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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/
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