```Date: Dec 11, 2012 1:44 AM
Author: Jose Carlos Santos
Subject: Re: Integral test

On 11/12/2012 00:57, Rotwang wrote:>> One of my students asked me today a question that I was unable to>> answer. Let _f_ be an analytical function from (0,+oo) into [1,+oo) and>> suppose that the integral of _f_ from 1 to +oo converges. Does it follow>> that the series sum_n f(n) converges? I don't think so, but I was unable>> to find a counter-example. Any ideas?>> (I'm assuming you mean [0, +oo) instead of [1, +oo).)>> Here's a proposal for constructing a counterexample; I don't know enough> analysis to be able to say whether the resulting function is analytic,> but it seems plausible.>> Let m(x) be an increasing analytic function from (0, oo) to [1, +oo),> such that m(x) > x for all x, and consider the function>> f(x) = (cos(2*pi*x)/2 + 1/2)^m(x + 1).>> Then f(n) = 1 for all n in N+. Note that the integral from n - 1/2 to n> + 1/2 of f(x) satisfies>>     int_{n - 1/2}^{n + 1/2} f(x) dx> <= int_{n - 1/2}^{n + 1/2} (cos(2*pi*x)/2 + 1/2)^m(n) dx>   = int_0^1 sin(pi*x)^{2*m(n)} dx> <= 1/2^(2n) + int_{1/2 - delta(n)}^{1/2 + delta(n)} dx>   = 1/2^(2n) + 2*delta(n)>> where delta(n) is chosen so that sin(pi/2 + pi*delta)^(2*m(n)) = 1/2.> Therefore, if we can choose m in such a way that delta(n) <= 1/2^n, for> example, then f will be integrable. To this end we can take>> m(x) = -log(2)/(2*log(cos(pi/2^{x + c})))>> where c > 0 is chosen so as to make m(x) sufficiently well-behaved for> small x.Thanks. I will check the details.Best regards,Jose Carlos Santos
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