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]

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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

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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]