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MathGroup Archive 1997

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Re: Re[a]>0 ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg6229] Re: [mg6198] Re[a]>0 ?
  • From: "C. Woll" <carlw at u.washington.edu>
  • Date: Fri, 28 Feb 1997 03:21:52 -0500
  • Sender: owner-wri-mathgroup at wolfram.com

On Thu, 27 Feb 1997, Jens Dreger wrote:

> Hi !
> 
> Can anyone tell me how I can make MMA take Re[a] for greater than 0 ?
> 
> 
> In[1]:= Integrate[E^(-a*x^2), {x, -Infinity, Infinity}]
> 
> Out[1]:= If[Re[a] > 0, Sqrt[Pi]/Sqrt[a], 
>           Integrate[E^(-a*x^2), {x, -Infinity, Infinity}]]
> 
> I would like to have just the answer "Sqrt[Pi]/Sqrt[a]", since I know
> that Re[a]>0 is true.
> 
> BTW: a/:Re[a]=1 works, but I don't want to specify the real part of a,
> just want to say it's greater than zero.
> 
> Thanks !
> 
> Jens.
> 
Hi Jens,

One idea is to just use the assumptions option of Integrate. Thus

Integrate[E^(-a*x^2), {x, -Infinity, Infinity},Assumptions->{Re[a]>0}]

ought to give you what you want. 

Another possible solution is force mathematica to think Re[a]>0. To do
this, get the FullForm of Re[a]>0, which is Greater[Re[a],0]. Thus, the
mathematica definition you need to override is connected with Greater. So,
try

Unprotect[Greater];
Re[a]>0 = True;
Protect[Greater];

and then try your integral again

Integrate[E^(-a*x^2), {x, -Infinity, Infinity}]

Both of the above approaches worked when I tried it.

Carl



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