Re: A wrong definite integral in 5.0?

*To*: mathgroup at smc.vnet.net*Subject*: [mg77250] Re: A wrong definite integral in 5.0?*From*: dimitris <dimmechan at yahoo.com>*Date*: Wed, 6 Jun 2007 06:56:15 -0400 (EDT)*References*: <f43hau$301$1@smc.vnet.net>

Hello. $VersionNumber 5.2 In[214]:= Integrate[Log[1 - 4*x*(1 - x)], {x, 0, 1}] Out[214]= -2 correct as you indicated. But let see if your version 5 can give the correct result. To this end I want you to try In[37]:= f[x_] = Log[1 - 4*x*(1 - x)] F[x_] = Integrate[f[x], x] (Limit[F[x], x -> 1, Direction -> 1] - Limit[F[x], x -> 1/2, Direction -> -1]) + (Limit[F[x], x -> 1/2, Direction -> 1] - Limit[F[x], x -> 0, Direction -> -1]) If the result is -2 then we finish. If not please send to me the result of Integrate[Log[1 - 4*x*(1 - x)],x] in your version! Dimitris / bolud-el-kotur : > I get this result in version 5.0, > > >Integrate[Log[1 - 4 x(1 - x)], {x, 0, 1}] > >-2 + I Pi > > and the same thing if I "declare" the singularity with {x,0,1/2,1}. > > Another way to look at the problem is computing, > > >Integrate[Log[1 - 4 x(1 - x)], {x, 0, 1/2}] > >-1 > > and > > >Integrate[Log[1 - 4 x(1 - x)], {x, 1/2, 1}] > >-1 + I Pi > > Since the integrand is symmetric about x=1/2, the result should have > been the same one (-1) in both cases, and the integral over [0,1] > should yield -2. > > A numerical approach, > > >NIntegrate[Log[1 - 4 x(1 - x)], {x, 0, 1/2, 1}, > MaxRecursion -> 100, SingularityDepth -> 20] > >-1.9999997086422834` > > gives the correct result, within the numerical accuracy required.