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
>