Re: A FullSimplify Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg41090] Re: A FullSimplify Problem
- From: bobhanlon at aol.com (Bob Hanlon)
- Date: Fri, 2 May 2003 03:58:10 -0400 (EDT)
- References: <b8o20k$p28$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
You probably need to help it...
e1=Integrate[Log[1-2a Cos[x]+a^2],{x,0,Pi}]//FullSimplify
e2 = FullSimplify[e1 //. {a_.*Log[x_]+a_.*Log[y_]+z___:>a*Log[x*y]+z,
a_.*Log[x_]-a_.*Log[y_]+z___:>a*Log[x/y]+z}]
Since
Log[-x]^2-Log[x]^2 ==
(Log[-x]-Log[x])(Log[-x]+Log[x])==
I*Pi*(I*Pi+2Log[x])==
-Pi^2+2*I*Pi*Log[x];
Then
e3 = Simplify[e2 /. Power[Log[Times[-1,Power[a,-1]]],2] :>
-Pi^2+2*I*Pi*Log[1/a]+Log[1/a]^2]
e4 = Simplify[
e3 /. {Log[Times[a,Power[Plus[1,a],-1]]] :>
Log[a]-Log[1+a],
Log[Plus[1,Power[a,-1]]] :>
Log[1+a]-Log[a]}]
e5 = e4 /. Log[x_]-Log[y_]:>Log[x/y]
Pi*Log[a^2]
Which is clearly equivalent to the requested form
Bob Hanlon
In article <b8o20k$p28$1 at smc.vnet.net>, "Ersek, Ted R"
<ErsekTR at navair.navy.mil> wrote:
<<
Subject: A FullSimplify Problem
From: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
To: mathgroup at smc.vnet.net
Date: Wed, 30 Apr 2003 08:37:40 +0000 (UTC)
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
>><BR><BR>