Let $M$ be a set of prime numbers of $\mathbb{Q}$ . The limit $$d(M)= \lim_{s\rightarrow 1^+} \frac{ \sum_{p \in M} p^{-s} }{ - \log(s-1)}$$ Where $p$ is a prime of $\mathbb{Q}$ is called Dirichlet Density of $M$. Also, the Natural density of $M$ is the limit $$ \delta(M)= \lim_{x\rightarrow \infty} \frac{ \# \{ p \in M : p \leq x\}}{ \# \{ p \in \mathbb{Q} : p \leq x\}} $$

Now, let $$M= \bigcup_{k=0}^{\infty}\{ p \mbox{ prime} : 10^k \leq p < 2\cdot10^k \}$$ Show that $\delta(M)$ not exists and $d(M)=\frac{\log(2)}{\log(10)}$

For $\delta(M)$ i can apply the Prime Number Theorem, but if $$M(x)= \# \{ p \in M : p \leq x \}= \sum_{p \in M, p\leq x}1 = \sum_{0 \leq k \leq \frac{\log(x)}{\log(10)}} \sum_{10^k \leq p < 2\cdot10^k, p \leq x} 1$$

I do not know how to follow...

  • $\begingroup$ Look at how $M(x)$ changes in the range $10^k < x < 2\cdot 10^k$, and how it changes in the range $2\cdot 10^k < x < 10^{k+1}$. In particular, estimate $M(x)/\pi(x)$ for $x = 10^k$ and $x = 2\cdot 10^k$. $\endgroup$ – Daniel Fischer Jul 15 '13 at 19:49
  • $\begingroup$ @DanielFischer $M(10^k)$? $\endgroup$ – P. M. O. Jul 15 '13 at 20:41
  • $\begingroup$ The $M(x)$ you defined in the second row from bottom. $\endgroup$ – Daniel Fischer Jul 15 '13 at 20:50
  • $\begingroup$ @DanielFischer $M(10^k)=\sum_{0 \leq k} \sum_{p=10^k}1=0? $ $\endgroup$ – P. M. O. Jul 15 '13 at 20:54
  • $\begingroup$ $M(x)= \# \{ p \in M : p \leq x \}$. Due to the construction of $M$, $M(10^k) \leqslant \pi(2\cdot 10^{k-1})$. $\endgroup$ – Daniel Fischer Jul 15 '13 at 20:56

Using $M(x)= \# \{ p \in M : p \leq x \}$, we observe that due to the definition of $M$, we have

$$M(10^k) \leqslant \pi(2\cdot 10^{k-1}), \text{ and } M(2\cdot 10^k) \geqslant \pi(2\cdot 10^k) - \pi(10^k).$$

Thus, supposing $k$ not too small, and using the prime number theorem, we obtain

$$\frac{M(10^k)}{\pi(10^k)} \leqslant \frac{\pi(2\cdot 10^{k-1})}{\pi(10^k)} \approx \frac15\left(1 + \frac{\log 5}{\log (2\cdot 10^{k-1})}\right) \approx \frac15$$


$$\frac{M(2\cdot 10^k)}{\pi(2\cdot 10^k)} \geqslant 1 - \frac{\pi(10^k)}{\pi(2\cdot 10^k)} \approx 1 - \frac12\left(1 + \frac{\log 2}{\log (10^k)}\right) \approx \frac12.$$

So the value of $\frac{M(x)}{\pi(x)}$ oscillates between $\approx \frac15$ (or smaller) and $\approx \frac12$, hence has no limit.

For the Dirichlet density, we use that

$$\sum_{p \leqslant x} \frac{1}{p} = \log \log x + M + O\biggl(\frac{1}{(\log x)^2}\biggr)\tag{1}$$

where $M$ is the Meissel-Mertens constant. De la Vallée Poussin's error bound in the prime number theorem gives a smaller remainder term, but that would give no advantage in our calculation. For $x > 2$, we obtain

$$\sum_{x < p \leqslant 2x} \frac{1}{p} = \log\biggl(1 + \frac{\log 2}{\log x}\biggr) + O\biggl(\frac{1}{(\log x)^2}\biggr) = \frac{\log 2}{\log x} + O\biggl(\frac{1}{(\log x)^2}\biggr)\tag{2}$$

