# Illogical failures of symbolic integration over Boole

Is there a way to work around integration over Boole being mysteriously flaky like examples show below? Is there a meaningful explanation for this behaviour?

First, rather obvious integral stays unevaluated:

Integrate[Boole[u^2 + v^2 < 1], {u,-1,1}, {v,-1,x}, Assumptions -> -1 < x < 1]

(* Integrate[Boole[u^2 + v^2 < 1], {u,-1,1}, {v,-1,x}, Assumptions -> -1 < x < 1] *)


While the following succeeds to produce a meaningful result:

Integrate[Boole[u^2 + v^2 < 1], {u,-1,1}, {v,-1,x}, Assumptions -> -1 < x < 0]

(* Pi + x Sqrt[1 - x^2] - ArcCos[x] *)


Why this succeeds particularly puzzles me, when the first one fails:

Integrate[Boole[u^2 + v^2 < 1], {u,-1,1}, {v,-1,x},
Assumptions -> -1 < x < Infinity] // FullSimplify

(* Pi + x Sqrt[1 - x^2] - ArcCos[x]             x <= 0
Pi                                           x >= 1
x Sqrt[1 - x^2] + ArcCos[x] + 2 ArcSin[x]    True *)

-
Whenever integration fails over a domain, I always do something like this: FullSimplify@ Integrate[Boole[u^2 + v^2 < 1], {u, -1, 1}, {v, -1, EulerGamma}] /. EulerGamma -> x –  RiemannZeta Sep 17 '13 at 17:11

It works if you do one integration at a time :

Integrate[Integrate[Boole[u^2 + v^2 < 1], {u, -1, 1}], {v, -1, x},
Assumptions -> -1 < x < 1]
(* 1/2 (\[Pi] + 2 x Sqrt[1 - x^2] + 2 ArcSin[x]) *)

-
It would also appear that swapping order to Integrate[Boole[u^2 + v^2 < 1], {v,-1,x}, {u,-1,1}, Assumptions -> -1 < x < 1] does the trick. Can someone explain why these tricks work? –  kirma Sep 17 '13 at 16:36
This answer doesn't give a perfect solution but... I hope your answer helps people stumbling to these same issues. :) –  kirma Sep 24 '13 at 20:31