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Today I found that in 1914, Littlewood proved that

(1) there are arbitrarily large values of $x$ for which

$$\pi(x)<li(x)-\frac{1}{3}\frac{\sqrt{x}}{\log(x)}\log(\log(\log(x)))$$

  • First: Is this true?

  • Second: Is (1) equivalent to:

(2) There are infinitely many $x$ such that

$$\pi(x)<li(x)-\frac{1}{3}\frac{\sqrt{x}}{\log(x)}\log(\log(\log(x)))$$?

Thanks!

Community
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Andrea
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    No, (2) does not necessarily imply (1). The infinitely many $x$ could lie in a bounded interval. – Dietrich Burde Feb 19 '15 at 14:11
  • But does "arbitrarily large $x$" mean that there's not a finite number of $x$ such that inequality holds? – Andrea Feb 19 '15 at 14:15
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    The point is that 1 implies 2, but the converse is not true. – quid Feb 19 '15 at 14:15
  • What's the difference between "arbitrarily large values" and "infinitely many values"? thanks – Andrea Feb 19 '15 at 14:17
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    $1/n$ takes infinitely many values in $[0,1]$, but not arbitrarily large values (for $n\in \mathbb{N}$). – Dietrich Burde Feb 19 '15 at 14:18
  • Ok, last question: are there infinitely arbitrarily large values of $x$? is this true? – Andrea Feb 19 '15 at 14:21
  • This is a bit tautological, and one would not say it like this, but it seems not wrong either. – quid Feb 19 '15 at 14:24
  • oh :P so what's a better statement for what I said before? or just saying "arbitrarily large values" implies that they are infinite? if yes, now I get your first comment and why that looks a bit tautological. – Andrea Feb 19 '15 at 14:33
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    Yes, arbitrarily large always implies (implicitly) that there are infinitely many. For each bound $M$ you get one $x_M$ greater than $M$ but then you get another $x_M'$ grreater than $x_M$ and so on; showing that in fact you have infinitely many larger than $M$. – quid Feb 19 '15 at 14:57

3 Answers3

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Concerning lower and upper bounds of $\pi(x)-li(x)$ Saouter and Demichel have shown in $2010$ that $$ \frac{-0.2x}{\log^3 x}-\frac{12x}{\log^4 x}-C_1-C_2\le \pi(x)-li(x)\le \frac{0.51x}{\log^3 x}-C_1 \quad \forall \; x\ge 355991, $$ with $$ C_1=li(2)-\frac{2}{\log 2}\left(1+\frac{1}{\log 2}+\frac{2}{\log^2 2}\right), $$

$$ C_2=\int_2^{e^8}\frac{48}{\log^5 t}dt-\frac{24}{\log^4 2}. $$

The best result for the difference is obtained if we believe in RH:

Theorem (Schoenfeld 1976): If the Riemann hypothesis holds, then for all $x\ge 2657$ we have $$ |\pi(x)-li(x)|\le \frac{1}{8\pi}\sqrt{x}\log x. $$

Dietrich Burde
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  • thanks, exactly what I was looking for! – Andrea Feb 19 '15 at 14:42
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    I am sorry but I do not understand the link bewteen the question an the answer at all. The question concerns a result showing that $\pi(x) < li(x) - g(x)$ with positive $g(x)$ of a certain size. This is in some sense exactly the opposite of the results recalled in the answer that can be used to show $\pi(x) > li(x) - g_1(x)$ yet never less than. – quid Feb 19 '15 at 14:53
  • You're right, I mean he answered to a question that I was going to ask.. – Andrea Feb 19 '15 at 14:57
  • @quid: I added the lower bound on $\pi(x)-li(x)$, too. I think these are the relevant estimates concerning $\pi(x)-li(x)$, right ? – Dietrich Burde Feb 19 '15 at 15:01
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    They are certainly relevant estiamates but different from the ones asked about. Your estimates give an upper bound on $|\pi(x) - li(x)|$ whiel the one in the question give a lower bound (with additional information in which direction). The result in the question is around sign-changes and oscillations of the remainder (not about bounding its size from above). // To make this more clear: if $\pi(x)$ were $0$ it would still fulfill the Littlewood estimate – quid Feb 19 '15 at 15:10
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    I see. The reference of the original paper of Hardy and Littelwood answers the question of course. I was assuming that the other estimates were also interesting in this context. – Dietrich Burde Feb 19 '15 at 15:23
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The result in the question is first proved by J.E. Littlewood in Comptes Rendus de l'Academie des Sciences, June 1914.

The result is expressed as:

$$\pi(x)-Li(x)<-K\frac{\sqrt{x}\log\log\log x}{\log x} $$

$$\pi(x)-Li(x)>K\frac{\sqrt{x}\log\log\log x}{\log x}. $$

He concludes that the inequality $\pi(x)<Li(x),$ "presumed by many authors for empirical reasons, cannot obtain for any value of $x$ however large."

He does not specify K.

His conclusion may be interpreted to mean that for any value of $x,$ however large, for which one sense of the inequality holds, one can find a larger $x$ for which the other sense holds, and so on.

Ingham's later proof is somewhat easier (The Distribution of Prime Numbers, Ch. V).

daniel
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The result you quote from Littlewood is stated and proved in that paper of Hardy and Littlewood (see theorem 5.8 and below, page 194).

Sary
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