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Say we have an integer $n$∊ℕ₀ & a sequence of $n+1$ real numbers $\alpha_k\in[0,\infty)∀k$, where $k=0\dots n$, and using $\log^{[k]}$ to denote $k$ functionings of the logarithm ($\log^{[0]}x\equiv x$, $\log^{[1]}x\equiv \log x$, $\log^{[2]}x\equiv \log\log x$, etc), I would conjecture that, $∀n, $ the integral $$\int_{e\uparrow\uparrow n}^\infty{dx\over\prod_{k=0}^n(\log^{[k]}x)^{\alpha_k}}$$ diverges, when $\alpha_k=1∀k\leq n$, or when with ascending $k$ the first $\alpha_k≠1$ is $<1$; and converges when with ascending $k$ the first $\alpha_k≠1$ is $>1$.

In the integral given the lower limit is chosen simply to keep the function in the denominator well clear of taking any argument that would result in a negative value being fed into the logarithm - the convergence|divergence of the integral is determined purely by the behaviour of the integrand as its argument $\to\infty$.

I am wondering whether this surmise is correct. My reasoning for supposing it is is that if the variable $y$ be substituted for $\log^{[n]}x$, then in the denominator we shall have successive orders of functioning of the exponential of $y$ from right to left ... but each raised to the power of its index $\alpha_k$ in order from left to right; and in the numerator we shall have the same product of the same factors, by reason of the chain rule, but each with unit exponent. So that considering the factors from left to right, the first one that does not completely cancel will be the first one at which $\alpha_k$ differs from unity; and also the one with the highest order of application of the exponential function: and if that $a_k$ is $<1$ the remnant will be in the numerator, and if $>1$, in the denominator. And the integral will diverge in the former case & converge in the latter, as subsequent remnants will be completely overruled, regardless of the size of their exponent, as an exponential of a variable always overrules a mere power of a variable, regardless of the relative sizes of the scaling of the exponential and the degree of the power ... and the comparison will be at least that. Finally, in the case of all the $\alpha_k$ till the last being $=1$, there will be complete cancellation of the exponentials; and we shall be left with $$\int_{e\uparrow\uparrow n}^\infty{dy\over y^{\alpha_n}} ,$$ the convergence|divergence of which depends on $\alpha_n$ in the well-familiar way.

I would also surmise that this theorem - if it indeed is one (and the question here is essentially whether it is one, and not merely a surmise, or incorrectly infererred) - translates into sum over integers. I'll refrain from fully explicating the logic of that surmise; but basically it's that if the correspondence between Σ & ∫ of $1/x$ holds by reason of the asymptotically-flat -ness of the logarithm, then it could reasonably be expected to hold when functions that are progressively yet asymptotically-flatter are factored-in.

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  • $\begingroup$ For clarification, is the theorem just showing how this integral ($n$ fixed) switches from divergent to convergent as one of the exponents changes -- or are you asserting that there is such a thing as an integral on "the very cusp of convergence"? $\endgroup$
    – RRL
    Dec 17, 2018 at 19:24
  • $\begingroup$ @RRL -- The thought behind it is ... we know that 1/x is divergent, but if we have 1/x^(1+ε) it becomes convergent, howsoever small ε might be ... so what else can we freight the denominator with & it still be divergent ... let's try a logarithm, as it increases slower than any power of x ... yes! it's still divergent ... but if we put a further (log)^ε on, no matter how small ε, it becomes convergent ... but could there possibly be just a tiny little bit more thst we could freight the denominator with and it still be divergent ... I know! let's try loglog(x) ... $\endgroup$ Dec 17, 2018 at 19:43
  • $\begingroup$ @RRL -- yes! it's still divergent! But if we put on even so much as a (loglog(x))^ε, no matter how small ε .... & so on & so on ... ¶ The idea came from noting how the sum of reciprocals is divergent; and then that the sum of reciprocals of primes only is still divergent and that's asymptotically equivalent to summing 1/x.logx, as primes of size x are on average logx apart ... but then! ... you get Brun's constant - the sum of twin primes, which is convergent & that is asymptotically sum of 1/x(logx)^2 ... so all this got me wondering just where exactly the cusp is tween con & div. $\endgroup$ Dec 17, 2018 at 19:54
  • $\begingroup$ @RRL -- SO I'm saying it looks to me like the cusp is at $$1/\prod_{k=0}^\infty\log^{[k]}(x) ;$$ and that if the exponent of any of those factors depart by the tiniest amount from unity it absolutely settles it one way or the other, utterly regardless of what any subsequent exponent might be. And I haven't found any explicit statement of what I have said here in any text or paper ... so I'm 'farming it out' here for confirmation ... or refutal, as the case may be! $\endgroup$ Dec 17, 2018 at 20:03
  • $\begingroup$ I think everything you say and prove here about the nature of this integral is correct (+1). My only concern is the meaning of "very cusp of convergence." $\endgroup$
    – RRL
    Dec 17, 2018 at 20:15

