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