>>>>> "Neilen" == Neilen Marais <nmarais at sun.ac.za> writes:
Neilen> Hi Ray
Neilen> On Fri, 08 Dec 2006 09:01:07 -0500, Raymond Toy wrote:
>>>>>>> "Neilen" == Neilen Marais <nmarais at sun.ac.za> writes:
Neilen> If I succesively take indefinite integrals and manually substitute the bounds I
Neilen> get the same answer, without any asksign questions (I omitted some of the
Neilen> intermediate results to save space):
>>
Neilen> (%i34) integrate(tmpp, x3);
Neilen> (%o34) (x2/2+x1/2)*sin((%pi*x2)/2+(%pi*x1)/2-%pi/4)*x3(%i35) ev(%, x3:1-x2-x1)
Neilen> - ev(%,x3:0);
>>
>> I did not investigate the exact reason for why you asked for the sign
>> of various things, but you should note that definite integrals are
>> often transformed to different forms to evaluate it. That is,
>> definite integration does not necessarily find the indefinite integral
>> and substitute in the limits, which is what you are doing here by
>> hand. If you do integrate(tmpp, x3, 0, 1-x1-x2), you will be asked
>> for the sign of x1+x2-1.
Neilen> Fair enough, I guess my real question is then, would I be making a mathematical
Neilen> mistake by substituting the limits without knowing the sign of x1+x2-1?
In this particular case, I don't think so. There's nothing special
about the integrand that knowing the sign would change.
>> If I do this:
>>
>> assume(x1>0, x2 > 0, x3 > 0, x1 < 1, 1-x1-x2> 0);
>>
>> the triple integral only asks for the sign of cos(%pi*x1/2) and
>> sin(%pi*x1/2). I guess that's a deficiency in maxima that it doesn't
>> know the sign of those.
Neilen> Would I be correct in guessing that Maxima would have been able to figure out
Neilen> what you passed in the assume() above if it had a built in multiple integral
Neilen> routine that could consider all the integral limits simultaneously?
I think that would be a simple matter of programming. :-) Someone
would have to modify the definite integration routines to inform
maxima about the range of the variables. I think it would be fairly
difficult for maxima to figure out that 1 - x1 - x2 > 0.
Ray