Re: Re: Defining Real Expressions

• To: mathgroup at smc.vnet.net
• Subject: [mg8597] Re: [mg8535] Re: Defining Real Expressions
• From: jpk at max.mpae.gwdg.de
• Date: Tue, 9 Sep 1997 03:07:15 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```> From raya at mech.ed.ac.uk Sun Sep  7 08:58:49 1997
> Date: Sat, 6 Sep 1997 23:16:07 -0400
> From: Raya Firsov-Khanin <raya at mech.ed.ac.uk>
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
> Subject: [mg8597] [mg8535] Re: Defining Real Expressions
>
> Marco Beleggia wrote:
>
> <I must evaluate an Integral in which I'd like to assign real values to
> <some parameters, but I don't know how to do that.
>
> <For example, in the following integral (a Fourier Transform):
>
> <f[x_,p_]=Integrate[y/(x^2+y^2) Exp[-I p y],{y,-Infinity,Infinity}],
>
> <p should be a real parameter. The output given by Mathematica is
> <conditioned to Im[p]==0, such as If[Im[p]==0,....,....], which is not
> <easy to handle, and I'd like to avoid this complication.
>
> <Is it possible ?
>
> The simplest way to deal with this is to declare
>
> p/:Im[p]=0
>
> Then, f[x, p] will give you an answer (if Arg[x^2]!=\[Pi]).
>
> -------
> 		              Raya Khanin
>

Hi,

look on the Assumptions Option of Integrate[] with some thing like

Integrate[
y/(x^2+y^2) Exp[-I p y],
{y,-Infinity,Infinity},
Assumptions->
{Im[p]==0,Im[x]==0}
]

Hope that helps
Jens

```

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