I'm going through the paper of D. Zagier on Short Proof of Prime Number Theorem. There it says in V that
$\Phi(s)=\int_1^\infty \frac{d\vartheta(x)}{x^s}$ . Can someone please explain in details why that is the case ?
And also how The Analytic Theorem is being applied here to the functions defined as f(t) and g(z) ?
MSE question link:
Newman's "Natural proof"(Analytic) of Prime Number Theorem (1980)
Edit: I have understood how the Analytic Theorem is being applied here but I am still unable to figure out about that integral. I saw a post mentioning that because $\vartheta(x)$ changes by $log(p)$ for primes only. I got that point but is their a way to formally proof that thing. Any relative reference is appreciated.
Thank you ,
Best Answer
This is a form of integration called Riemann-Stieltjes integration. In short, we define
$$ \int_a^b f(x) d g(x) = \sum f(c_j) \big( g(x_{j + 1}) - g(x_j) \big) $$
for partitions $a = x_0 < \cdots < x_n = b$ and points $c_j \in [x_j, x_{j + 1}]$. This is a weighted generalization of typical Riemann integration.
Concretely, we can think of this sort of integral as adding $f(x) \Delta(g(x))$ whenever $g(x)$ changes.
Here, $\vartheta(x) = \sum_{p \leq x} \log p$ is the first Chebyshev function. Observe that this changes only on intervals $(p - \epsilon, p + \epsilon)$ when $p$ is prime (for sufficiently small $\epsilon$). Thus a Riemann-Stieltjes integral of the form
$$ \int_a^b f(x) d \vartheta(x)$$
for a continuous function $f$ on an interval $[a, b]$ (and where $p_1, \ldots, p_k$ are the primes in the interval $[a, b]$) will be the limit of partition terms that look like
$$ \sum_{j = 1}^k f(\alpha_j) \big( \vartheta(p_j + \epsilon) - \vartheta(p_j - \epsilon) \big) \to \sum_{j = 1}^k f(p_j) \log p_j.$$
Here, I write $\alpha_j$ to mean some number in $(p_j - \epsilon, p_j + \epsilon)$. As I've assumed $f$ is continuous, the limit as $\epsilon \to 0$ tends to $f(p_j)$. And the point is that for all $\epsilon > 0$, $\vartheta(p_j + \epsilon) - \vartheta(p_j - \epsilon) = \log p_j$ exactly. As $\vartheta(x)$ only changes around $p_j$, there is no other contribution to the integral.
More explicitly, I have given an explicit limit of partitions of the interval $[a, b]$, consisting of $2\epsilon$ boxes around primes; and I claim that taking refinements of this partition (i.e. taking $\epsilon \to 0$) converges.
Applying this to $f(x) = x^{-s}$ and to the integral $\int_1^B x^{-s} d \vartheta(x)$ shows that
$$ \int_1^B \frac{d \vartheta(x)}{x^s} = \sum_{1 \leq p \leq B} \frac{\log p}{p^s}. $$
Taking the improper integral amounts to taking the limit as $B \to \infty$, which is equal to $\sum (\log p)/p^s$ as long as $s > 1$, which guarantees absolute convergence.