Ytukyg

Say we have an integer $n$∊ℕ₀ & a sequence of $n+1$ real numbers $alpha_kin[0,infty)∀k$, where $k=0dots n$, and using $log^{[k]}$ to denote $k$ functionings of the logarithm ($log^{[0]}xequiv x$, $log^{[1]}xequiv log x$, $log^{[2]}xequiv loglog x$, etc), I would conjecture that, $∀n, $ the integral $$int_{euparrowuparrow n}^infty{dxoverprod_{k=0}^n(log^{[k]}x)^{alpha_k}}$$ diverges, when $alpha_k=1∀kleq n$, 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 $toinfty$.

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 $$int_{euparrowuparrow n}^infty{dyover y^{alpha_n}} ,$$ 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.

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.