Re: Improper integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg44614] Re: [mg44608] Improper integral*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Tue, 18 Nov 2003 06:41:43 -0500 (EST)*References*: <200311170838.DAA01254@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 17 Nov 2003, at 17:38, Jean-Claude Poujade wrote: > 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 > No it is the same, and it is correct. Presumably the reason why you think the answer should be zero is: In[21]:= Integrate[x/(1 + x^2), {x, -a, a}] Out[21]= 0 But Integrate[x/(1 + x^2), {x, -Infinity, Infinity}] is not just the limit of the above as a->Infinity. What has to be true is that the limits of Integrate[x/(1 + x^2), {x, a, b}] must exist as a -> -Infinity and b->Infinity independently of one another. This is of course not true. If you defined the infinite integral in a different way you could end up with all sorts of contradictions. For example, observe that: Limit[Integrate[x/(1 + x^2), {x, -a, 2*a}], a -> Infinity] Log[2] and so on. Andrzej Kozlowski Chiba, Japan http://www.mimuw.edu.pl/~akoz/

**References**:**Improper integral***From:*poujadej@yahoo.fr (Jean-Claude Poujade)