Rotwang
Posts:
1,678
From:
Swansea
Registered:
7/26/06


Re: Integral test
Posted:
Dec 10, 2012 7:57 PM


On 10/12/2012 21:05, José Carlos Santos wrote: > Hi all, > > 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.
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