Re: Integrate modified in version 6?

*To*: mathgroup at smc.vnet.net*Subject*: [mg78113] Re: Integrate modified in version 6?*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Sat, 23 Jun 2007 07:08:11 -0400 (EDT)*References*: <200706200930.FAA09746@smc.vnet.net> <f5dgdj$nlh$1@smc.vnet.net>

Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > My version of Mathematica 6.0 gives: > > Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}] > > (1/2)*(MeijerG[{{-(1/2), 0}, {1}}, {{0, 0, 1/2}, {}}, 1] - > MeijerG[{{-(1/2), 0}, {1, 1}}, {{0, 0, 1/2, 1/2}, {}}, 1]/Sqrt > [Pi]) > > I don't know whether this answer is right or wrong (Mathematica takes > for ever to evaluate this numerically) but one thing is clear: it > certainly is not -Infinity. How is that clear? Also note that Bob H. and Curtis O. both report getting -Infinity as the result in version 6 (which I don't have). FWIW, using version 5.1, FullSimplify applied to those two MeijerG terms yields Indeterminate and ComplexInfinity, resp. My guess is that the output from version 6 involving MeijerG might be pure nonsense. If so, maybe that's why it takes "for ever" to evaluate numerically. In any event, version 5.1 gets a nice answer easily: In[1]:= Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}] Out[1]= -1 - 2*Catalan + Pi*(1/4 + Log[2]) In[2]:= N[%] Out[2]= 0.131053 In[3]:= NIntegrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}] Out[3]= 0.131053 and so the behavior of version 6 on this particular integral is clearly not as good. David W. Cantrell

**References**:**Integrate modified in version 6?***From:*dimitris <dimmechan@yahoo.com>