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