Controlling evaluation or Re: Plot[] acting funny (help)

*To*: mathgroup at yoda.ncsa.uiuc.edu*Subject*: Controlling evaluation or Re: Plot[] acting funny (help)*From*: jacobson at cello.hpl.hp.com*Date*: Wed, 23 Jan 91 09:49:09 PST

antonyc at tybalt.caltech.edu (Bill T. Cat) asks about the following plot Plot[NIntegrate[((Sin[t x])^2),{x,0,Pi}],{t,0,10}] I think the problem was confusion caused by Mathematica drawing the x axis at y=1. Adding PlotRange->All makes it look as expected. Now for my question: I observed that this integral can easily be done analytically. So I tried plotting Plot[Release[Integrate[((Sin[t x])^2),{x,0,Pi}]],{t,0,10}, PlotRange->All] and this works just fine, and a lot faster, but gives a bunch of error messages, since Integrate[((Sin[t x])^2),{x,0,Pi}] evaluates to Pi Cos[Pi t] Sin[Pi t] -- - ------------------- 2 2 t so there is a division by zero at t=0. But there is also a Sin[Pi t] in the numerator, so the limit as t->0 is well defined. Now what I'd like to do is, in a one-liner, make it do the integral and wrap a limit around it, something like Plot[Release[Limit[Integrate[((Sin[tt x])^2),{x,0,Pi}]],tt->t],{t,0,10}, PlotRange->All] but this doesn't quite work, as it takes the limit too soon and I still get the division by zero. I want Plot to see on each iteratation Limit[Pi/2 - (Cos[Pi*tt]*Sin[Pi*tt])/(2*tt),tt-> <<current value of t>>] Any ideas about how to do this? (No replies please about Limit being slow, or about using Off[], etc. This is a question about controlling evaluation.) -- David Jacobson