Let $f(x)=x\cdot(\log x)^x$ for $x\geq 2$, then integrating $\log f(x)=\int_2^x f'(t)/f(t)dt$, it is easy to prove the first statement of following, and directly if we put $x=e^H_n$ and add $H_n$ the second ($e>2$)

Lemma. For $x\geq 2$, $$x\log \log x=2\log\log 2 +Li(x)+\int_2^x\log\log tdt$$ where $Li(x)$ is the logarithmic integral $\int_2^xdt/\log t$. And thus $$H_n+e^{H_n}\log (H_n)=H_n+2\log\log 2+Li(e^{H_n})+\int_2^{e^{H_n}}\log\log tdt$$ where $H_n$ is the nth harmonic number $1+1/2+\cdots +1/n$.

Now my question is

Question. Can you continue previous computations to prove a reasonable improve of $$\sigma(n)\leq \text{something}\leq \text{RHS of second statement in previous lemma},$$ where $\sigma(n)$ denote the sum of the positive divisors of $n$, $\sum_{d|n}d$.

Thanks is advance, my only goal is edit the best possible post in this Math Stack Exchange, I apologize to bring this question here because I believe that this community can made contributions. Sorry by my english.


My question as is built, tell us how take an ingredient from an equivalence of Riemann Hypothesis, $\sigma(n)$, and other ingredient, $Li(x)$, from another equivalence of RH and put these together, well, in a not understandable way. Topics as certain type of abundant numbers, Robin approach to RH, the order of growth of $\sigma(n)$ or $H_n$, and the relationship between $Li(x)$ and Riemann Hypothesis can find in the literature. In [4] (I have not an open source to provide it) we can read an integral equal to $\sigma(n)$.

Simply you can read the first paragraph of the section Initial Ideas, page 343 of [2] (given as free paper), and Lagarias equivalence in the last paragraph before the section Other Zeta and L-functions in page 347 to know two equivalences with Riemann Hypothesis.

Thus (via [3]) Riemann Hypothesis holds if and only if $$\sigma(n)\leq H_n+2\log\log 2+Li(e^{H_n})+\int_2^{e^{H_n}}\log\log t dt.$$


[1] Wikipedia.

[2] J. Brian Conrey, The Riemann Hypothesis, Notices of the AMS, MARCH 2003, Volume 50, Number 3 (www.ams.org/notices).

[3] Jeffrey C. Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis, The American Mathematical Monthly Vol. 109, No. 6 (Jun. -Jul., 2002), pp. 534-543.

[4] George Purdy, An Integral Equal to $\sigma(n)$, Problems and Solutions, Problem E 1850 [1966, 82] American Mathematical Monthly Vol. 74 N. 5 MAY 1967, p. 594-595.

  • $\begingroup$ I've made the change in the misprint and deleted the first comment, to avoid confusion, it is better . Thanks. $\endgroup$ – user243301 Sep 10 '15 at 6:45

$\int \limits_{2}^{x} \ln \ln t dt = t \ln \ln t-Li(t)|_{2}^{x} = x \ln \ln x-2 \ln \ln 2+Li(2)-Li(x)$

So $\int \limits_{2}^{x} \ln \ln t dt+ 2 \ln \ln 2+Li(x)+\ln x = \ln x+ Li(2)+x \ln \ln x $, put $x=e^{H_n}$ to get $H_n + Li(2) +e^{H_n} \ln H_n$

So we arrive at $ Li(2) + H_n +e^{H_n} \ln H_n \geq \sigma(n)$ without the $Li(2) \approx 1.04516 $ its the famous C.Lagarias inequality which is true for all $n\geq 1$ iff the R.H. is true.

So nothing is new under the sun, this is known from the 80's.

  • $\begingroup$ Many thanks for your contribution, I am going to study it. As I've said using [3] we get the last paragraph, I that I am asking is if we can do more analysis about it, deduce a more interesting claim than mine from such RHS, if that is possible. $\endgroup$ – user243301 Oct 13 '17 at 18:54

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