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In this paper is written that the prime number theorem in the form $\psi(x) = ( 1 + o(1) ) x$ is elementary equivalent to $$\sum_{n \le x } \frac{\Lambda(n)}{n} = \log x - \gamma + o(1) $$

I started to prove this as follows. By partial summation,

\begin{align*} \sum_{n \le x } \frac{\Lambda(n)}{n} &= \psi(x) \frac{1}{x} - \int_1^x \psi(t) d\left( \frac{1}{t} \right)\\ &= 1 + o(1) + \int_1^x \frac{t + o(1) t}{t^2} dt\\ & = 1 + \log x + \text{something} \end{align*} In order to treat the $o(1)$ inside the integral correctly, it seems natural to split the integral from $0$ to $x_0$ and from $x_0$ up to $x$, where $x_0$ is such that I can bound the $o(1)$ somehow. I cannot make it work however. I also think that I am on the wrong path because I do not see how the $\gamma$ will appear. Can somebody give a hint about how to prove the statement at the top of the page? I think that "elementary equivalent" means no complex analysis.

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    $\begingroup$ I don't think that as it stands, the claim is true; the given form of the prime number theorem allows the case $\psi(x)=x+\frac x{\log \log x}$, which would not give the desired result. $\endgroup$ – Laertes Jun 5 '15 at 16:23
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By Perron's formula we have $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+\infty}-\frac{\zeta'}{\zeta}\left(s+1\right)\frac{x^{s}}{s}ds $$ with $x\notin\mathbb{N} $. We have a double pole at $s=0 $ with residue $\log\left(x\right)-\gamma $ and it is possible to show (with the same tecniques used to show the prime number theorem in the form $\psi\left(x\right)=x+O\left(x\exp\left(-c_{1}\log^{1/2}\left(x\right)\right)\right) $) that the contribute of the other poles is infinitesimal. Then $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}=\log\left(x\right)-\gamma+o\left(1\right). $$

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  • $\begingroup$ thank you for this answer, but I think "elementary equivalent" in this context means no complex analysis (I stated this in my question) $\endgroup$ – wiskundeliefhebber Jun 6 '15 at 20:41
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    $\begingroup$ @wiskundeliefhebber I don't think it is possible prove it only by partial summation and the prime number theorem in the form $\psi\left(x\right)=x\left(1+o\left(1\right)\right)$. We need some precise bound about the error. In fact in the paper we find -...is "elementary" equivalent...-. $\endgroup$ – Marco Cantarini Jun 6 '15 at 20:51
  • $\begingroup$ I am not yet 100% convinced yet. I found out after posting the question that there was already a related post: math.stackexchange.com/questions/76778/…, where it is also claimed that it is not equivalent. On the other hand it is claimed to be equivalent in exercise 3.10 in these lecture notes: math.uiuc.edu/~hildebr/ant $\endgroup$ – wiskundeliefhebber Jun 6 '15 at 21:28
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    $\begingroup$ This result IS equivalent to PNT. But to get gamma you need a precise bound of the error term, not only $o(1)$. $\endgroup$ – Marco Cantarini Jun 6 '15 at 21:44
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The summation by parts yields $$\sum_{n \leqslant x} \frac{\Lambda(n)}{n} = \frac{\psi(x)}{x} + \int_1^x \frac{\psi(t)}{t^2}\,dt = \log x + \frac{\psi(x)}{x} + \int_1^x \frac{\psi(t)-t}{t^2}\,dt\,.$$ Thus it is easy to see that under the hypothesis $\psi(x) \sim x$ the limit $$\lim_{x\to \infty}\; \Biggl(\log x - \sum_{n \leqslant x} \frac{\Lambda(n)}{n}\Biggr) \tag{1}$$ exists if and only if $$\lim_{x \to \infty} \int_1^{x} \frac{\psi(t) - t}{t^2}\,dt \tag{2}$$ exists. It is easy to see that if $(1)$ exists, then the limit is $\gamma$, so it would suffice to prove the existence of $(1)$ without caring about its value.

Actually, one step in Newman's proof of the prime number theorem is to show that the existence of $(2)$ implies $\psi(x) \sim x$ (see e.g. here; although Zagier's exposition of Newman's proof uses $\vartheta$ rather than $\psi$, the argument is practically identical for both choices), so the existence of $(2)$ implies that of $(1)$ without further assumptions. And since the existence of $(1)$ implies $\psi(x) \sim x$ via summation by parts, we see that $(1)$ and $(2)$ are elementarily equivalent. However, on the face of it, $(2)$ seems to be stronger than $\psi(x) \sim x$, and there's no obvious direct way to obtain $(2)$ from $\psi(x) \sim x$. Plugging $\psi(t) = t + o(t)$ into the integral cannot yield more than $$\int_1^x \frac{\psi(t) - t}{t^2}\,dt \in o(\log x)\,.$$

We need a different angle of attack. Until yesterday evening, I was not aware of a direct elementary way to show that $\psi(x) \sim x$ implies the existence of $(1)$, and my answer would have been to first show (using Landau's proof) that $\psi(x) \sim x$ implies the convergence of $$\sum_{n = 1}^{\infty} \frac{\mu(n)}{n}$$ and then use that to reach the goal. This route is fun, but it is also rather long. But yesterday evening, I happened to see this Math Overflow answer by Vesselin Dimitrov sketching a short elementary and direct proof of $$\psi(x) \sim x \implies \sum_{n \leqslant x} \frac{\Lambda(n)}{n} = \log x - \gamma + o(1)\,.$$ This is very much a fun proof too, so let's fill in the details.

