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 ÎÎ³Ï?Î±Ï?Îµ: > 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 \!\({x, 0, \ > Ã°}\). \!\(\*ButtonBox[\"More...\", ButtonStyle->\"RefGuideLinkText\", > \ > ButtonFrame->None, ButtonData:>\"Integrate::idiv\"]\)" > 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 > > > Ã?/Ã? Bob Hanlon Ã?Ã£Ã±Ã¡Ã¸Ã¥: > > 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 > > >