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. $\lim_{n\to\infty} \frac{f(x)}{g(x)}=1$.

  • 2
    $\begingroup$ Could you give a link to this article you speak of, please? $\endgroup$ – J. M. is a poor mathematician Sep 26 '10 at 6:09
  • $\begingroup$ Maybe you are interested in this? $\endgroup$ – draks ... Mar 16 '12 at 13:16

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.


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