 Reciprocal-prime sum in residue class one modulo four

The progression version of the Mertens second theorem gives ${\sum }_{\begin{array}{c}p\le y\\ p\equiv 1(\mathrm{mod}4)\end{array}}1/p=\frac{1}{2}\log \log y+c+O(1/\log y)$. Subtracting from the all-prime sum shows that the forbidden primes $p\equiv 3(\mathrm{mod}4)$ contribute half the logarithmic density. This is the source of the square root in a half-dimensional interval sieve.