Re: Defining Real expressions

• To: mathgroup at smc.vnet.net
• Subject: [mg8527] Re: [mg8494] Defining Real expressions
• From: "C. Woll" <carlw at u.washington.edu>
• Date: Thu, 4 Sep 1997 02:20:29 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Marco,

In general, it is hard to make mma assume certain properties of
parameters. However, Integrate has an option Assumptions that will do what
you want. Thus

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

will return an expression with no If statement.

Carl Woll
Dept of Physics
U of Washington

On Tue, 2 Sep 1997, 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 ?
>
> Thanks
> --
> Marco Beleggia
>
>

```

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