from $(1)$, so

$$\sum_{10^k < x \leqslant 2\cdot 10^k} \frac{1}{p} = \frac{A}{k} + R(k)\tag{3}$$

with $A = \frac{\log 2}{\log 10}$ and $R(k) \in O(k^{-2})$ for $k \geqslant 1$. Define

$$S(x) = \sum_{\substack{p\in M \\ p \leqslant x}} \frac{1}{p}.$$

For $x > 20$ pick $K$ such that $2\cdot 10^K \leqslant x < 2\cdot 10^{K+1}$. Then

$$S(x) = \sum_{k = 1}^K \Biggl(\frac{A}{k} + R(k)\Biggr) + O(K^{-1}) = A\log K + (A\gamma + C) + O(K^{-1})$$

with $C = \sum_{k = 1}^{\infty} R(k)$, since

$$\sum_{10^{K+1} < p \leqslant x} \frac{1}{p} \in O(K^{-1})$$

by $(2)$, and $C - \sum_{k = 1}^K R(k) = \sum_{k = K+1}^{\infty} R(k) \in O(K^{-1})$ too. With

$$K = \biggl\lfloor \frac{\log \frac{x}{2}}{\log 10}\biggr\rfloor = \frac{\log x - \log 2}{\log 10}\cdot \Bigl( 1 + O\bigl((\log x)^{-1}\bigr)\Bigr)$$

we thus obtain

$$S(x) = A\log \log x + B + O\bigl((\log x)^{-1}\bigr),$$

where $B = A\gamma + C - A\log \log 10$. Since $S(x) = 0$ for $x < 11$, we can thus write

$$\sum_{p \in M} \frac{1}{p^{1+\varepsilon}} = \varepsilon \int_e^{\infty} \frac{S(x)}{x^{1+\varepsilon}}\,dx = A\varepsilon \int_e^{\infty} \frac{\log \log x}{x^{1+\varepsilon}}\,dx + B\varepsilon\int_e^{\infty} \frac{dx}{x^{1+\varepsilon}} + O\Biggl(\varepsilon\int_e^{\infty} \frac{dx}{x^{1+\varepsilon}\log x}\Biggr).$$

For the first term, we calculate

\begin{align} \varepsilon \int_e^{\infty} \frac{\log \log x}{x^{1+\varepsilon}}\,dx &= \varepsilon \int_1^{\infty} e^{-\varepsilon t}\log t\,dt \tag{$x = e^t$}\\ &= \int_{\varepsilon}^{\infty} e^{-u}(\log u - \log \varepsilon)\,du \tag{$u = \varepsilon t$} \\ &= \log \frac{1}{\varepsilon}\cdot \Biggl(1 - \int_0^{\varepsilon} e^{-u}\,du\Biggr) + \Gamma'(1) -\int_0^{\varepsilon} e^{-u}\log u\,du \\ &= \log \frac{1}{\varepsilon} + \Gamma'(1) + O(\varepsilon \lvert\log \varepsilon\rvert). \end{align}

The second term is $B e^{-\varepsilon}$, and to bound the error term, we note that

$$\int_e^{\infty} \frac{dx}{x^{\varepsilon}(x\log x)} = \underbrace{\frac{\log \log x}{x^{\varepsilon}}\biggr\rvert_e^{\infty}}_{{}=0} + \varepsilon\int_e^{\infty} \frac{\log \log x}{x^{1+\varepsilon}}\,dx,$$

so reusing the previous result we finally get

$$\sum_{p\in M} \frac{1}{p^{1+\varepsilon}} = A\log \frac{1}{\varepsilon} + A\Gamma'(1) + B + O(\varepsilon\lvert\log\varepsilon\rvert),$$

which shows

$$d(M) = A = \frac{\log 2}{\log 10}.$$

  • $\begingroup$ Thanks,you know how to do $d(M)$? $\endgroup$ – P. M. O. Jul 15 '13 at 21:54
  • $\begingroup$ Not off the top of my head. I need to think about that. $\endgroup$ – Daniel Fischer Jul 15 '13 at 21:57

This is a non-trivial result, I may give you an idea of proof but first you should look at this.


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