I'm studying the book `Multiplicative Number Theory I. Classical Theory' by Hugh L. Montgomery and Robert C. Vaughan, and I don't understant a step of the proof for Perron's Formula(in Section 5.1) with $\rm{si}$-remainder, which states that $$\sum_{n\le x}'a(n) = \frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}\alpha(s)\cdot\frac{x^s}{s}ds+R$$ where $\alpha(s) = \sum_{n=1}^{\infty}a(n)n^{-s}$ has abscissa of absolute convergence $\sigma_a$, $\sigma>\max\{0, \sigma_a\}$, and $$R = \frac{1}{\pi}\sum_{x/2<n<x}a_n{\rm si}\left(T\log\frac{x}{n}\right)-\frac{1}{\pi}\sum_{x<n<2x}a_n{\rm si}\left(T\log\frac{n}{x}\right) + O\left(\frac{4^\sigma+x^\sigma}{T}\sum_{n=1}^{\infty}\frac{|a_n|}{n^\sigma}\right).$$ In the proof, the authors first change sum and integral so that $$\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}\alpha(s)\cdot\frac{x^s}{s}ds = \sum_{n=1}^{\infty}\frac{a_n}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}\left(\frac{x}{n}\right)^s\frac{ds}{s}.$$ Then the authors says that it suffices to show $$\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT}y^s\frac{ds}{s} = \begin{cases}1+O(y^\sigma/T) &\text{if }y\ge2\\ 1+ \frac{1}{\pi}{\rm si}(T\log y)+O(2^\sigma/T) &\text{if }1\le y\le 2\\ -\frac{1}{\pi}{\rm si}(T\log y)+O(2^\sigma/T) &\text{if }1/2\le y\le 1\\ O(y^\sigma/T) &\text{if }y\le 1/2\end{cases}$$ for $\sigma>0$.
What I don't understand is how this suffices for the case $\frac{1}{2}\le y\le 2$. The error term $2^\sigma$ cannot be bounded by $(4^\sigma+x^\sigma)/n^\sigma$ as long as my calculation is correct. We have the condition $x/2\le n\le 2x$, but this did not help. The only thing worked is to change the last error to $O((4x)^\sigma/T\sum_{n=1}^{\infty}|a_n|/n^\sigma)$. Is this a typo? Or there is a way to justify this error bound?
In practice, you can always assume $x\ge4$ and $\sigma$ is fixed, so the last term eventually becomes
$$ \ll{x^\sigma\over T}\sum_{n\ge1}{|a_n|\over n^\sigma}, $$
which is sufficient for applications.