
Re: Integral test
Posted:
Dec 18, 2012 4:42 AM


José Carlos Santos <jcsantos@fc.up.pt> writes: > 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 counterexample. 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 wellbehaved for > > small x. > > Thanks. I will check the details.
Well, it's not into, for a start.
Phil  I'm not saying that google groups censors my posts, but there's a strong link between me saying "google groups sucks" in articles, and them disappearing.
Oh  I guess I might be saying that google groups censors my posts.

