Re: Re: integration problem
- To: mathgroup at smc.vnet.net
- Subject: [mg16319] Re: [mg16264] Re: integration problem
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Sun, 7 Mar 1999 01:05:43 -0500
- Sender: owner-wri-mathgroup at wolfram.com
On Fri, Mar 5, 1999, James Escamilla <jescamilla at ti.com> wrote: > Michel Gosse wrote: >> >> 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 > >My calculator (a TI-89) returns LN(256/27)/5 for >Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 1}]. > >It also solves Integrate[1/(2*x + Sqrt[3*x + 1]), x] as: >-LN((ABS(2*SQRT(3*X+1)-1))^3/(243*(ABS(SQRT(3*X+1)+2))^3* >(ABS(SQRT(3*X+1)+2*X))^5))/10 > >If Mathematica can't solve this and my calculator can, >I'd say it's a BIG BUG in Mathematica. Mathematica can actually compute this integral, and even gets the right answer, provided the package <<Calculus`Limit` is loaded. Without the package we get: In[1]:= Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 1}] Out[1]= ? But after loading the package: In[2]:= <<Calculus`Limit` In[3]:= FullSimplify[Integrate[1/(2*x + Sqrt[3*x + 1]), {x, 0, 1}]] Out[3]= 2 3/5 Log[2 (-) ] 3 Note that this is exactly the answer your calculator got! The reason for this is that mathematica can actually find the indefinte integral In[7]:= Integrate[1/(2*x+Sqrt[3*x+1]),x] Out[7]= 4 1 1 - ArcTanh[- Sqrt[1 + 3 x]] - - ArcTanh[2 Sqrt[1 + 3 x]] + 5 2 5 2 1 - Log[-1 + x] + -- Log[1 + 4 x] 5 10 However, unless the Limit package is loaded Mathematica can't calsulate the limit of this last expre4ssion as x->1. It is usually a good idea to load in the Limit package in such type of situation. Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp/ http://eri2.tuins.ac.jp/