Re: Why no simplification ?

*Subject*: [mg2582] Re: Why no simplification ?*From*: groskyd at gv.ssi1.com (David Rosky)*Date*: Mon, 27 Nov 1995 21:31:10 -0500*Approved*: usenet@wri.com*Distribution*: local*Newsgroups*: wri.mathgroup*Organization*: Silicon Systems, Inc.

In <48h4bc$rcr at ralph.vnet.net>, crobc at epix.net (Christopher R. Carlen) writes: > >I evaluated the integral > >Integrate[ Sqrt[4t^2 + 4 + t^-2], {t, 1, E} ] > >by hand, which of course works out to E^2. By adding the terms over the >common denominator t^2, a perfect square trinomial results in the >numerator. Taking the square root yields just Integrate[ (2t^2 + 1)/t, >{t, 1, E} ] which is valid for t > 0 . > >Now I can understand that Sqrt[ x^2 ] really can't be simplified to x, >but is really |x| . But in the case of a perfect square trinomial like >the above, which is always positive, why doesn't Mathematica recognize >this ? > >Because it fails to recognize this simplification, the output is a big >mess, containing another integral. But anyone can see that this is >really quite a simple integral. > >Anyone have any ideas how to get this integral to come out in a more >simple manner ? > I entered the above integral into Mathcad (which I also use) and it was able to recognize the simplification and produced the result e^2. It also generated the indefinite integral as t^2 + ln(t). Mathcad uses the Maple symbolic processor. In Mathematica, the Simplify function also failed to see it. I assume that Integrate probably calls Simplify on the integrand, but I thought I would check nevertheless. You may want to send this example directly to WRI at their suggestions or support address. It appears to be a valid oversight. Regards, David Rosky (groskyd at gv.ssi1.com)