Natural density implies Dirichlet density Let $M$ be a set of prime ideals of a number field $K$ . The limit
$$d(M)= \lim_{s\rightarrow 1^+} \frac{ \sum_{P \in M} N(P)^{-s} }{  \sum_{P} N(P)^{-s}}$$
Where $P$ is a prime of $K$ and $N$ denotes the norm of $K/ \mathbb{Q}$ is called Dirichlet Density of $M$. Also, the Natural density is the limit
$$ \delta(M)= \lim_{x\rightarrow \infty} \frac{ \# \{ P \in M :  N(P) \leq x\}}{  \# \{ P :  N(P) \leq x\}} $$
Show that the Natural denstity implies the Dirichhlet density.
Also, $d(M)=\delta(M)$.
Any help? I'm trying for days to prove this exercise.
 A: Here is a proof when $K=Q$, for general number field proof will be analogous. 
$\# \{ p\in M :p\leq x\}=M(x)$. Let's define $a_n=1$ when $n=p$ prime, and $0$ otherwise. So, the sum 
$\displaystyle\sum_{p\in M,\\p\leq x}\frac{1}{p^s}=\sum_{n\leq x}\frac{a_n}{n^s}=\frac{M(x)}{x^s}+s\int_{1}^{x}\frac{M(t)}{t^{s+1}}dt$.
Now suppose 
$\delta(M)= \lim_{x\rightarrow \infty} \frac{ \# \{ p \in M : p \leq x\}}{  \# \{ p :  p \leq x\}}$ exists and equals to $R$. That means by the Prime Number Theorem
$\lim_{x\to\infty}\frac{M(x)\log x}{x}=R$ or $M(x)=\frac{Rx}{\log x}+o(\frac{x}{\log x})$
Now plug in this formula of $M(x)$ into above summation. We have,
$d(M)=\lim_{s\to1+}\frac{\lim_{x\to\infty}\frac{M(x)}{x^s}+s\int_{1}^{x}\frac{M(t)}{t^{s+1}}dt}{-\log (s-1)}$,
as, $\sum_{p}\frac{1}{p^s}=-\log (s-1) + O(1)$.
Now  using L'hospital rule we can have that,
$d(M)=\lim_{s\to1+}(s-1)\int_{1}^{\infty}\frac{M(t)\log t}{t^{s+1}}dt$
So, $\displaystyle\\d(M)=\lim_{s\to1+}(s-1)(R\int_{1}^{\infty}\frac{dt}{t^{s}}+o(\int_{1}^{\infty}\frac{dt}{t^{s}}))$
We know that, $\lim_{s\to1+}(s-1)\int_{1}^{\infty}\frac{dt}{t^{s}}=1$ as $\epsilon\to0$.
so $d(M)=R+o(1)=R$
For general case just "switch all $p$'s by $N(P)$'s", and use Prime Ideal Theorem instead of PNT.
A: It suffices to prove the theorem for $K=Q$. For general number field proof will be analogous. Let
$M(x)=\# \{ p\in M :p\leq x\}$. Now suppose
$\delta(M)= \lim_{x\rightarrow \infty} \frac{ \# \{ p \in M : p \leq x\}}{  \# \{ p :  p \leq x\}}=\frac{M(x)}{\pi(x)}$ exists and equals to $R$. 
Since $\lim_{x\to\infty}\frac{M(x)}{x}=0=\lim_{x\to\infty}\frac{\pi(x)}{x}$, the Dirichlet series
$$\sum_{p\in M}\frac{1}{p^s}=s\int_1^\infty\frac{M(x)}{x^{s+1}}dx,\quad\sum_p\frac{1}{p^s}=s\int_1^\infty\frac{\pi(x)}{x^{s+1}}dx,\quad s>1$$
In order to prove our theorem, we need a lemma. 
lemma
Let $A(x),B(x)$ be non-negative integrable function and the limit $\lim_{x\to\infty}\frac{A(x)}{B(x)}=R$.
If
$$\lim_{s\to 1^+}\int_1^\infty\frac{B(x)}{x^{s+1}}dx=+\infty$$
then we have that
$$\lim_{s\to1^+}\frac{\int_1^\infty\frac{A(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R.$$
proof
Since $\lim_{x\to\infty}\frac{A(x)}{B(x)}=R$, for every $\epsilon>0$, there exists a constant $C=C(\epsilon)$ such that $|R(x):=A(x)-RB(x)|\leq C+\epsilon B(x)$ for all $x\geq1$. 
Hence
$$\frac{\int_1^\infty\frac{A(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R+
  \frac{\int_1^\infty\frac{A(x)-RB(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R+
  \frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}.$$
  Hence, it suffices to prove
  $$\lim_{s\to 1^+}\frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=0$$
  Since
  $$0\leq \left|\int_1^\infty\frac{R(x)}{x^{s+1}}dx\right|\leq \int_1^\infty\frac{|R(x)|}{x^{s+1}}dx\leq \int_1^\infty\frac{C+\epsilon B(x)}{x^{s+1}}dx=\frac{C}{s}+\epsilon\int_1^\infty\frac{B(x)}{x^{s+1}}dx$$
We have that
  $$\limsup_{s\to1^+}\frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}\leq \epsilon$$
  Hence
$$\lim_{s\to 1^+}\frac{\int_1^\infty\frac{R(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=0$$
  and
  $$\lim_{s\to 1^+}\frac{\int_1^\infty\frac{A(x)}{x^{s+1}}dx}{\int_1^\infty\frac{B(x)}{x^{s+1}}dx}=R$$
Now, the proof is complete.
