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Re: sum of integrals over patial intervals != integral

• To: mathgroup at smc.vnet.net
• Subject: [mg71820] Re: sum of integrals over patial intervals != integral
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Thu, 30 Nov 2006 06:05:17 -0500 (EST)
• References: <ekh5hc$rn5$1@smc.vnet.net><ekjer5$cc2$1@smc.vnet.net>

I know that the subject of symbolic definite integration is very
difficult and sometimes the implementated algorithms fail to give the
desired results especially when the integrand contains branch cuts.

Although I do consider Mathematica as the best in this field among the
CAS I have worked isn't strange (at least!) that the same version of
Mathematica for different platforms give completely different results
(Mac->right vs Win->false)?

Dimitris

Î?/Î? dimitris έγÏ?αÏ?ε:
> In[4]:=
> $Version
> Out[4]=
> "5.2 for Microsoft Windows (June 20, 2005)"
>
> In[2]:=
> f[x_] := Log[Sin[x]^2]*Tan[x];
>
> In[3]:=
> Integrate[f[x], {x, 0, Pi}]
> Integrate::"idiv" : "Integral of \
> \!\(\(\(Log[\(\(\(Sin[x]\)\^2\)\)]\)\)\\ \(\(Tan[x]\)\)\) does not
> converge \
> on \!\({x, 0, \
> ð}\). \!\(\*ButtonBox[\"More...\", ButtonStyle->\"RefGuideLinkText\",
> \
> ButtonFrame->None, ButtonData:>\"Integrate::idiv\"]\)"
> Out[3]=
> Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}]
>
> But
>
> In[5]:=
> Integrate[f[x], {x, 0, Pi/2, Pi}]
> Out[5]=
> 0
>
> In[9]:=
> Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi},
>   GenerateConditions -> False]
> Out[9]=
> 0
>
>
> �/� Bob Hanlon �ãñáøå:
> > Works in my version:
> >
> > $Version
> >
> > 5.2 for Mac OS X (June 20, 2005)
> >
> > f[x_]:=Log[Sin[x]^2]Tan[x];
> >
> > Integrate[f[x],{x,0,Pi}]
> >
> > 0
> >
> >
> > Bob Hanlon
> >
> > ---- Peter Pein <petsie at dordos.net> wrote:
> > > Dear group,
> > >
> > > I wanted Mathematica to show, that for f[x_]:=Log[Sin[x]^2]Tan[x],
> > > Integrate[f[x],{x,0,Pi}]==0, because f[x]+f[Pi-x]==0.
> > >
> > > Mathematica says Integrate[f[x],{x,0,Pi}] does not converge, but
> > > Integrate[f[x],{x,0,Pi/2}] and Integrate[f[x],{x,Pi/2,Pi}] evaluate to
> > > -Pi^2/12 resp. P^2/12 and the sum is zero. The more general integral
> > > Integrate[f[x],{x,0,z},Assumptions->Pi/2<z<=Pi] evaluates explicitly (?).
> > >
> > > What did I do wrong?
> > > http://people.freenet.de/Peter_Berlin/Mathe/komisch.nb
> > > 
> > > TIA,
> > > Peter
> > >