Tao Section 5: summing the Type I envelope, as the argument gives it

Summing the Type I pointwise envelope, with the constants the argument gives. Let $q\ge4$ with $(a,q)=1$, let $4\alpha=\frac aq+\beta$ with $|\beta|\le q^{-2}$, let $x>0$ and $U,V\ge40$ with $UV\le\frac x4$, and let $W$ be any nonnegative function satisfying, for every positive odd $d\le UV$,

(the second alternative dropped when the sine vanishes). Then

This is the combinatorial half of the source's Type I estimate, separated from the analysis that produces the envelope, and with the two constants its own chain yields. Writing $C=4(\log2)\log2x$, the range of $d$ splits as follows. For $d\le\frac q2$ one has $\|4d\alpha\|_{\mathbb R/\mathbb Z}\ge\frac1q-\frac{q/2}{q^2}=\frac1{2q}$, hence $\frac{1}{|\sin(2\pi d\alpha)|}\le2q$ and $W(d)\le\min(2qC,\frac{C}{|\sin|})$, and the odd-restricted Vinogradov lemma bounds that contribution by $2qC+\frac1\pi Cq\log4q$. Each subsequent block $2jq+\frac q2<d\le2(j+1)q+\frac q2$, $0\le j\le\frac{UV}{2q}-\frac14$, has length exactly $2q$, so the same lemma applies with prefactor $2$ and the first alternative frozen at the block's left endpoint, giving $2\frac{x}{2jq+q/2}\log x+4C+\frac4\pi Cq\log4q$. Summing the harmonic part by the integral test — including its $j=0$ term — produces the first term, and the rest is bounded by $\frac1\pi C(2UV+5q)(\pi+\log4q)$, which is at most $1.78(UV+\frac52q)(8+\log q)\log2x$ since $\frac{8\log2}\pi\le1.78$ and $\pi+\log4q\le8+\log q$.

Formalization Note $W$ is an arbitrary nonnegative function on $\mathbb N$, constrained only on the positive odd $d\le UV$, so the statement is exactly the passage from the envelope to the two terms and says nothing about exponential sums. The index set is the platform's theorem51Divisors U V, and the sine is written $\sin(\pi\cdot2\alpha\cdot d)$.