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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/



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