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/