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
>