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).