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I'm trying to evaluate an integral $ \iint_D \frac{1}{1+(x^2+y^2)^2}dxdy$ given the region $ D $ with the following inequalities: $ 1\leq x^2+y^2\leq 4 \text{, } x \geq 0\text{ and } y \geq 0$.

I decided to use the following syntax:

Integrate[1/(1+(x^2+y^2)^2)Boole[1<=x^2+y^2<=4],{x,0,Infinity},{y,0,Infinity}]

What happened was that Mathematica keeps performing calculations, but it never executes the computation. I've been letting it do the job for the past 5 minutes, but still no sign of a solution.

For fun, I did run

Integrate[(1+(x^2+y^2)^2)Boole[1<=x^2+y^2<=4],{x,0,Infinity},{y,0,Infinity}]

which actually yielded $ 6 \pi $!

Is my syntax OK and the Mathematica is unable to perform such computation (which I highly doubt), or is it something wrong with my syntax?

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    $\begingroup$ Integrate[1/(1 + (x^2 + y^2)^2), {x, y} ∈ Annulus[{0, 0}, {1, 2}, {0, π/2}]] returns -π (π - 4 ArcTan[4])/16 quickly, tho. $\endgroup$ Nov 9, 2017 at 15:49

1 Answer 1

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After changing to polar coordinates, the integral is basic:

Integrate[ρ/(1 + (ρ^2)^2), 
          {ρ, 1, 2}, {θ, 0, π/2}]

(* -(1/16) π(π - 4 ArcTan[4]) *)
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