First, we define $$\delta(x) := \sup\; \biggl\{ \biggl\lvert \frac{\psi(y)}{y} - 1\biggr\rvert : y \geqslant x\biggr\}$$ for $x\geqslant 1$. Then $\delta$ is nonicreasing, and $\delta(x) \to 0$ for $x \to \infty$ is just our assumption $\psi(x) \sim x$. I note that $\lvert\delta(x)\rvert \leqslant 1$ for all $x \geqslant 1$. Then we will need a function $\omega \colon [1,\infty) \to [1,\infty)$ such that $\lim\limits_{x\to\infty} \omega(x) = \infty$ and $$\lim_{x\to \infty} \delta(x)\log \omega(x) = 0\,.$$ We can take $\omega(x) = 1/\delta(x)$ for example. Next, following Chebyshev, let $$T(x) := \sum_{n \leqslant x} \log n = \log \bigl(\lfloor x\rfloor!\bigr)\,.$$ From Euler's sum formula, summation by parts, or Stirling's formula one easily obtains $$\lvert T(x) - (x\log x - x)\rvert \leqslant 1 + \log x$$ for $x \geqslant 1$, and writing Legendre's formula in terms of the von Mangoldt function, we have $$T(x) = \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{x}{n}\biggr\rfloor\,.$$ Now put $z = x\omega(x)$ and by splitting the sum at $x$ find \begin{align} T(z) &= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor + \sum_{x <n \leqslant z} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor \\ &= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor +\sum_{x < n \leqslant z} \Lambda(n)\sum_{k \leqslant z/n} 1 \\ &= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor + \sum_{k < \omega(x)} \sum_{x < n \leqslant z/k} \Lambda(n) \\ &= \sum_{n \leqslant x} \Lambda(n)\biggl\lfloor \frac{z}{n}\biggr\rfloor + \sum_{k \leqslant \omega(x)} \Biggl(\psi\biggl(\frac{z}{k}\biggr) - \psi(x)\Biggr) \\ &= z\sum_{n \leqslant x} \frac{\Lambda(n)}{n} + \sum_{k \leqslant \omega(x)} \Biggl(\psi\biggl(\frac{z}{k}\biggr) - \psi(x)\Biggr) - \sum_{n \leqslant x} \Lambda(n)\biggl\lbrace \frac{z}{n}\biggr\rbrace\,. \end{align} Dividing by $z$ and rearranging then yields \begin{align} \Biggl\lvert \sum_{n \leqslant x} \frac{\Lambda(n)}{n} &{}- \log x + \gamma \Biggr\rvert \\ &= \Biggl\lvert \frac{T(z)}{z} - \log x + \gamma + \frac{\lfloor \omega(x)\rfloor}{\omega(x)}\cdot \frac{\psi(x)}{x} - \sum_{k \leqslant \omega(x)} \frac{1}{z}\psi\biggl(\frac{z}{k}\biggr) + \frac{1}{z}\sum_{n \leqslant x} \Lambda(n)\biggr\lbrace \frac{z}{n}\biggr\rbrace\Biggr\rvert \\ &\leqslant \Bigl\lvert \log z - 1 - \log x + \gamma + \frac{\lfloor \omega(x)\rfloor}{\omega(x)} - \sum_{k \leqslant \omega(x)} \frac{1}{k}\Biggr\rvert \\ &\qquad + \frac{1 + \log z}{z} + \frac{\lfloor\omega(x)\rfloor}{\omega(x)}\delta(x) + \sum_{k \leqslant \omega(x)} \frac{1}{k}\delta\biggl(\frac{z}{k}\biggr) + \frac{\psi(x)}{z} \\ &\leqslant \biggl\lvert 1 - \frac{\lfloor \omega(x)\rfloor}{\omega(x)}\biggr\rvert + \Biggl\lvert \log \omega(x) + \gamma - \sum_{k \leqslant \omega(x)} \frac{1}{k}\Biggr\rvert \\ &\qquad + \frac{1+\log z}{z} + \delta(x)(2+\log \omega(x)) + \frac{2}{\omega(x)} \\ &\leqslant \frac{4}{\omega(x)} + \frac{1 + \log x + \log \omega(x)}{x\omega(x)} + \delta(x)(2 + \log \omega(x))\,. \end{align} The right hand side tends to $0$ for $x \to \infty$ by the constraints imposed on $\omega$.

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