pi \left(-1+\frac{1}{1+R^2}+\text{Log}\left[1+R^2\right]\right)$ which diverges logarithmically as $R \to \infty$ $\endgroup$ – Dr. Wolfgang Hintze Sep 4 '14 at 8:39

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