I am reading a few papers about counting smooth numbers in the interval $[x, x+\sqrt{x}]$, including the work of [Harman](https://pdfs.semanticscholar.org/04d4/801dc56a086f989df6f9176cefc62b0650de.pdf), and [Matomaki](https://pdfs.semanticscholar.org/229f/f5bc99e60c25365425bf0c6d42ec40ed8eaf.pdf).


Both authors mentioned that the Dirichlet polynomial techniques break down when the length of the interval is about $\sqrt{x}$. Why is that the case? In particular, does this defect come from the method itself or the same barrier applies to all natural methods (e.g. sieve methods) to attack this question?



Counting smooth numbers in short intervals
Suppose that $F(s) = \sum\_{n \sim x} a\_n n^{-s}$ where $a\_n = 1$ if $n$ is $x^{\varepsilon}$ smooth and $a\_n = 0$ otherwise. We want to understand the distribution of the $a\_n$'s in a short interval of length $h$. Then by the usual Perron formula, 
$$
\sum\_{x < n < x + h} a\_n = \frac{1}{2\pi i} \int\_{-T}^{T} F(1 + it) \cdot \frac{(x + h)^{1 + it} - x^{1 + it}}{1 + it} dt + O(x / T). 
$$


After a few simplifications we see that we roughly have to choose $T = x / h$ and that the main integral simplifies to something like


$$
h \int\_{-x/h}^{x/h} F(1 + it) x^{it} dt
$$


At this point most methods take absolute values. However once we take absolute values the best bounds for $F$ that we can hope for is $|F(1 + it)| \ll 1/\sqrt{x}$. This corresponds to having perfect square-root cancellation at every point. In such a (most optimistic scenario), the bound that we get for the above is at best


$$
h \times \frac{\sqrt{x}}{h} = \sqrt{x}
$$


This means that as soon as we take absolute values in the Perron integral we forfeit the possibility to go into intervals shorter than $\sqrt{x}$. To go into shorter intervals one needs to somehow exploit cancellations in the Perron integral. At this stage one is rather advised to use other Fourier techniques (for instance exponential sums). 


