Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1997
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1997

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Defining Real expressions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg8522] Re: [mg8494] Defining Real expressions
  • From: Daniel Lichtblau <danl>
  • Date: Thu, 4 Sep 1997 02:20:20 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

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


Maybe something along these lines?

In[29]:= Integrate[y/(x^2+y^2) Exp[-I p y],
	{y,-Infinity,Infinity},
	Assumptions-> Im[p]==0 && Im[x]==0]
         -I Pi Sign[p]
Out[29]= -------------
                2  2
          Sqrt[p  x ]
         E

Another possibility is to assume that the "natural" assumptions to make
are the ones you have in mind (and also that the integration code has
the same idea of "natural" as you have).

In[30]:= Integrate[y/(x^2+y^2) Exp[-I p y],
	{y,-Infinity,Infinity},GenerateConditions->False]
         -I Pi Sign[p]
Out[30]= -------------
                2  2
          Sqrt[p  x ]
         E

Daniel Lichtblau
Wolfram Research
danl at wolfram.com


  • Prev by Date: Q. re results from Regression[ ]
  • Next by Date: RE: Re: Several variab
  • Previous by thread: Re: Defining Real expressions
  • Next by thread: Re: Defining Real expressions