From: [Permission pending] Date: Wed, 14 Dec 1994 16:20:34 -0500 To: rusin@washington.math.niu.edu Subject: Re: a very simple asymptotics question Newsgroups: sci.math Hi, Thanks for your comments. As to the function you were alluding to, it was pointed out to me that the integral of 1/(x (log x)^2) converges at 0. Indeed, its antiderivative is 1/log x --- actually the negative of that. [sig deleted - djr] ============================================================================== Date: Wed, 14 Dec 94 15:48:52 CST From: rusin (Dave Rusin) To: [Permission pending] Subject: Re: a very simple asymptotics question Actually the function you quoted is _not_ sufficient, as you can't evaluate the antiderivative at x=0. But (as I realized after posting) you can get away with a minor variant: f(x)=1/[x (log x - 1)^2 ]. dave ============================================================================== Date: Wed, 14 Dec 94 17:32:30 -0500 From: [Permission pending] To: rusin@math.niu.edu Subject: Re: a very simple asymptotics question Yes, you're right. Actually, I saw your follow-up posts after I responded to you, and thus was unaware at the time of your corrections. Thanks a bundle, [Permission pending]