Prime Number theorem and the prime counting function

Question

Could someone please help me understand this proof given in an article by William Miller. It is supposed to follow from the prime number theorem that given $A(x)$ the sum of all primes less than or equal to $x$ and $\theta(x)$ the sum of the logarithm of all primes less than or equal to $x$,

$$A(x)\sim \frac{x^2}{2\log x} \ \ \ \rm and \ \ \ \theta(x) \sim x,$$

the following identity is used:

$$\theta(x) = \int_1^x \log(t)\mathrm{d}(\pi(t)),$$

where $\pi(t)$ is the prime counting function. I don't understand why this is. Here $\sim$ means asymptotic to i.e. $\lim_{n\to\infty} \frac{f(x)}{g(x)}=1$.

Could you give a link to this article you speak of, please? – J. M. ain't a mathematician Sep 26 '10 at 06:09 — J. M. ain't a mathematician, Sep 26 '10 at 06:09
Maybe you are interested in this? – draks ... Mar 16 '12 at 13:16 — draks ..., Mar 16 '12 at 13:16

score 5 · Answer 1 · answered Sep 26 '10 at 06:45

The notation $$\int \log(t)\ d(\pi(t))$$ is an example of the Stieltjes integral. Some like this (I don't) when discussing summation, as integrating with a step function $f(t)$ inside the $d$ notation is essentially a summation. Here $\pi(t)$ is constant, save at prime values of $t$ where it jumps by $1$. This means that $$\int_a^b g(t)\ d(\pi(t))=\sum_{p\ \mathrm{prime}}g(p)\times 1.$$ Here the summation is over all primes in the interval from $a$ to $b$ (I can't remember whay you do about the endpoints; one reason I don't like the notation). If you like you can think that $$\int_a^b g(t)\ d(\pi(t))=\int_a^b g(t)\pi'(t)\ dt$$ where $\pi'$ is seen as a generalized function, so as a bunch of delta-functions at the primes. Again one has to do the right thing (or is it the left thing? :-)) at the endpoints.

Prime Number theorem and the prime counting function

1 Answers1