       Re: Re: Integrate question

```  It represents the same mathematica object, so you will get the same

Integrate[1/6/u^3, {u, -1, 2}]
Integrate::idiv:Integral of 1/u3 does not converge on {-1,2}. >>
Integrate[1/(6*u^3), {u, -1, 2}]

What else did you expect? It's the same and therefore 1/16 it is
equally "unacceptable" or "non-acceptable" as an answer as before.
The integral does not converge in any usual mathematical sense : as
Riemann or Lebesgue integral or even in the sense of "Principal
value". However, that does not mean that 1/16 is not "acceptable"
answer for any purpose anyone can imagine. It just does not represent
any of the quantities people normally use integrals for, such as an
area, or an average value etc.  For example, you consider the
divergent series  1-1+1-1+1 -...., then there is a context in which
the answer 1/2 is an or rather the "acceptable" value of this sum,
but in most usual contexts this is a divergent series and thus its
sum has no value. There are some situations where taking 1/2 as the
Even something like  1+1 = 5 is acceptable, if you are doing
arithmetic modulo 3. All mathematics makes sense (or does not as the
case may be) in some precisely defined context. Without a context you
are just playing with meaningless symbols.

Andrzej Kozlowski

On 19 Oct 2007, at 18:09, Oskar Itzinger wrote:

> Would setting
>
> u=3 x^2 - 1
>
> and integrating
>
> (1/6)/u^3 on [-1,2]
>
> be acceptable?
>
> /oskar
>
> "Oskar Itzinger" <oskar at opec.org> wrote in message
> news:ff77b9\$ncr\$1 at smc.vnet.net...
>> Hmm, from Mathematica 2.1 Help:
>>
>> Integrate can evaluate definite integrals whenever the correct
>> result can
> be
>> found by taking limits
>> of the indefinite form at the endpoints.
>>
>> ?
>>
>> /oskar
>>
>> "Andrzej Kozlowski" <akoz at mimuw.edu.pl> wrote in message
>> news:ff4hpp\$k0e\$1 at smc.vnet.net...
>>>
>>> On 16 Oct 2007, at 16:28, Oskar Itzinger wrote:
>>>
>>>> Mathematica 5.2 under IRIX complains that
>>>>
>>>> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
>>>>
>>>> doesn't converge on [0,1].
>>>>
>>>> However, Mathematica 2.1 under Windows gives the corrrect answer,
>>>> (1/16).
>>>>
>>>> When did Mathematica lose the ability to do said integral?
>>>>
>>>> Thanks.
>>>>
>>>>
>>>>
>>>
>>>
>>> The reason is that Mathematica 2.1 was wrong and Mathematica 5.2 is
>>> much more careful and right. What Mathematica 2.1 did here was
>>> simply:
>>>
>>> Subtract @@ (Integrate[x/(3 x^2 - 1)^3, x] /. {{x -> 1}, {x -> 0}})
>>>   1/16
>>>
>>> in other words, it applied the Newton-Leibnitz rule in a mindless
>>> way. Later versions are more intelligent and see that the
>>> singularity at
>>>   =CE=B1 = Last[x /. Solve[3*x^2 - 1 == 0, x]]
>>>   1/Sqrt
>>>
>>> One can also see this graphically (of course!):
>>>
>>> Plot[x/(3 x^2 - 1)^3, {x, 0, 1}]
>>>
>>>
>>> the integral still might exist in the sense of Cauchy PrincipalValue
>>> but we see that it does not:
>>>
>>> Integrate[x/(3*x^2 - 1)^3, {x, 0, 1},  PrincipalValue -> True]
>>> Integrate::idiv:Integral of x/(3 x^2-1)3 does not converge on
>>> {0,1}. >>
>>> Integrate[x/(3*x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
>>>
>>> If you still don't beleive it, you can do it "by hand":
>>>
>>> int = FullSimplify[Integrate[x/(3*x^2 - 1)^3,
>>>           {x, 0, 1/Sqrt - =CE=B5}] +
>>>         Integrate[x/(3*x^2 - 1)^3,
>>>           {x, 0, 1/Sqrt + =CE=B5}], =CE=B5 > 0]
>>> (9*=CE=B5^2*(3*=CE=B5^4 - 8*=CE=B5^2 + 5) - 4)/
>>>     (18*(3*=CE=B5^3 - 4*=CE=B5)^2)
>>>
>>> Limit[int, =CE=B5 -> 0]
>>> -=E2=88=9E
>>>
>>>
>>> Andrzej Kozlowski
>>>
>>>
>>>
>>>
>>
>>
>>
>
>
>

```

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