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MathGroup Archive 1991

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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


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