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MathGroup Archive 2006

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Re: Integral problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65665] Re: [mg65634] Integral problem
  • From: "Carl K. Woll" <carlw at wolfram.com>
  • Date: Wed, 12 Apr 2006 06:00:26 -0400 (EDT)
  • References: <200604110804.EAA11299@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

ivan.svaljek at gmail.com wrote:
> I've tried running an integral expression through mathematica 5.0 and
> came up with this:
> 
> http://aspspider.net/isvaljek/mathematica/index.html
> 
> First two are from mathematica, and the last one is from a site running
> web mathematica.

The first set of input/output was:

In[1]:=
Integrate[x^3/(x^8-2),x]

Out[1]=
(Log[Sqrt[2] - x^4] - Log[Sqrt[2] + x^4])/(8*Sqrt[2])

The second output (webMathematica) gave

(Log[x^4 - Sqrt[2]] - Log[Sqrt[2] + x^4])/(8*Sqrt[2])

I guess the question is why the argument of the first log changes sign? 
Note that (up to branch issues)

Log[Sqrt[2]-x^4] == Log[(-1)(x^4-Sqrt[2])] == Log[-1]+Log[x^4-Sqrt[2]]

So, the difference between the two expressions is a constant, albeit a 
complex constant (Log[-1]==Pi I). Since an indefinite integral is only 
defined up to a constant, both answers are correct.

A simpler manifestation of this issue:

In[2]:=
D[Log[x-1],x]//Simplify
D[Log[1-x],x]//Simplify

Out[2]=
1/(-1+x)

Out[3]=
1/(-1+x)

Both Log[x-1] and Log[1-x] have a derivative equal to 1/(x-1), so they 
must both be indefinite integrals of 1/(x-1).

Carl Woll
Wolfram Research


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