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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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