Re: Integrate modified in version 6?

*To*: mathgroup at smc.vnet.net*Subject*: [mg78142] Re: Integrate modified in version 6?*From*: "Dana DeLouis" <dana.del at gmail.com>*Date*: Sat, 23 Jun 2007 07:23:04 -0400 (EDT)

> The good old (?) Mathematica 5.2 returns > Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}] > -1 - 2*Catalan + Pi*(1/4 + Log[2]) Hi. I get -Infinity also using 6.0. It appears if one uses "TrigToExp", one gets the same answer as 5.2's. equ = z*(ArcSin[z]/(1 + z)^2); Integrate[equ, {z, 0, 1}] -Infinity NIntegrate[equ, {z, 0, 1}] 0.13105306534661287 (* Using TrigToExp to get same answer as 5.2 *) Integrate[TrigToExp[equ], {z, 0, 1}] -1 - 2*Catalan + Pi*(1/4 + Log[2]) N[%] 0.13105306534661243 (* Redoing calculation which "appears" to indicate they are similar *) N[NIntegrate[equ, {z, 0, 1}, WorkingPrecision -> 20]] 0.1310530653466124 -- Dana DeLouis Using 6.0, but having to use 5.2 Help files. "dimitris" <dimmechan at yahoo.com> wrote in message news:f5asi1$9t6$1 at smc.vnet.net... >I don't have 6 to be more rigorous but > based on some integrals post in another > forum by Vladimir Bondarenko I am quite > sure that something has change in the integration > algorithm for definite integrals... > > Say for example the integral Integrate[z ArcSin[z]/(1+z)^2, {z, 0, > 1}]. > > In Mathematica 6 we have > > Integrate[z ArcSin[z]/(1+z)^2, {z, 0, 1}] > -Infinity > > which is far away from truth. > > The good old (?) Mathematica 5.2 returns > > In[3]:= > Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}] > {N[%], NIntegrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]} > > Out[3]= > -1 - 2*Catalan + Pi*(1/4 + Log[2]) > Out[4]= > {0.13105306534661265, 0.1310530653479215} > <snip>