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~