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 >