Re: Re: Re: integration problem
- To: mathgroup at smc.vnet.net
- Subject: [mg16346] Re: [mg16318] Re: [mg16232] Re: [mg16172] integration problem
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Thu, 11 Mar 1999 02:16:31 -0500
- Sender: owner-wri-mathgroup at wolfram.com
As I explained in an earlier message (which has not yet, at the moment when I am writing it, passed the censor I mean moderator :) all you need need to do is to load the package <<Calculus`Limit` Because of the very general method mathemtica uses in this case you will get complex numbers in the answer: In[1]:= Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 2}] Out[2]= 1 1 - (-2 I Pi - 4 ArcTanh[-] + ArcTanh[2]) + 5 2 1 Sqrt[7] - (4 ArcTanh[-------] - ArcTanh[2 Sqrt[7]] + Log[3]) + 5 2 1 327680 -- (4 I Pi + 2 ArcTanh[4] - Log[------]) + 10 81 1 327680 -- (4 I Pi - 2 ArcTanh[4] + Log[------]) 10 81 Actually the imaginary part is zero. You can get a (relatively) simple real answer by using: In[2]:= FullSimplify[ ComplexExpand[Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 2}], TargetFunctions->{Re,Im}]] Out[2]= 1 1 -- Log[- (2741 + 1036 Sqrt[7])] 10 9 >This problem gets even stranger: > >Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 1}] = Infinity > >Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 1, 2}] = -Infinity > >Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 2}] > >?!?(Integrate::"idiv" : > "Integral of ?!?(1?/?(?(2?? x?) + ? at ?(1 + ?(3?? x?)?)?)?) does not ? >converge on ?!?({0, 2}?)."?) > >Looks serious. > >Kevin > > >-----Original Message----- >From: Richard Finley <rfinley at medicine.umsmed.edu> To: mathgroup at smc.vnet.net >To: mathgroup at smc.vnet.net >Subject: [mg16346] [mg16318] [mg16232] Re: [mg16172] integration problem > > >>Michel, >> >>I can only speculate as to what is going wrong. If you rationalize the >>denominator of your function you get: >> >>1/(2 x + Sqrt[ 3 x + 1] ) == 2 x/((x-1)(4 x + 1) - Sqrt[3 x + 1]/((x - >>1)(4 x + 1) or the difference of two functions which each have a >>singularity at x = 1 so one can no longer assume the integral of the >>difference is the difference of the integrals. I suspect that Mathematica >is >>doing this transformation prior to evaluating the integral?? It doesn't >>do it in every case because I have tried other similar examples which give >>the correct answer. Another example which gives the wrong answer in a >>similar situation is: >> >>In(1) = Integrate[1/(1+Sqrt[x+1]),{x,0,1}] >>Out(1) = -2 + Log[4] + 2( Sqrt[2] - Log[1 + Sqrt[2]) >>In(2) = %//N >>Out(2) = 0.451974 >>In(3) = Integrate[(-1 + Sqrt[x+1])/x , {x,0,1}] >>Out(3) = Sum::div : Sum does not converge. ..... >>etc, etc.... >>In(4) = NIntegrate[(-1 + Sqrt[x+1])/x , {x,0,1}] >>Out[4] = 0.451974 >> >>Perhaps someone from Wolfram can comment on the reasons for this and if it >>will be corrected in the next release?? >> >>regards, RF >> >>>>> Michel Gosse <michel.gosse at interpc.fr> 03/02/99 12:13AM >>> >>Hello >>Mathematica 3.01 returns infinity for the calculus : >>Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 1}] >>but when i evaluate : >>NIntegrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 1}] >>it returns 0.449, which seems good. >>What is the problem with the integrate function ? >>Regards >> >> >> > Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp/ http://eri2.tuins.ac.jp/