# Integrating over a region without singularity gives an error indicating the function has singularity in that region

I have the following 2D region over which I want to integrate a function:

r1  = Sqrt[mη^2 + (Sqrt[w1^2 - mπ^2] + Sqrt[w2^2 - mπ^2])^2] // Expand;
r2  = Sqrt[mη^2 + (Sqrt[w1^2 - mπ^2] - Sqrt[w2^2 - mπ^2])^2] // Expand;
mη  = 0.547;
mηp = 0.958;
mπ  = 0.137;

RegionPlot[mηp - w1 - w2 < Re[r1] && mηp - w1 - w2 > Re[r2],
{w1, .1, .25}, {w2, .1, .25}, BoundaryStyle -> Blue, FrameLabel -> {"w1", "w2"}]


Now I want to integrate the following function over this region:

function[w1_, w2_]=Abs[1.05133+ (6.16152 (0.656093- 1.916 w1 - 1.916 w2)
(0.958- 1. w1 - 1. w2))/(0.921055- 1.916 w1 - 1.916 w2) -
(10.1147 (0.656093- 1.916 w1 - 1.916 w2) (0.958- 1. w1 - 1. w2))/
(1.57895- 1.916 w1 - 1.916 w2) - 55.0594 (-((0.479 w1 (-0.618555 + 1.916 w1))/
(0.0307393+ 1.916 w1)) - ( 0.479 w2 (-0.618555 + 1.916 w2))/(0.0307393+ 1.916 w2))]^2


When I plot the function it doesn't have any singularity:

Plot3D[function[w1,w2],{w1, 0.137, 0.2445}, {w2, 0.137, 0.2445}, BoxRatios -> {1, 1, 4},
RegionFunction -> Function[{w1, w2, z}, mηp - w1 - w2 < Re[r1] && mηp - w1 - w2 > Re[r2]]]


But when I use the following command:

NIntegrate[function[w1,w2] Boole[Re[r2]< mηp - w1 - w2 <
Re[r1]], {w1, 0.137, 0.2445}, {w2, 0.137, 0.2445}]


Mathematica says it has a singularity. Why?

-
Mma is not saying it has a singularity: Numerical integration converging too slowly; suspect one of the \ following: singularity, value of the integration is 0, highly \ oscillatory integrand, or WorkingPrecision too small –  belisarius Sep 6 '12 at 13:29
what should I do to solve this problem? Because I know this integral should have an answer about 0.0000872 –  Soodeh Z. Sep 6 '12 at 13:34
I don't know am I right or not but it seems that over this region the shape of the function is common(It doesn't have singularity, It is not highly osccillating or...), then what is the problem? With Boole does it calculate the function in the region only? Because I know out of this region function has singularity. –  Soodeh Z. Sep 6 '12 at 13:45
@soodeh What's the point of defining function = Abs[...]^2 ? Why not function = (...)^2 ? –  b.gatessucks Sep 6 '12 at 13:45
@b.gatessucks You are right. I can omit it. –  Soodeh Z. Sep 6 '12 at 13:46