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The Selberg upper-bound sieve replaces an exact but long inclusion-exclusion expansion by a short nonnegative square, and then minimizes its main term. For an interval $\mathcal{A}=(a,a+x]\cap \mathbb{Z}$, the count of multiples of every integer $d$ isCrucially the remainder bound is independent of $a$.

Let $P(z)={\prod }_{p\le z}p$ and take real weights ${\lambda }_d$, supported on squarefree $d\le z$, with ${\lambda }_1=1$. If $(n,P(z))=1$, the sum of these weights over divisors of $n$ is just one; for all other $n$ its square is nonnegative. ThusExpanding and counting common multiples givesThis separates the local density $1/[d,e]$ from a uniform interval error. The weights will be chosen to minimize $Q$.

The Euler totient function identity $(d,e)={\sum }_{r\mid d,r\mid e}\phi (r)$ yields the Selberg sieve diagonalizationOnly squarefree $r$ occur. Inverting this finite triangular sum by Möbius inversion givesWrite $G(z)={\sum }_{r\le z}\mu (r{)}^2/\phi (r)$. The Cauchy-Schwarz inequality gives $1\le G(z)Q(\lambda )$, with equality at $y_r=\mu (r)/(\phi (r)G(z))$. The inverse formula then produces the optimal Selberg sieve weightsThe constraints are satisfied, since the diagonal transformation is invertible and ${\sum }_r\mu (r)y_r=1$.

We need the leading constant, not just $G(z)\gg \log z$. Here is an elementary proof of the Selberg sieve denominator asymptotic. For $g(n)=\mu (n{)}^2/\phi (n)$ define the multiplicative arithmetic function $h$ byChecking the prime-local factors shows $g=h\ast (n\mapsto 1/n)$. Also ${\sum }_n|h(n)|\log (2n)<\infty$: the local factors for absolute values differ from one by $O(p^{-2})$, and their logarithmic first moments sum $O((\log p)/p^2)$. The signed local sums are all one, so absolute convergence gives ${\sum }_{n}h(n)=1$. Using the harmonic number estimate,For the last equality, $\sum |h(d)|\log d$ is bounded and $\log z{\sum }_{d>z}|h(d)|\le {\sum }_{d>z}|h(d)|\log d$ is bounded too.

The optimal Selberg weights have modulus at most one. To prove this here, for squarefree $d$ use $d/\phi (d)={\sum }_{e\mid d}1/\phi (e)$. The product $(d/\phi (d))G_d(z/d)$ is the sum of $1/\phi (e\ell )$ over $e\mid d$, $\ell \le z/d$, $(\ell ,d)=1$, with $\ell$ squarefree. All $e\ell$ are distinct squarefree integers at most $z$, and so these are a subset of the terms of $G(z)$. The displayed weight formula now gives $|{\lambda }_d|\le 1$. ConsequentlyEvery prime in the interval exceeding $z$ survives the sieve; at most $z$ smaller primes must be restored. Choose $z=\lfloor \sqrt{x}/(\log x{)}^2\rfloor$. Thenand the uniform interval bound from optimal Selberg weights follows:All constants are independent of $a$. Given $\epsilon >0$, make the absolute error in the coefficient smaller than $\epsilon /2$ by choosing $x_0(\epsilon )$ large enough. Uniformly for every $a>0$ and $x>x_0(\epsilon )$,

For all interval origins $a\ge 0$, the number of multiples of any $d$ in $(a,a+x]$ is $x/d+O(1)$ with an absolute error bound. The Selberg sieve diagonalization minimizes ${\sum }_{d,e\le z}{\lambda }_d{\lambda }_e/[d,e]$, subject to ${\lambda }_1=1$, at $1/G(z)$. The optimal Selberg weights have modulus at most one, so the total remainder is $O(z^2)$. Restoring the at most $z$ small primes gives $\pi (a+x)-\pi (a)\le z+x/G(z)+O(z^2)$. Take $z=\lfloor \sqrt{x}/(\log x{)}^2\rfloor$ and use the Selberg sieve denominator asymptotic. The resulting $o(1)$ is independent of $a$.