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MathGroup Archive 2007

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Re: Boole does not work with symbolic limits in Integrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73128] Re: [mg73102] Boole does not work with symbolic limits in Integrate
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sun, 4 Feb 2007 06:57:15 -0500 (EST)
  • Reply-to: hanlonr at cox.net

Integrate[2 u Boole[a < u], {u, 0, x},Assumptions->{0<a<x}]

-a^2 + x^2

Simplify[Integrate[2 u Boole[a < u], {u, 0, x},Assumptions->{a<x}],a<x]

Piecewise[{{x^2, (a <= 0 && x > 0) || (a < 0 && x < 0)}, {-a^2 + x^2, a > 0}}]

Assuming[{0<a<x},Integrate[2 u Boole[a < u], {u, 0, x}]]

-a^2 + x^2

Simplify[Assuming[{a<x},Integrate[2 u Boole[a < u], {u, 0, x}]],a<x]

Piecewise[{{x^2, (a <= 0 && x > 0) || (a < 0 && x < 0)}, {-a^2 + x^2, a > 0}}]


Bob Hanlon

---- Mitch Murphy <mitch at mix5.com> wrote: 
> 
> hey hey
> 
> use of the boole indicator function inside integrate only works for  
> numerical arguments, symbolic limits of integration just returns the  
> unevaluated integral.
> 
> simple example ...
> 
> 	Integrate[2 u Boole[a < u], {u, 0, x}]
> 	
> 	out[] = Integrate[2 u Boole[a < u], {u, 0, x}]
> 
> i would expect ...
> 
> 	Integrate[2 u , {u, a, x}]
> 
> 	out[] = x^2 - a^2
> 
> but it does work with numerical limits of integration ...
> 
> 	Integrate[2 u Boole[a < u], {u, 0, 2}]
> 	
> 	out[] = Piecewise[{{4, a <= 0}, {4 - a^2, 0 < a < 2 }]
> 
> 
> anybody know a trick to make this work ? i've tried a bunch of  
> variations using evaluate[], logicalexpand[], piecewiseexpand[],  
> assumptions->x Reals, ..., but nothing works.
> 
> cheers,
> Mitch
> 


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