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Re: A FullSimplify Problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg41093] Re: A FullSimplify Problem
  • From: "Dana DeLouis" <delouis at bellsouth.net>
  • Date: Fri, 2 May 2003 03:58:32 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Here is what I found so far.  If I used â?¦

Integrate[Log[1 - 2*a*Cos[x] + a^2], {x, 0, Pi}, 
  PrincipalValue -> True, Assumptions -> {{a, x} â?? Reals}]

I get an output based on the condition of 'a'.

If[1/a + a >= 2 || 2 + 1/a + a <= 0 || 1/a + a == 0 || 
   Im[1/a + a] != 0, Pi*(-Log[2] + Log[1 + a^2] + 
    Log[1 + Sqrt[1 - (4*a^2)/(1 + a^2)^2]]), 
  Integrate[Log[1 + a^2 - 2*a*Cos[x]], {x, 0, Pi}]]

If I ignore the Im part, the rest of the condition appears to be a complicated way of saying "a<>0".

InequalitySolve[1/a + a >= 2 || 2 + 1/a + a <= 0 || 
   1/a + a == 0, a]

a < 0 || a > 0


If I look at the equation derived from above, and try to simplify it, I get...

FullSimplify[Pi*(-Log[2] + Log[1 + a^2] + 
    Log[1 + Sqrt[1 - (4*a^2)/(1 + a^2)^2]]), {a â?? Reals}]

Pi*Log[(1/2)*(1 + a^2 + Abs[-1 + a^2])]

This equation behaves just like the equation you are expecting.
Apparently, Mathematica wants to use a version of a^2 instead of Abs[a].
I've tried to simplify this to your equation, but had no luck.  At least it had a Abs[ ] in it!  :>)

If I give it that a <-1, or  a > 1, then the two equations are very similar, and close to what we want...


FullSimplify[Pi*(-Log[2] + Log[1 + a^2] + 
    Log[1 + Sqrt[1 - (4*a^2)/(1 + a^2)^2]]), 
  {a â?? Reals, a < -1}]

Pi*Log[a^2]

Here, it would be better to use your version of  2*Pi*Log[Abs[a]] 

FullSimplify[Pi*(-Log[2] + Log[1 + a^2] + 
    Log[1 + Sqrt[1 - (4*a^2)/(1 + a^2)^2]]), 
  {a â?? Reals, a > 1}]

2*Pi*Log[a]

This is as close as I could get it to your version.

It looks like Mathematica is not doing a very good job here.


-- 
Dana DeLouis 
Windows XP
$VersionNumber -> 4.2
= = = = = = = = = = = = = = = = = 
 
 
"Ersek, Ted R" <ErsekTR at navair.navy.mil> wrote in message news:b8o20k$p28$1 at smc.vnet.net...
> 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
> 
>



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