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Topic: Integral test
Replies: 17   Last Post: Dec 20, 2012 12:29 PM

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Rotwang

Posts: 1,678
From: Swansea
Registered: 7/26/06
Re: Integral test
Posted: Dec 10, 2012 7:57 PM
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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 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.

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