see inline.

On Fri, Jun 29, 2012 at 5:04 AM, Vinzent Steinberg
<vinzent.steinb...@gmail.com> wrote:
> 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?
>

Yes. This is a special case of the so-called Selberg integral (named
for Atle Selberg) ,  see
http://en.wikipedia.org/wiki/Selberg_integral

The special case above has the value: (G is the gamma function)

S_3(beta+1,1,gamma=1/2) = \prod_{j=1}^3  \frac{ G(beta+1+(j-1)gamma)
G(1+(j-1)gamma) G(1+j gamma) }

{ G(beta+2+(3+j-2)gamma) G(1+gamma) }


Kjetil



> Vinzent
>
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