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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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