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MathGroup Archive 2002

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Re: Simple integral problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32677] Re: Simple integral problem
  • From: "Dave Snead" <dsnead6 at charter.net>
  • Date: Wed, 6 Feb 2002 03:41:30 -0500 (EST)
  • References: <a3f0id$b2r$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

It looks like a bug since the antiderivative of
x(1 - x)Log[1 + 1/x]
is (for constant c)
(5*x)/6 - x^2/6 - (x^2*(-3 + 2*x)*Log[1 + 1/x])/6 -  (5*Log[1 + x])/6 + c
which evaluates to
c + 2/3 - (2*Log[2])/3 when x is 1
and to
c as x approaches 0.
So the value 2/3 - (2*Log[2])/3  (or 0.204569...)  looks correct.

--Dave Snead

Kresimir Kumericki <kkumer at phy.hr> wrote in message
news:a3f0id$b2r$1 at smc.vnet.net...
> I tried this simple-looking integral in Mathematica 4.0 for Unix:
>
>  Integrate[x(1 - x)Log[1 + 1/x], {x, 0, 1}]
>
> and I got
>
>  -37/36 -2 Log[2]/3      ( = -1.48988 )
>
> which seems to be wrong because:
>
>  NIntegrate[x*(1 - x)*Log[1 + 1/x], {x, 0, 1}]  = 0.204569
>
> If perform undefinite Integral[ ... , x]  and than do Limit[..]s,
> I get the right result.
> If I put together the argument of Log: 1+1/x -> (x+1)/x,
> I again get the right result: 2/3 -2 Log[2]/3.
>
> Is this some bug or am I missing something? I have to do a lot of similar
> integrals these days so I would be glad if someone could shed some light
on
> this. Thanks,
>
> --
> Kresimir Kumericki    kkumer at phy.hr    http://www.phy.hr/~kkumer/
> "Fizika svemira" - http://eskola.hfd.hr/fizika_svemira/svemir.html
>



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