       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=Integrate[E^(-v^2/vth^2),{v,-Infinity,+Infinity}]
>
>and i obtai a complex result:
>
>Out=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
>

```

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