Tao Section 5: the integral test for the Type I block sum $\sum_j x/(2jq+q/2)$

Let $x\ge0$, $q>0$ and $M$ be reals, and let $J$ be a natural number with $J\le\frac{M}{2q}-\frac14$. Then

This is the block sum that appears when the Type I sum of the source's minor-arc theorem is cut into blocks $2jq+\tfrac q2<d\le 2(j+1)q+\tfrac q2$ of length $2q$: on the $j$-th block the weight $x/d$ is bounded by its value at the left endpoint, and what remains is exactly the sum above, with $M=UV$ the length of the divisor range. The bound is the integral test applied to $\sum_j\frac1{4j+1}$, whose partial sums are $1+\tfrac14\log(4J+1)$ up to the first term.

Deviation from the source The source's display asserts the same bound without the additive $4$, namely $\sum_j\frac{x}{2jq+q/2}\le\frac{x}{2q}\log\bigl(\frac{2M}{q}+4\bigr)$. That cannot hold in general: the $j=0$ term alone is $\frac{x}{q/2}=\frac{x}{2q}\cdot4$, so the claimed right-hand side is already exceeded whenever $\log(\frac{2M}{q}+4)<4$, that is whenever $M/q<\tfrac12(e^4-4)\approx25.3$. For a concrete instance take $q=1$, $M=10$, so that $J=4$: the left-hand side is $2.8937\,x$ while the source's right-hand side is $\tfrac x2\log24=1.5890\,x$. The additive $4$ above is what the integral test actually gives, and the source's downstream estimate has room for it in its remaining terms; the correction is recorded here rather than propagated silently.

Formalization Note The summation index runs over Finset.range (J+1), i.e. $0\le j\le J$, and the hypothesis $J\le\frac{M}{2q}-\frac14$ is what the source's condition on the block index gives. No lower bound on $M$ is assumed: the hypothesis on $J$ already forces $M\ge q/2$, so the logarithm is taken at a value $\ge5$.