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MathGroup Archive 1998

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Re: Can I get ComplexExpand to really work?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg14582] Re: Can I get ComplexExpand to really work?
  • From: Alan Lewis <alanlewis at home.com>
  • Date: Fri, 30 Oct 1998 03:07:49 -0500
  • Organization: @Home Network
  • References: <719f5p$lc6@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,
  If you set the option GenerateConditions->False inside the integral,
you will not get the conditional output.

Alan 

Topher Cawlfield wrote:
> 
> Hi,
> 
> I'm having lots of problems getting Mathematica to make simplifying
> assumptions.  It always seems to want to produce horribly complex
> results because it assumes every variable is complex.  I wish I had
> better control of that.  In fact, it would also be nice if I could
> assure Mathematica that certain variables were positive as well.
> 
> It sounds like the function ComplexExpand should do the trick, at least
> by assuming that variables are real unless otherwise specified.  But it
> doesn't really seem to work for me.  Here's an example:
> 
> ComplexExpand[Integrate[E^(I a x^2), {x, -Infinity, Infinity}]]
> 
> produces:
> 
> If[Im[a] == 0, Sqrt[Pi/2] (1 + I Sign[a]) / (a^2)^(1/4), Integrate[E^(I
> a x^2), {x, -Infinity, Infinity}]]
> 
> But if it really was assuming that 'a' was real, then it should know
> that Im[a] == 0!  If I could also tell it that 'a' was positive, the
> answer would be:
> 
> (1 + I) Sqrt[Pi/a]  or better still, Sqrt[2 Pi I / a]
> 
> This is much simpler, and is the answer I want.
> 
> Of course, my real application of this problem is much more complicated,
> but ultimately comes down to doing that integral (several times over).
> The right answer should be just about that simple, but instead
> Mathematica gives me about 5 pages of output.
> 
> Is there any hope of getting reasonable symbolic results here?
> 
>  - Topher Cawlfield


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