 Square-root logarithmic sum over primes

Put $A(t)={\sum }_{p\le t}(\log p)/p$. The Mertens first theorem gives $A(t)=\log t+O(1)$. Partial summation yieldsThe main terms sum to $2\sqrt{\log x}$ up to a constant. The error integral is bounded because ${\int }_2^{\infty }dt/(t(\log t{)}^{3/2})$ converges. Thus a prime sum with a fractional logarithmic weight follows from an elementary bounded-error estimate, without the Prime number theorem.