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MathGroup Archive 2006

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Re: Unexpected condition on convergence of integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67325] Re: Unexpected condition on convergence of integral
  • From: "Scout" <Scout at nodomain.com>
  • Date: Sun, 18 Jun 2006 05:13:21 -0400 (EDT)
  • References: <e6tc9d$dtn$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Andrew Moylan" <andrew.moylan at anu.edu.au>
news:e6tc9d$dtn$1 at smc.vnet.net...
> Integrate[1 / (1 + (x - y)^2), {y, -Infinity, Infinity}] yields If[Im[x] 
> != 0, Pi, ...], but (I think) this integral should converge to Pi 
> unconditionally (try it for some real values of x).
>
> Using the option "Assumptions -> Im[x] == 0" in Integrate[] further yields 
> the following warning: "Integral of [bla] does not converge on {-Infinity, 
> Infinity}".
>
> Can anyone shed some light on why Integrate can't determine that this 
> integral converges for all x?
>

Mathematica 4 gives the "expected" value:

    In[1]:= $Version
    Out[1]= "4.0 for Microsoft Windows (April 21, 1999)"
    In[2]:= Integrate[1/(1 + (x - y)^2), {y, -Infinity, Infinity}]
    Out[2]= Pi

As reported in the Help,
 "When an integrand depends on a parameter, the indefinite integral should 
be considered valid for GENERIC values of the parameter.(...)
   By contrast, when you ask for a definite integral, Mathematica tries to 
return a result that is ALWAYS valid, ... ".

So, you can calculate the indefinite integral of your function and then 
replace the extreme values:
    In[1]:=$Version
    Out[1]=5.2 for Microsoft Windows (June 20, 2005)

    In[2]:= Integrate[1/(1+(x-y)^2),y]
    Out[2]= -ArcTan[x-y]

    In[3]:=( %2 /. y-> Infinity ) - ( %2 /. y-> -Infinity )
    Out[3]= Pi

Bye,
    ~Scout~


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