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
>

```

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