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
>
>