# How can one deduce an approximation for the density function of prime numbers from this Euler's theorem?

The author of Riemann's Zeta Function, H.M.Edwards, says:

1. According to Euler, $$\sum_{p when $$x\longrightarrow\infty$$.

2. $$\log(\log(x))=\int_{1}^{\log(x)} \frac{du}{u}=\int_{e}^{x} \frac{dv}{v\log(v)}$$

so (1) says that the integral of $$\frac{1}{v}$$ relative to the measure $$\frac{dv}{\log(v)}$$ diverges in the same way as the integral of $$\frac{1}{v}$$ relative to the point mesaure which assigns weight $$1$$ to primes and weight $$0$$ to all other points. In this sense, (1) can be regarded as saying that the density of primes is roughly $$\frac{1}{\log(v)}$$.

And this is what the author says. I know that density formula for a number $$x$$ gives us the probability for a number $$y of being prime, but I don't how the author identified $$\frac{1}{\log(v)}$$ as the density formula (it's an approximation of the actual density formula, I know it). He implies that it is a trivial reasoning, but I can't see it.

I think it could be because of the language: I'm a Spanish student, so my English is not too good and this is the first time that I read something like "the integral of $$\frac{1}{v}$$ relative to the measure $$\frac{dv}{\log(v)}$$ diverges in the same way…". I am not clear what it means, so I want to use this question for two things: to find out how the author identified the density formula, and to learn some technical English vocabulary.

Thanks.

• Hi Sergio, and welcome to the MathOverflow. Your question is nicely written and well motivated: however is not a research level question, thus it will be probably closed very quickly, as MathOverflow is devoted exclusively to such kind of Q&A. It is better suited for the Math.SE, where you are already a member, so consider to post it there for a better reception and possibly an answer. Feb 1 at 17:03
• Please use a high-level tag like "nt.number-theory". I added this tag now. Feb 1 at 17:16
• Language note: You almost certainly mean "deduce" (to derive as a logical conclusion) rather than "deduct" (to subtract or remove) in the title. I have edited accordingly. Feb 1 at 17:20

## 2 Answers

What Edwards is trying to say is this: the Prime Number Theorem, i.e. the relation $$\pi(x)\sim x/\log x$$, implies Euler's earlier result that $$\sum_{p. The Prime Number Theorem says that the density of primes around $$x$$ is $$1/\log x$$, and Euler's result is a weak version of it (meaning an easier-to-prove consequence).

Let $$d \pi$$ be the measure on $$\mathbb{R}_{>0}$$ which is a delta function of size $$1$$ at the primes. In other words, $$\int_{y}^x d \pi = \pi(x) - \pi(y)$$. The prime number theorem is that $$\pi(x)$$ is approximately the logarithmic integral, in other words, that $$\int^x d \pi \sim \int^x \tfrac{dt}{\log t}. \qquad (1)$$ I've left out the lower bounds of the integrals because they don't matter for asymptotics.

A common perspective in Analytic Number Theory is to think about between different weighted averages of $$d \pi$$ and try to show that they are close to the corresponding weighted averages of $$\tfrac{dt}{\log t}$$. Another such relation is that $$\int^x \tfrac{d \pi}{t} = \int^x \tfrac{dt}{t \log t}+O(1). \qquad (2)$$ Relation (2) was proved by Merten, but anticipated by Euler, who wrote $$\sum_p \tfrac{1}{p} = \log \log \infty. \qquad (3)$$ Edwards writes (page 2 in my edition) that it is not clear exactly what Euler meant by (3) -- it could have been (2), but $$\int^x \tfrac{d \pi}{t} \sim \int^x \tfrac{dt}{t \log t}$$ is also a reasonable interpretation -- but Edwards is pointing out (3) as possibly the historically first statement saying that $$d \pi$$ is approximately $$\tfrac{dt}{\log t}$$.

At this part in the book, Edwards is discussing the history of the notion that $$d \pi$$ and $$\tfrac{dt}{\log t}$$ are close in any sense, before getting into the different ways in which one might define closeness. Of course, the technically hard part of analytic prime number theory is working out the relations between these different notions of $$$$approximately the same".