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 >