1 Answer 1

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My interpretation of on "the very cusp of convergence" is a function with a divergent integral such that any asymptotically smaller function has a convergent integral.

To argue there can be no integral on the very cusp of convergence in this sense, consider a positive function $f : [1,\infty) \to \mathbb{R}^+$ where $\int_1^\infty f(x) \, dx = +\infty$.

Taking $g(x) = \int_1^x f(t) \, dt$, we have for $\beta > \alpha > 1$,

$$\left|\int_{\alpha}^\beta \frac{f(x)}{g(x)} \, dx \right| \geqslant \frac{1}{g(\beta)}\int_{\alpha}^\beta f(x) \, dx = \frac{g(\beta)- g(\alpha)}{g(\beta)} = 1 - \frac{g(\alpha)}{g(\beta)}$$

Since $g(\alpha)/g(\beta) \to 0$ as $\beta \to +\infty$ with $\alpha$ fixed, there exists for any $\alpha$, no matter how large, a $\beta $ such that the RHS is greater than $1/2$. Since the Cauchy criterion is violated and the integrand is positive, we have

$$\int_{1}^\infty \frac{f(x)}{g(x)} \, dx = + \infty$$

However, since $g(x) \to +\infty$ as $x \to +\infty$ we have $f(x) \geqslant \frac{f(x)}{g(x)}$ for all sufficiently large $x$.

Hence, given a divergent integral there always is another divergent integral with an asymptotically subordinate integrand.

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  • $\begingroup$ Are you saying it's sort of like an open set ... similar to the way you can't exhibit the least element in (0,1]? $\endgroup$ Dec 17, 2018 at 20:22
  • $\begingroup$ Yes -- this notion of "no slowest" divergent series and improper integral is discussed in various places -- for example, here. $\endgroup$
    – RRL
    Dec 17, 2018 at 20:29
  • $\begingroup$ So no matter how close a function might be to this 'cusp', as I call, it, that function divided by its own integral will still be divergent!? $\endgroup$ Dec 17, 2018 at 20:30
  • $\begingroup$ Well! It's certainly extremely interesting this business of there being no 'least divergent' sum or integral! Looks like refutal afterall then! I am surprised, TBPH ... and I can't imagine what in the world increasing-to-∞ function you could put in the denominator of my function & it still be divergent! Well ... as you've pointed out - it's own integral! But ... isn't that kind of what it is, sort of: the limit to infinity of doing precisely that? $\endgroup$ Dec 17, 2018 at 20:39
  • $\begingroup$ I'm actually genuinely grateful for this contribution & very much appreciate it. One ought not to become too fond of one's own results afterall; unless perhaps they be ironcladly proven! ¶ In your first paragraph I think you've put 'convergent' where you mean di -vergent. It doesn't throw me atall, as I get what you're saying ... but someone else might be abitt. $\endgroup$ Dec 17, 2018 at 20:42

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