Am Montag, 25. Juni 2012 21:14:38 UTC+2 schrieb kjetil1001:
>
> > Actually you can work around this limitation by using piecewise 
> functions: 
> > 
> > In [12]: myabs = lambda x: Piecewise((x, x>=0), (-x, x<0)) 
> > 
> > In [13]: integrate( (t1*t2*t3)**beta * myabs((t1-t2)*(t1-t3)*(t2-t3)), 
> > (t1,0,1),(t2,0,1),(t3,0,1)).doit() 
> > Out[13]: 0 
>
> But that answer 0 is wrong! 
>

You are right, it should not be 0, this is a bug. Do you know the correct 
value?

Vinzent 

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