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Re: Integral test
Posted:
Dec 18, 2012 4:42 AM
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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 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.
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
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Oh - I guess I might be saying that google groups censors my posts.
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