       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:= \$Version
Out= "4.0 for Microsoft Windows (April 21, 1999)"
In:= Integrate[1/(1 + (x - y)^2), {y, -Infinity, Infinity}]
Out= 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:=\$Version
Out=5.2 for Microsoft Windows (June 20, 2005)

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

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

Bye,
~Scout~

```

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