Finite summation of the inverse of prime numbers. Show that $\sum\limits_{p \leqslant x}1/p = \frac{\pi(x)}{x} + \int_2^x \frac{\pi(u)}{u^2} du.$

Question

Show that $$\displaystyle\sum\limits_{p \leqslant x}1/p = \dfrac{\pi(x)}{x} + \int_2^x \dfrac{\pi(u)}{u^2} du.$$

In the equation above, $\pi(x)$ denotes the prime counting function. To get started, how do we deal with the integral $\displaystyle\int_2^x \dfrac{\pi(u)}{u^2} du$ on the right hand side? Not really sure what to with it because of $\pi(u)$ in the numerator.

In my number theory class we were given the Riemann Zeta Function: $$\zeta(s) = \displaystyle\sum\limits_{n=1}^\infty \dfrac{1}{n^s}: s \in \mathbb{C}, \operatorname{Re}(s) > 1.$$

We proved $\zeta(s) = \displaystyle\prod\limits_{p \text{ prime}} \left(1-\dfrac{1}{p^s} \right)^{-1}$ and that $\displaystyle\sum\limits_{p \leqslant y}^\infty \dfrac{1}{p} > \log(\log y) - 1,$ $y \in \mathbb{Z}^{>0},$ but not much more than that (that I can think would be useful).

Some hints or pointers in the right direction appreciated.

Answer 1

the first thing to start with is the following, we call primes in order $p_1,p_2,\cdots \cdots$. Let $p_n$ the greatest prime less then or equal to $x$ (so as a consequence $\pi(x)=n$) then:

$$\begin{align}\int_2^x \dfrac{\pi(u)}{u^2} du&=\sum_{i=1}^{n-1} \int_{p_i}^{p_{i+1}} \dfrac{\pi(u)}{u^2} du+\int_{p_n}^x \dfrac{\pi(u)}{u^2} du\\ \\ &=\sum_{i=1}^{n-1} i\int_{p_i}^{p_{i+1}} \dfrac{1}{u^2} du+n\int_{p_n}^x \dfrac{1}{u^2} du \\ \\ &=\sum_{i=1}^{n-1} i\left(\frac{1}{p_i}-\frac{1}{p_{i+1}}\right)+n\left(\frac{1}{p_n}-\frac{1}{x}\right)\end{align}$$

The next steps are protected in the following bloc

$$\begin{align}\int_2^x \dfrac{\pi(u)}{u^2}&=\sum_{i=1}^{n-1}\frac{i}{p_i}-\sum_{i=1}^{n-1}\frac{i}{p_{i+1}}+\frac{n}{p_n}-\frac{n}{x}\\ \\ &=\sum_{i=1}^{n-1}\frac{i}{p_i}-\sum_{i=1}^{n-1}\frac{i-1}{p_{i}}+\frac{n-1}{p_n}+\frac{n}{p_n}-\frac{n}{x}\\ \\ &=\sum_{i=1}^{n-1}\left(\frac{i}{p_i}-\frac{i-1}{p_{i}}\right)+\frac{1}{p_n}-\frac{n}{x}\\ \\ &=\sum_{i=1}^{n-1}\frac{1}{p_i}+\frac{1}{p_n}-\frac{n}{x}\\ \\ &=\sum_{p\leq x}\frac{1}{p}-\frac{\pi(x)}{x}\end{align}$$

I think you can continue from here

Answer 2

Using the Stieltjes Integral and Integration by Parts yields $$ \begin{align} \sum_{p\le n}\frac1p &=\int_1^n\frac1x\,\mathrm{d}\pi(x)\\ &=\frac{\pi(n)}{n}-\frac{\pi(1)}{1}+\int_1^n\frac{\pi(x)}{x^2}\,\mathrm{d}x\\ &=\frac{\pi(n)}{n}+\int_2^n\frac{\pi(x)}{x^2}\,\mathrm{d}x \end{align} $$

Answer 3

The most rapid way is to use the Abel's summation formula. If we put $a\left(k\right)=1 $ if $k=p$ is a prime number and $0$ otherwise, we get $$\sum_{p\leq n}\frac{1}{p}=\sum_{k\leq n}\frac{a\left(k\right)}{k}=\frac{\pi\left(n\right)}{n}+\int_{2}^{n}\frac{\pi\left(t\right)}{t^{2}}dt. $$

Answer 4

$$\sum_{p\le x} \frac1p = \sum_{p\le x} (\frac1x\ +\ \int_p^x \frac1{u^2}du)= \frac{\pi(x)}{x} + \int_1^x \frac{\pi(u)}{u^2}du$$