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Let $S(x)$ be continuous, differentiable, and such that $S(x)=O(x/\log x)$. Let $J(x)=\int_x^{\infty} \frac{S(y)(1+\log y)}{y^2\log^2 y}dy$ and let $K(x)=\int_x^{\infty}\frac{S(y)}{y^2}dy$. Let $K(2)>J(2)>0$, and let $J(x)>0$ for $x>2$. Does it follow that $K(x)>0$ for $x>2$? If this is not the case, could someone supply such an $S$ for which this fails?

Context: Integrals like this are closely related to integrals that appear in the study of sum of prime reciprocals in Rosser and Schoenfeld’s well known paper on formulas for some functions of prime numbers. The positivity of an integral like $J$ is equivalent to the Riemann Hypothesis, and the positivity of an integral like $K$ is conjectured to be equivalent.

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  • $\begingroup$ What does it mean to say "The positivity of an integral like $J$ is equivalent to the Riemann Hypothesis"? Can we get a precise statement? Citation? $\endgroup$ – Stopple Apr 5 '19 at 23:12
  • $\begingroup$ @Stopple $J(x)>0$ for $x\geq 3$ iff the Riemann Hypothesis is true. $\endgroup$ – EGME Apr 6 '19 at 8:08
  • $\begingroup$ @Stopple More precisely and specifically, in $J$, $S$ is not just any function: in the known RH-equivalent statement, $S(x)$ is $x-\theta(x)$. There is no citation, this is unpublished work, and the authors have not posted it either. I haven’t asked why. You can actually replace $x-\theta(x)$ by a continuous differentiable function and obtain a similar integral which remains positive iff RH holds; hence my question. $\endgroup$ – EGME Apr 6 '19 at 8:58
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An integration by parts shows that $$ K(x) = \frac{\log^2 x}{1+\log x} J(x) + \int_x^{\infty} J(y) \frac{\log^2 y+2\log y}{y(1+\log y)^2}\, dy >0 . $$ (I'm also assuming here that $S(x)/x^2$ is integrable, but we need this anyway to make $K$ well defined.)

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  • $\begingroup$ Thank you ... It seems it should work the other way around too, if I am not mistaken, right? Namely, if you assume the positivity of $K$, then the positivity of $J$ should follow? I haven’t worked it out ... $\endgroup$ – EGME Apr 6 '19 at 14:17
  • $\begingroup$ Right, now I worked it out ... it’s correct right? Thanks $\endgroup$ – EGME Apr 6 '19 at 16:37
  • $\begingroup$ @EGME: The calculation should be correct, as far as I can see (write the integrand in the integral defining $K$ as $-J'(y)\log^2y/(1+\log y)$). Also, the converse will be false: if $S$ is negative somewhere, followed immediately by a positive shape that almost exactly cancels that negative part, then you can make $K>0, J<0$ (since the extra $1/\log y$ will tip the almost cancellation towards the negative side when computing $J$). $\endgroup$ – Christian Remling Apr 6 '19 at 17:04

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