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The continuous Buchstab function satisfies $w(u)=1/u$ for $1\le u\le 2$ and $\frac{d}{du}(uw(u))=w(u-1)$ for $u>2$. It describes the density of rough numbers.

A delay differential equation relates a derivative at the current argument to values at earlier arguments. For example, the Buchstab function obeys $\frac{d}{du}(uw(u))=w(u-1)$ for $u>2$.

The Buchstab function takes values both above and below $e^{-\gamma }$ on every interval of length one in $[1,\infty )$. The equation $\frac{d}{du}(u(w(u)-e^{-\gamma }))=w(u-1)-e^{-\gamma }$ propagates any fixed sign forwards; the rapid decay to zero then forces an impossible identically zero tail.

Use the convention that a z-sieved number is a positive integer with no prime factor strictly below $z$; $1$ is included. Such integers are also called rough numbers. Using instead exclusion of primes at most $z$ changes an endpoint convention, not the following fixed-$u$ asymptotic.

The Buchstab function is the continuous function $w:[1,\infty )\to \mathbb{R}$ determined bySuccessive integration over intervals of length one determines the Buchstab function uniquely from its initial values.

A standard fixed-$u$ form of Buchstab theorem states that, for each fixed $u>1$, as $z\to \infty$,For $1<u\le 2$, this agrees with the Prime number theorem, since apart from an immaterial endpoint the z-sieved numbers in this range are $1$ and the primes from $z$ to $z^u$. The restriction $u>1$ matters: at $u=1$ the count is bounded, so the displayed asymptotic does not hold there.

Put $c=e^{-\gamma }$, where $\gamma$ is the Euler--Mascheroni constant, and let $f(u)=w(u)-c$. The Buchstab function equation becomesThe given Gamma function bound implies $uf(u)\to 0$. Indeed,so $u/\Gamma (u+1)\to 0$.

Suppose first that $f\ge 0$ throughout an interval $[u,u+1]$, with $u\ge 1$. The delay differential equation and integration successively on $[u+1,u+2]$, $[u+2,u+3]$, and so on propagate nonnegativity to the entire tail $[u,\infty )$. In particular, $vf(v)$ is nonnegative and nondecreasing for $v\ge u+1$. Its limit is zero, so it is identically zero there. Thus $f=0$ on that tail.

The delay differential equation then propagates this equality backwards: if $f=0$ for $v\ge a$, then $f(v-1)=0$ for $v>\max (a,2)$. Repeating finitely many times would give $f=0$ on $(1,2)$, contradicting $f(v)=1/v-c$ there. Hence every interval $[u,u+1]$ contains a point where $f<0$.

If instead $f\le 0$ on $[u,u+1]$, the same propagation makes $vf(v)$ nonpositive and nonincreasing on the tail. Its limit zero again forces it to vanish, giving the same contradiction. Thus every such interval also contains a point where $f>0$. The oscillation of the Buchstab function is therefore strict on every unit interval: