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
>
>