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 > >