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,