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MathGroup Archive 2007

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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.



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