|
[Date Index]
[Thread Index]
[Author Index]
Re: Integration error? Integrate[1/(x^3-1)]?
- To: mathgroup at smc.vnet.net
- Subject: [mg121716] Re: Integration error? Integrate[1/(x^3-1)]?
- From: Heike Gramberg <heike.gramberg at gmail.com>
- Date: Mon, 26 Sep 2011 20:07:18 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201109260816.EAA08704@smc.vnet.net>
An indefinite integral is unique up to an additive constant. Since
Log[1-x] is equal to Log[x-1]
plus or minus Pi I (try for example Log[1-x] /. x->4 which should give
Pi I + Log[3]) the two
answers are equally valid.
Heike.
On 26 Sep 2011, at 10:16, Travis Ayres wrote:
> The indefinite integral of 1/(x^3-1) with respect to x.
> In input form, put into Mathematica 8:
> Integrate[1/(x^3 - 1), x]
> Gives result:
> -(ArcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3]) + (1/3)*Log[1 - x] - (1/6)*
> Log[1 + x + x^2]
>
> Computing online with the Wolfram integrator:
>
http://integrals.wolfram.com/index.jsp?expr=1%2F%28x^3-1%29&random=false
> Gives:
> -(ArcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3]) + Log[-1 + x]/3 - Log[1 + x +
> x^2]/6
>
> Look at the (1/3)*Log[1-x] term.
> Mathematica 8 gives me Log[1-x], the online integrator gives the
> answer Log[x-1].
> The answers are exactly the same in all other terms. I ran across this
> because I was trying the tutorials, and I noticed my answer was
> different than the result in the tutorial even.
>
> Is this an error?
> Thanks all!
>
Prev by Date:
Re: More FinancialData glitches
Next by Date:
Constrain locator
Previous by thread:
Integration error? Integrate[1/(x^3-1)]?
Next by thread:
Re: Integration error? Integrate[1/(x^3-1)]?
|