Re: Improper integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg44624] Re: Improper integral*From*: vb at cybertester.com (Vladimir Bondarenko)*Date*: Tue, 18 Nov 2003 06:41:54 -0500 (EST)*References*: <bpa2sm$1eh$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

poujadej at yahoo.fr (Jean-Claude Poujade) wrote in message news:<bpa2sm$1eh$1 at smc.vnet.net>... > Bonjour le groupe, > > I'm not a mathematician and I wonder why Mathematica doesn't return 0 > for this doubly infinite improper integral : > > In[1]:=$Version > Out[1]=4.1 for Microsoft Windows (November 2, 2000) > > In[2]:=Integrate[x/(1+x^2),{x,-Infinity,Infinity},PrincipalValue->True] > Integrate::idiv[...]does not converge[...] > Out[2]:=Integrate[x/(1+x^2),{x,-Infinity,Infinity},PrincipalValue->True] > > maybe it's different with Mathematica 5.0 ? > --- > jcp Bonjour, JCP> why Mathematica doesn't return 0 This is a bug. JCP> maybe it's different with Mathematica 5.0 ? No news, good news ;) In[1] := $Version Out[1] = 5.0 for Microsoft Windows (June 11, 2003) In[2] := Integrate[x/(1 + x^2), {x, -Infinity, Infinity}, \ PrincipalValue -> True] Integrate::"idiv" : Integral of x/(1 + x^2) does not converge on {-Infinity, Infinity} Integrate[x/(1 + x^2), {x, -Infinity, Infinity}, PrincipalValue -> True] My prediction for you is that, by a subtle reason, this behaviour will be fixed in Mathematica 6 or earlier. Cheers, Vladimir Bondarenko http://www.cybertester.com/