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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 11, 2012 12:47 PM
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On 11/12/2012 00:57, Rotwang wrote:
> 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.


Sorry, this didn't work on account of my stupidity. Here's a (hopefully)
corrected version:

Let m(x) be an increasing analytic function from (1, +oo) to [1, +oo),
and consider

f(x) = (cos(2*pi*x)/2 + 1/2)^m(x + 1)

so that f(n) = 1 for all n in N+. 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

Let delta(n) be such that sin(pi*(1/2 +/- delta))^{2*m(n)} =
cos(pi*delta)^{2*m(n)} = 1/2^n. The above integral can be split into the
part with 1/2 - delta < x < 1/2 + delta, where the integrand is <= 1,
and the remaining part where the integrand is <= 1/2^n. Hence

int_{n - 1/2}^{n + 1/2} f(x) dx
<= 1/2^n + 2*delta.

If we can choose m such that delta(n) <= 1/2^n, then f will be
integrable. Equality holds if m is given by

m(x) = -x*log(2)/(2*log(cos(pi/2^x))).

Since f is a nonincreasing function of m, any m larger than the above
will do; we can take

m(x) = -x*log(2)/(2*log(cos(pi/2^{x + c})))

where c > 0 is chosen to make m(x) analytic near x = 1.


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