[Math] Proof of Chebyshev’s theorem

analytic-number-theoryasymptoticsinequalitynumber theory

(a) Show that $\int_2^x\frac{\pi(t)}{t^2}dt=\sum_{p\leq x }\frac{1}{p}+o(1)\sim\log\log x.$

(b) Let $\rho(x)$ be the ratio of the two functions involved in the prime number theorem:

$$\rho(x)=\frac{\pi(x)}{x/ \log x}$$
Show that for no $\delta>0$ is there a $T=T(\delta)$ such that $\rho(x)>1+\delta$ for all $x>T$, nor is there a $T$ such that $\rho(x)<1-\delta$ for all $x>T$. This means that
$$\lim \inf \rho(x)\leq 1 \leq \lim \sup \rho(x),$$
so that if $\lim \rho(x)$ exists, it must have the value $1$.

I don't know how to apply (a) to (b), and I couldn't find any sources related to such a proof. Could you give me a proof on that?

Best Answer

(a) $\to$ (b). I show one half of part (b), namely $\liminf \rho(x) \leqslant 1$. The other half is analogous.

Assume that there exists $\delta >0$ and $T \geqslant 2$ such that $\rho(t) \geqslant 1+\delta$ for all $t \geqslant T$. That is, $$ \pi(t) \geqslant (1+ \delta) \frac{t}{\ln t}, \tag{$\ast$} $$ for all $t \geqslant T$. Therefore, $$ \begin{align*} \int_2^x \frac{\pi(t)}{t^2} \, dt & \geqslant \int_T^x \frac{\pi(t)}{t^2} \, dt \\ &\stackrel{(\ast)}{\geqslant} \int_T^x (1 + \delta) \frac{t}{\ln t} \cdot \frac{1}{t^2} \, dt \\ &=(1 + \delta) \int_T^x \frac{1}{t \, \ln t} \, dt \\ &= (1 + \delta) \cdot \left. \ln \ln t \right\vert_{t=T}^{t=x} \\ &= (1 + \delta) \cdot (\ln \ln x - \ln \ln T) \\ &\sim (1 + \delta) \cdot \ln \ln x, \end{align*} $$ for large $x$. This clearly contradicts part (a). It thus follows that $\liminf \rho(x) \leqslant 1$.

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