Abel summation recovers the prime-counting function from theta

Define a sequence by $a_n=\log n$ when $n$ is prime and $a_n=0$ otherwise, for $2\le n\le N$. Then [L2] shows that its partial sums satisfy $A_m:={\sum }_{2\le n\le m}{a}_n=\theta (m)$ for every integer $m\in [2,N]$, while [L1] gives ${\sum }_{2\le n\le N}\frac{a_n}{\log n}=\pi (N)\mathrm{.}$

Apply [L3] to the finite sum in step 1.1 with $b_n=1\mathrm{/}\log n$. This yields $\pi (N)=\frac{\theta (N)}{\log N}-{\sum }_{n=2}^{N-1}\theta (n)(\frac{1}{\log (n+1)}-\frac{1}{\log n})\mathrm{.}$

For each integer $n$ with $2\le n\le N-1$, the function $f(t)=1\mathrm{/}\log t$ has derivative given by [L4], so $\frac{1}{\log (n+1)}-\frac{1}{\log n}={\int }_n^{n+1}{f}^{\mathrm{\prime }}(t)dt=-{\int }_n^{n+1}\frac{dt}{t{\log }^{2}t}\mathrm{.}$ Since $\theta (t)=\theta (n)$ on $[n,n+1)$, step 2.1 becomes $\pi (N)=\frac{\theta (N)}{\log N}+{\int }_2^N\frac{\theta (t)}{t{\log }^{2}t}dt\mathrm{.}$

Because $N=\lfloor x\rfloor$, there are no integers in $(N,x]$. Hence $\pi (x)=\pi (N)$ and $\theta (t)=\theta (N)=\theta (x)$ for every $t\in [N,x]$. Using [L4], we compute ${\int }_N^x\frac{\theta (t)}{t{\log }^{2}t}dt=\theta (x){\int }_N^x\frac{dt}{t{\log }^{2}t}=\theta (x)(\frac{1}{\log N}-\frac{1}{\log x})\mathrm{.}$ Adding this identity to step 3.1 gives $\pi (x)=\pi (N)=\frac{\theta (x)}{\log x}+{\int }_2^x\frac{\theta (t)}{t{\log }^{2}t}dt\mathrm{.}$