Re: A FullSimplify Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg41087] Re: [mg41050] A FullSimplify Problem
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Thu, 1 May 2003 05:00:49 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
This seems unlikely since Mathematica 4.2 returns
FullSimplify[Integrate[Log[1-2a Cos[x]+a^2],{x,0,Pi}],{a>1}]
4 ¥ð Log[a]
which is double the right answer. With the assumption a<-1 the answer
is even worse (if returning a complex number when a real one is
expected is "worse" than returning twice the right answer):
FullSimplify[Integrate[Log[1 - 2*a*Cos[x] + a^2],
{x, 0, Pi}], {a < -1}]
Out[7]=
Pi*(2*I*Pi + Log[1/a^2])
Of course this does not prove that the unsimplified answer returned by
Mathematica is wrong, in fact numerical tests suggest that it is
probably right and the problem may be with FullSimplify. But I would
not count on it.
Andrzej Kozlowski
Yokohama, Japan
http://www.mimuw.edu.pl/~akoz/
http://platon.c.u-tokyo.ac.jp/andrzej/
On Wednesday, April 30, 2003, at 05:22 pm, Ersek, Ted R wrote:
> At http://mathworld.wolfram.com/LeibnizIntegralRule.html
> I learned that
> Integrate[Log[1-2a Cos[x]+a^2],{x,0,Pi}]
> = 2*Pi*Log[Abs[a]]
>
> Mathematica knows how to do this integral, but gives a much more
> complicated
> result. Can anyone explain how to use FullSimplify and other
> transformations to show that the complicated result Mathematica gives
> is
> equivalent to the answer above?
>
> Thanks,
> Ted Ersek
>
>
>
>