# Understanding Newman's proof of the prime number theorem

I am trying to understand D.J. Newman's proof of the prime number theorem, as presented by D. Zagier. I am not too familiar with analysis, and so there are some things I don't understand.

1. In (III), what does Zagier mean by "$\theta(x)$ changes by $O(\log x)$ if $x$ changes by $O(1)$?" I know the definition of Big O notation, but not what this grammatical construction means.

2. In (V), I am having trouble parsing the integral $$\int_1^\infty \frac{d \theta(x)}{x^s}.$$ What does the notation $d \theta(x)$ mean? Is $\theta(x)$ the variable of integration in some sense? If so, how? (This may be a naive question.)

3. In the lower bound for $\theta(x)$ (on p. 707), in the bounds for the first summation -- $x^{1-\epsilon} \leq p \leq xx$ -- is $xx$ a typo, supposed to be $x$, or is it correct? If it's correct, why is the first inequality true?

4. I don't understand why the lower bound for $\theta(x)$, $$\theta(x) \geq (1-\epsilon) \log x (\pi(x) + O(x^{1-\epsilon})),$$ is good enough. As $\epsilon \rightarrow 0$, $O(x^{1-\epsilon})$ tends to $O(x)$, and based on my interpretation of the previous summations, I expect this error term to be negative. Since $\pi(x) \leq x$, I basically don't understand why this is saying more than that $\theta(x) \geq 0$.

Thank you!

• For #$1$, $\theta(x+a)-\theta(x)=\sum_{x\leq p\leq {x+a}}\ln(p)<a\ln(x+a)<c\ln(x)$, for some constant c – Ethan Jun 16 '13 at 5:37
• For #$2$, its a stieltjes integral, read here en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral – Ethan Jun 16 '13 at 5:38
• For #$3$, Yes it is a typo – Ethan Jun 16 '13 at 5:38
• For #$4$, If you're not comfortable doing it this way, try summation by parts, you should get the result pretty fast, $$\sum_{\substack{n\leq x\\n\in \text{ S}}}f(n)g(n)=f([x])\sum_{\substack{ n\leq x \\ n\in \text{S}}}g(n)-\sum_{n\leq x-1}(\sum_{\substack{k\leq n \\ k\in \text{S}}}g(k))(f(n+1)-f(n))$$ Let $\text{S}$ be the set of primes less then or equal to x, then let $f(n)=\ln(n)$ – Ethan Jun 16 '13 at 5:39
• @Ethan, why not roll these comments up into an answer? – Gerry Myerson Jun 16 '13 at 5:59