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

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

The second output (webMathematica) gave

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

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

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

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

Out=
1/(-1+x)

Out=
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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