$\sum_{p}\frac{1}{p^s}$ how can i solve this? This Exercise is in text Tom M Apostol. Analytic number theory page246.

$$s\int_1^\infty\frac{\pi(x)}{x^sx}dx = \sum_{p}\frac{1}{p^s}$$ where the sum is extended over all primes. ($x^sx$ means that $x^s$ times $x$.)

To use abel's identity, I  tried that the sum is extended over all positive integers.
But I can't find any idea. How can I solve this?
 A: Hints:
Realize that $\pi(x)=k$ is constant for $p_k \leq x < p_{k+1}$, so write as a sum
$$s\int_2^\infty \frac{\pi(x)}{x^{s+1}} dx = \sum_{k=1}^\infty \int_{p_k}^{p_{l+1}} \frac{ks}{x^{s+1}} dx.$$
You should be able to compute the integral and get a sum in the form:
$$\sum_{k=1}^\infty k\left[g(k)-g(k+1) \right].$$
Split the sum up and shift indices on the right one. (Alternatively, use Abel's identity, aka summation by parts.) Once you do this, you should get the result. One issue you need to fix/deal with is that I assumed a lower bound of $2$ instead of $1$. You need to fix this.
A: Define a function $f: \mathbb{Z} \to \mathbb{Z}$ as follows: $f(n) =1$ if $n$ is prime and $0$ else. Then you have the following:
$$\sum_p \frac{1}{p^s} = \sum_{n=1}^\infty \frac{f(n)}{n^s}$$
Now we can apply Abel's summation formula. Note here that by our definition of $f$, $\sum_{i=1}^{n} f(i) = \pi (n)$. Thus, proceed using Abel's sum formula:
$$\sum_{p \leq t} \frac{1}{p^s} = \frac{\pi (t)}{t^s} + s \int_{1}^{t} \frac{\pi(x)}{x^{s+1}} dx$$
Where $t$ is some integer. Thus, the result follows if we allow that $t\to \infty$, and by the prime number theorem, the term on the outside of the integral will tend to $0$, and the result follows immediately.
