Re: Integrating UnitSteps

*To*: mathgroup at smc.vnet.net*Subject*: [mg48799] Re: Integrating UnitSteps*From*: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig-de>*Date*: Wed, 16 Jun 2004 07:49:00 -0400 (EDT)*Organization*: Uni Leipzig*References*: <cap3m9$cab$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, Integrate[UnitStep[x - 1]/x^2, {x, b, Infinity}, Assumptions -> {b > 0}] Regards Jens "BZ" <BZ at caradhras.net> schrieb im Newsbeitrag news:cap3m9$cab$1 at smc.vnet.net... > Hi guys! > > I'm trying to integrate a function that has a discontinuity at a > single point. I'm using UnitStep to do this, but it doesn't work very > well. To illustrate this, a simple example (my real function is much > more complicated than this): > > In[1]:= Integrate[1/x^2, {x, b, Infinity}] > > 1 > Out[1]= - > b > > Ok, so far so good, but now let's add a discontinuity at x=1: > > In[2]:= Integrate[UnitStep[x - 1]/x^2, {x, b, Infinity}] > > UnitStep[-1 + x] > Out[2]= If[b < 1, 1, Integrate[----------------, {x, b, Infinity}]] > 2 > x > > Which is correct, in principle. However, I'm trying to get an > explicit expression for b>1: > > In[3]:= FullSimplify[%, b > 1] > > UnitStep[-1 + x] > Out[3]= If[b < 1, 1, Integrate[----------------, {x, b, Infinity}]] > 2 > x > > Why isn't this expression simplified? Why doesn't Mathematica > evaluate the Integration inside the If[] (the UnitStep is 1 there > anyway)? Should I be using UnitSteps at all for these kinds of > functions? > > -- > BZ >