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MathGroup Archive 2002

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Re: Simple integral problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32660] Re: [mg32650] Simple integral problem
  • From: BobHanlon at aol.com
  • Date: Sat, 2 Feb 2002 01:19:36 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 2/1/02 5:40:01 PM, kkumer at phy.hr writes:

>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.

I find that Integrate is more reliable if the integrand is 
simplified as much as practical prior to integration.

Integrate[Expand[
    x(1-x)Log[1+1/x]],{x,0,1}]

(-(2/3))*(-1 + Log[2])

%==Integrate[FunctionExpand[
      x(1-x)Log[1+1/x]],{x,0,1}]

True

and, as you noted,

%%==Integrate[x(1-x)Log[
        Together[1+1/x]],{x,0,1}]

True


Bob Hanlon
Chantilly, VA  USA


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