I am teaching a graduate course in Complex Analysis and I am covering Newman's proof of the prime number theorem. I have been using the simplified version in the papers of Zagier and Korevaar. However, I ran into a problem. Both papers rely on this theorem:

Theorem Let $f:[0,\infty)\rightarrow\mathbb{C}$ be bounded and locally integrable and let $$ g(z):=\int_{0}^{\infty}f(t)e^{-tz}dt,\quad\operatorname{Re}z>0. $$ Assume that for every $z\in\mathbb{C}$ with $\operatorname{Re}z=0$ there exists $r_{z}>0$ such that $g$ can be extended holomorphically to $B(z,r_{z} )$. Then the generalized Riemann integral \begin{equation} \int_{0}^{\infty}f(t)\,dt \label{pn1} \end{equation} is well-defined and equals $g(0)$.

This theorem is used to prove that the generalized Riemann integral $$ \int_{1}^{\infty}\frac{\theta(x)-x}{x^{2}}dx $$ converges. Here, $$ \theta(x):=\sum_{p\text{ prime}\leq x}\log p,\quad x\in\mathbb{R}. $$ Everything is fine up to this point. Then the authors use the convergence of this integral to prove that \begin{equation} \lim_{x\rightarrow\infty}\frac{\theta(x)}{x}=1. \label{pn limit theta}% \end{equation} Their proof is as follows: Assume by contradiction that $$ \limsup_{x\rightarrow\infty}\frac{\theta(x)}{x}>1. $$ There there exists an increasing sequence $x_{n}\rightarrow\infty$ such that $\theta(x_{n})>(1+\varepsilon)x_{n}$ for all $n\in\mathbb{N}$ and for some $0<\varepsilon<1$. Since $\theta$ is increasing, if $x>x_{n}$, $\theta (x)\geq\theta(x_{n})>(1+\varepsilon)x_{n}$, and so \begin{align*} \int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{\theta(x)-x}{x^{2}}dx & \geq \int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{(1+\varepsilon)x_{n}-x}{x^{2}}dx\\ & =\int_{1}^{(1+\varepsilon)}\frac{(1+\varepsilon)-s}{s^{2}}ds>0 \end{align*} where we made the change of variables $x=x_{n}s$ so $dx=x_{n}ds$. Since $x_{n}\rightarrow\infty$, by selecting a subsequence we can assume that $x_{n+1}\geq2x_{n}$ for all $n$. Hence, by summing all the disjoint integrals on the left-hand side we obtain that $$ \int_{\bigcup(x_{n},(1+\varepsilon)x_{n},)}\frac{\theta(x)-x}{x^{2}}% dx=\sum_{n=1}^{\infty}\int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{\theta (x)-x}{x^{2}}dx\\=\sum_{n=1}^{\infty}\int_{1}^{(1+\varepsilon)}\frac {(1+\varepsilon)-s}{s^{2}}ds=\infty. $$ The papers claim that this fact contradicts the fact that the integral converges and proves that $$ \limsup_{x\rightarrow\infty}\frac{\theta(x)}{x}\leq1. $$ However, this is not the case since all we know is that $$ \lim_{T\rightarrow\infty}\int_{1}^{T}\frac{\theta(x)-x}{x^{2}}dx=\ell \in\mathbb{R}% $$ but this does not prevent that $$ \int_{1}^{\infty}\frac{(\theta(x)-x)^{+}}{x^{2}}dx=\int_{1}^{\infty}% \frac{(\theta(x)-x)^{-}}{x^{2}}dx=\infty. $$ The typical example is $$ \int_{1}^{\infty}\frac{\sin x}{x}dx, $$ which exist as an improper Riemann integral but not as Lebesgue integral. Am I missing something? If not, is there a correct proof?


1 Answer 1


It doesn't seem to me that either article follows the line of reasoning as you have presented it. Indeed, we do not take the integral over the union of integrals $(x_n,(1+\varepsilon)x_n)$. We do get that the integral over those intervals is infinite, but you correctly note this does not give a contradiction. Instead, the argument goes as follows.

Let me denote $$F(T)=\int_{1}^{T}\frac{\theta(x)-x}{x^{2}}dx,\quad C=\int_{1}^{(1+\varepsilon)}\frac{(1+\varepsilon)-s}{s^{2}}ds$$ so that $F(T)$ converges and $C$ is a positive constant. Since $F(T)$ converges, it is in particular Cauchy, so for large enough $x,y$ we have $|F(x)-F(y)|<C$. But for any $x_n$ we have $F((1+\varepsilon)x_n)-F(x_n)\geq C$ by the calculation you present, and for large $x_n$ this is the desired contradiction.

  • $\begingroup$ You are right. Thanks! $\endgroup$
    – Gio67
    Mar 31, 2020 at 13:51

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