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Re: Re: Re: Integral confusion

  • To: mathgroup at smc.vnet.net
  • Subject: [mg107338] Re: [mg107316] Re: [mg107297] Re: Integral confusion
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 9 Feb 2010 07:59:21 -0500 (EST)
  • References: <hkm7d8$os0$1@smc.vnet.net> <201002080836.DAA29853@smc.vnet.net> <op.u7tlcde4tgfoz2@bobbys-imac.local> <201002090744.CAA10501@smc.vnet.net>

The general reason for that is that Mathematica does not compute 
indefinite integrals as you would do by hand but by means of a general 
algorithm, know as the Risch algorithm (more exactly, using various 
extensions of the Risch algorithm developed since Risch's original work 
in 1970).
This algorithm is purely algebraic and works over the complex numbers, 
in other words, you cannot expect to obtain "real looking" answers from 
real integrals (but they will be "real" up to a possible complex 
constant).
This is why you can't expect answers returned by Mathematica to be the 
same as you obtain by hand in general. Some computer algebra programs, 
when given a function to integrate, first attempt a few heuristic 
methods to try to get an answer of the kind that a user would be likely 
to get by hand and if these methods do not work switch to the Risch 
algorithm. Other programs do not bother with the heuristics and they go 
straight into the Risch algorithm. I am not sure what Mathematica does 
but it seems to me it is closer to the second type.

Also, there are many cases when the Risch algorithm can find the 
integral of a function in terms of elementary functions but Mathematica 
returns an answer in terms of special functions. Such answers often can 
be obtained much more quickly and can have other advantages, but again 
they will look rather different from the answers obtained by hand (which 
usually do not involve special functions).

In general it is unrealistic to expect computer programs, which work by 
means of general algorithms, to return the same answers as human beings 
obtain by hand using various heuristic tricks.

Andrzej Kozlowski



On 9 Feb 2010, at 08:44, Nasser M. Abbasi wrote:

> The "weird" thing is not what specific constant of integration it returned,
> but the fact that it did return one at all.
>
> Integrate[x^2, x] gives x^3/2. not x^3/2+Pi*Sqrt[-1]  or x^3/2+99 even
> though both are correct.
>
>> I don't know Integrate's steps in obtaining the result, but I'm sure it
>> didn't add extraneous constants deliberately.
>
> Ofcourse. I am sure of that too. But the final result is NOT what one would
> normally see in a textbook or expect or do when solving by hand, even though
> as I said, it is correct. It is just not "normal" looking.
>
> I am sorry, I did not mean to say anything bad about Mathematica when I said
> this result was "weird" even though it is correct.
>
> --Nasser
>
> ----- Original Message -----
> From: "DrMajorBob" <btreat1 at austin.rr.com>
> To: "Nasser M. Abbasi" <nma at 12000.org>; <mathgroup at smc.vnet.net>
> Sent: Monday, February 08, 2010 9:54 AM
> Subject: [mg107316] Re: [mg107297] Re: Integral confusion
>
>
>>> But Mathematica does (normally?) return results for indefinite integrals
>>> without an explicit constant of integration.
>>
>> What's "explicit" or weird about the constants of integration in this
>> problem?
>>
>> I don't know Integrate's steps in obtaining the result, but I'm sure it
>> didn't add extraneous constants deliberately.
>>
>> Bobby
>>
>> On Mon, 08 Feb 2010 02:36:16 -0600, Nasser M. Abbasi <nma at 12000.org>
>> wrote:
>>
>>>
>>> "Jon Joseph" <josco.jon at gmail.com> wrote in message
>>> news:hkm7d8$os0$1 at smc.vnet.net...
>>>> All: Is this integral wrong? If not could someone explain the minus sign
>>>> inside the log?
>>>>
>>>> Integrate[1/(x + 1) - 1/(x + 6), x] // Simplify
>>>>
>>>> log(-2 (x + 1)) - log(2 (x + 6))
>>>>
>>>> Thanks, Jon.
>>>>
>>>
>>> Well, lets see:
>>>
>>> log(-2 (x + 1)) - log(2 (x + 6))
>>>          = log(-2)+log(1+x) -log(2)-log(x+6)
>>>          = log(-1)+log(2)+log(1+x)-log(2)-log(x+6)
>>>          = log(-1)+ log(1+x) - log(x+6)
>>>
>>> but log(-1) = Sqrt[-1]*Pi
>>>
>>> so result is
>>>
>>>         Sqrt[-1]*Pi + log(1+x) - log(x+6)
>>>
>>> But we all know that the result should be
>>>
>>>         log(1+x) - log(x+6)
>>>
>>> So, an extra term, Sqrt[-1]*Pi term pops up. But this term is a
>>> constant, so
>>> its derivative is zero, i.e. a constant of integration.
>>>
>>> Since
>>>
>>> D[log(-2 (x + 1)) - log(2 (x + 6)),x] will give back
>>>
>>>               1/(x + 1) - 1/(x + 6)
>>>
>>> So, in theory, the answer given by Mathematica is NOT wrong.
>>>
>>> But Mathematica does (normally?) return results for indefinite =
integrals
>>> without an explicit constant of integration. So I am not sure why it=

>>> does in
>>> this case, and if it it does, why did not pick this constant? Why =
not
>>> C[1]
>>> as it does for DSolve[]?
>>>
>>> So, if I have to guess, I'd say this result is at least very weired, =
but
>>> mathematically it is not wrong?
>>>
>>> --Nasser
>>>
>>>
>>>
>>
>>
>> --
>> DrMajorBob at yahoo.com
>
>



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