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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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