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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