Mertens' second theorem for primes

Apply [L2] to the sequence $a_n=(\log n)\mathrm{/}n$ on primes and $a_n=0$ otherwise, with $b_n=1\mathrm{/}\log n$. Exactly as in Abel summation recovers the prime-counting function from theta, this gives ${\sum }_{p\le x}\frac{1}{p}=\frac{A(x)}{\log x}+{\int }_2^x\frac{A(t)}{t{\log }^{2}t}dt\mathrm{.}$

By [L1], write $A(t)=\log t+R(t)$ with $R(t)=O(1)$. Substituting into step 1.1 yields ${\sum }_{p\le x}\frac{1}{p}=1+\frac{R(x)}{\log x}+{\int }_2^x\frac{dt}{t\log t}+{\int }_2^x\frac{R(t)}{t{\log }^{2}t}dt\mathrm{.}$ Since ${\int }_2^{x}dt\mathrm{/}(t\log t)=\log \log x-\log \log 2$, we obtain ${\sum }_{p\le x}\frac{1}{p}=\log \log x+(1-\log \log 2)+{\int }_2^x\frac{R(t)}{t{\log }^{2}t}dt+O(1\mathrm{/}\log x)\mathrm{.}$

Because $R$ is bounded and ${\int }_x^{\mathrm{\infty }}\frac{dt}{t{\log }^{2}t}=\frac{1}{\log x},$ the improper integral ${\int }_2^{\mathrm{\infty }}\frac{R(t)}{t{\log }^{2}t}dt$ converges, and replacing the upper limit $x$ by $\mathrm{\infty }$ changes step 2.1 by only $O(1\mathrm{/}\log x)$. Therefore ${\sum }_{p\le x}\frac{1}{p}=\log \log x+B_1+O(1\mathrm{/}\log x),$ where $B_1:=1-\log \log 2+{\int }_2^{\mathrm{\infty }}\frac{R(t)}{t{\log }^{2}t}dt\mathrm{.}$ This constant is exactly the limit in The Meissel-Mertens constant.

The displayed asymptotic implies ${\sum }_{p\le x}\frac{1}{p}-\log \log x\to B_1$ as $x\to \mathrm{\infty }$, so the definition of The Meissel-Mertens constant is well posed.