Re: INDEFINITE INTEGRALS

*To*: mathgroup at smc.vnet.net*Subject*: [mg18096] Re: INDEFINITE INTEGRALS*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Thu, 17 Jun 1999 12:26:35 -0400*Organization*: Universitaet Leipzig*References*: <7jubb0$576@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Jerry, whats wron with ArcTan[x/Sqrt[a]]/Sqrt[a] it is the correct result. The Log[-1+x]-Log[1+x] is simply a expanded version from the relation ArcTan[x]==I Log[(1-I x)/(1+I x)]/x (Abram. Stegun 4.4.28) Regards Jens Blimbaum Jerry DLPC wrote: > > In order to see how well Mathematica does at integration, I tried > several types of integrals where the values of the parameters changed the > results. I compared the integrals for specific values of the parameters and > then symbolically with assumptions. For example: > > Integrate[1/(1+x^2),x] vs. Integrate[1/(-1+x^2),x] and got the > correct results of > > ArcTan[x] and 1/2 (Log[-1+x] -Log[1+x].> > However, if i try the same thing except symbolically, namely > Integrate[1/(a+x^2),x,Assumptions->{a>0}] and > Integrate[1/(a+x^2),x,Assumptions->{a<0}] I got > back ArcTan[x/Sqrt[a]]/Sqrt[a] for both. > I tried variations on this theme using other functions as 1/(x > Sqrt[a+x^2]), Sqrt[a x^2 + c], 1/Sqrt[c+a x^2], where I integrated them > symbolically and then assumed a = -1 and +1. In all cases I got the same > results as above. > > Am I doing this wrong or is something in Mathematica incomplete? > The Mathematica book explicitly states that it checks for variation of the > parameters, but these results dont seem to support that. (Mathematica > 3.01) > > Thanks. Jerry Blimbaum NSWC Panama City, Florida