Re: Re: A little problem with positive numbers

*To*: mathgroup at smc.vnet.net*Subject*: [mg18150] Re: [mg18088] Re: A little problem with positive numbers*From*: BobHanlon at aol.com*Date*: Fri, 18 Jun 1999 00:51:48 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Nicola, I'm running v4; however, I think this will give the same result in v3. Integrate[E^(-v^2/vth^2), {v, -Infinity, +Infinity}, Assumptions -> {vth > 0}] Sqrt[Pi]*vth Bob Hanlon In a message dated 6/17/99 4:42:11 PM, nicola at Linuz.sns.it writes: >On Mon, 14 Jun 1999, Murray Eisenberg wrote: > >> What, exactly, are you trying to do? And what version of Mathematica? >> >> The new Version 4 "Support for assumptions in Simplify, FunctionExpand >> and related functions" may do what you need. >> > >Really im using version 3. >Im tryng to calculate something like >(in my case is a bit more complex but the problem >is the same): > >In[1]=Integrate[E^(-v^2/vth^2),{v,-Infinity,+Infinity}] > >and i obtai a complex result: > >Out[1]=If[Re[vth^2]>0,Sqrt[Pi]Sqrt[vth^2],Integrate[E^(-v^2/vth^2), >{v,-Infinity,+Infinity})]] > >I would like to communicate to Math that >im sure vth is a positive number. > >Bye and thank you for any help >