Smooth divisor-square sieve asymptotic (${\sum }_{n\in I}{F}_X(n)\sim c_{\varphi }X/\log X$)

For a real smooth cutoff $\varphi$ supported on $[-1/3,1/3]$ with $\varphi (0)=1$, define $F_X(n)=({\sum }_{d\mid n}\mu (d)\varphi (\log d/\log X){)}^2$. Its sum over any interval of length $X$ is $c_{\varphi }X/\log X+O_{\varphi }(X/{\log }^{2}X+X^{2/3})$, uniformly in the interval location. Here $c_{\varphi }$ is the derivative energy constant for a smooth sieve cutoff. Expanding the square yields an interval-counting error $O(X^{2/3})$ and an Euler product for a smoothed divisor-square correlation; its zeta function pole provides the main term. Every prime exceeding $X^{1/3}$ has weight one, giving the short-interval prime upper bound from a smooth divisor weight.

For $\mathrm{Re}z,\mathrm{Re}{z}^{\prime }>0$, the series ${\sum }_{d,e}\mu (d)\mu (e)d^{-z}{e}^{-z^{\prime }}/[d,e]$ has local factor $1-p^{-1-z}-p^{-1-z^{\prime }}+p^{-1-z-z^{\prime }}$. Dividing by the displayed zeta function ratio leaves a holomorphic function $H$ near zero with $H(0,0)=1$. The local factor of $H$ is $(1-a-b+c)(1-c)/((1-a)(1-b))$, where $a=p^{-1-z}$, $b=p^{-1-z^{\prime }}$, $c=p^{-1-z-z^{\prime }}$, and equals one exactly at zero. With $z=(1+it)/\log X$, the zeta function poles produce the kernel $(1+it)(1+i{t}^{\prime })/((2+i(t+t^{\prime }))\log X)$ that determines the smooth divisor-square sieve asymptotic.

Write $e^u\varphi (u)=\int \psi (t)e^{-iut}dt$. The double integral $\iint \psi (t)\psi (t^{\prime })(1+it)(1+i{t}^{\prime })/(2+i(t+t^{\prime }))dtd{t}^{\prime }$ equals the displayed constant. Express the denominator as a Laplace integral and each differentiated Fourier factor as $-{\varphi }^{\prime }(u)$, then use Fubini's theorem. For a real cutoff the resulting square is positive even though the Fourier transform factors have no complex conjugation. The Cauchy-Schwarz inequality and $\varphi (0)=1$, $\varphi (1/3)=0$ give $c_{\varphi }\ge 3$.